Circuit Theory By U A Bakshi A V Bakshi 2n5i4y

<2

v(t) = 20 t i(t) For 2 < t < 4, v(t) is

= Cdd~l)

=0.5x 10-6 x20 =1xt0·5A= lOI!A

... for0 <2

con.~tant.

v(t) = 40 V

... for2 <4

dv(t) _ -6 i(t) = Cdt = O.::>XlO xO = 0 A

... for2 <4

For 4 < t < 8, v(t) is a ramp with slope =

0 ~~ = - 10 8

v(t) = -10 t + 80

... for4 <8

i(t) = C d~~t) = 0.5x t0-6 x(-10) =- 5 IJ.A.

... for4<1<8

The current waveform is shown in the Fig. 1.17(a). ~I)

10~A I---.

Fig. 1.17(a) NutP. : For 4 < t < 8 , the equation of the line is y = mt + C where m = slope = -10. It gives equation y = - 10 t + C. Now point (8, 0) is on the line so substituting in the equation,

0 = -10x8+C

i.e. C = 80

Thus equation of line is y = - 10 t + 80 where y = i(t), as used above.

1.7 Analysis of Series Circuits A series circuit is one in which several resistances are connected one after the other. Such connection is also called end to end connection or cascade connection. There is only one path for the flow of current. Copyrgh!ed matcri~

Circuit Theory

1 ·13

Basic Circuit Analysis & Network Reduction Techniques

1.7.1 Resistors In Series Consider the resistances shown in the Fig. 1.18.

....._ R2

•

.I.

.I.

?-- V,---t-- v2

.:II.

•

~.(!.

I

v3-•

-

Vvolts

Fig. 1.18 A series circuit

The resistance R 1 , R 2 and R3 are said to be in series.. The combination is connected across a source of voltage V volts. Naturally the current flowing through all of them is same indicated as I amperes. e.g. the chain of small lights, used for the decoration purposes is good example of series combination.

Now let us study the voltage distribution. Let v,, V2 and V3 be the voltages across the terminals of resistances R1, R2 and R3 respectively.

v

Then, Now accord ing to Ohm's law,

=

v 1 + v2

+

v3

V1 = IR 1 , V2 =IR2 , V 3 = 1 R 3

Current through aU of them is same i.e. I

Applying Ohm's law to overall circuit, V = I Rc'
Rcq = R1+R2 +R3 i.e. total or equ ivalent resistance of the series circuit is arithmetic sum of the resistances connected in series. For n resistances in series,

1.7.1.1 Characteristics of Series Circuits 1) The same current flows through each resistance. 2) The supply voltage V is the sum of the individual voltage drops across the resistances. V = Vt + V2 +..... + Vn

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Circuit Theory


3) The eqwvalcnt resistance is equal to the sum of the individual l'eSistances. 4) The equivalent resistance is the largest of all the individual resistances.

i.e

1.7.2 Inductors in Series Consider the Fig. 1.19 (a). Two inductors L 1 and L.2 are coMe<:ted in series. The cunents flowing through l. 1 and l.z nre i 1 and iz while voltages developed across L 1 and L 2 nrc Vl t and VL2 respectively. The cqlrivalcnt circuit is shown in the Fig. 1.19 (b).

vL • o-o--~'----oo Leo (b)

(a)

Fig. 1.19 Inductors in series

di

VL = Lcqdt

We have,

For series combination,

and

Yt

~

Vu • Yt.z

..

di di di Lcq - = Lt Ji + Lz dt dt

..

L

di

cq

..

di dl = (Lt + L,• )ift

L,"

~

Lt + L2

That means. equivalent inductance of the series combination of two ;nductances is the · sum oi inductances coMccted in series. Key Point : The total equivalent inductance of the series circuit is sum of the inductances connected in series.

For n inductances in series,

l.

L,q =

1-t+~+~+ ... +t.,

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Circuit Theory

Basic Circuit Analysis & Networ11 Reduction Techniques

1 • 15

1.7.3 Capacitors in Series Consider the Fig. 1.20 (a). Two capacitors C1 ;md C2 arc connected in series. The currents flowing through and voltages developed across C1 and C2 are i 1, i 2 and Vc 1 and Vcz respectively. The equivalent circuit is shown in the Fig. 1.20 (b).

•

Vel

·w cl

il

vc

Vc2

12

' ·~~-

i

• ' y

•

~

•

ci!Q

(a)

(b)

Fig. 1.20 Capacitors In series I

We have,

Vcz =

c2

For se.ries combina lion, i

~

i1

~

Vc

~

Vc 1 + Vcz

iz

I

Ji 2

--

d1

while

I V = Ceq

I .

J1d1

--

and

I

-c cq

Jidt -~

= it = i2

But I

I )' Jidl J idt = ( -+ C1I -Cz __ Ceq I

Ceq

Ceq

I

= -+-

c, c2

= c1c1c2 +C2

That means, reciprocal of equivalent capacitor of the series combination is the sum of the reciprocal of individual capacitances.

Key Point : Tlu rtciproall of the total equivalent CJJPIIcitor of tlu series combination is tht sum of 1/u rtciproco/s of the individual azpacilors, connected in str~. For n capacitors in series,

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Circ:uit Theory

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Basic: Circuit Analysis & Networt Reduc:tion Tec:hniques

Example 1.2 : Twv o:apacittmctS C 1 and C2 of values of ~0 p.F and 5 p.F, respectivtly are comzccted in series. What is tile equivaltmt capacitance of lht combination ? [Aprii/May-20031

Solution :

C 1 = 10 11F

Ceq = ..

and

Cz = Sj1F

C 1C 7 = lOx 10· 6 xSxlO - 6 = 3 .333 x 10 _ 6 Ct +C2 lOx 10-6 +Sx to- 6

= 3.33 )IF

1.8 Analysis of Parallel Circuits The parallel circuit is one in which several resistances a.rc connected across one another in such a way that one terminal of each is connected to form a junction point while the remaining ends are also ed to form another junction point.

1.B.1 Resistors in Parallel Consider a parallel cil'cuit shown in the Fig. 1.21. In the parallel connection shown, the three resistances R 1 , R 2 and R 3 are connected in parallel and combination is connected across a source of voltage 'V ',

R, • v ~

In parallel circuit current ing through each resistance is different. Let total current drawn is say ' I ' as shown. There are 3 paths for thi• current, one + v through Rt, second through R2 and third through R3. Depending upon the values of R, , Rz and R3 the + appropriate fraction of total current es through v them. These individual currents are shown as I t ,lz and Fig. 1.21 A para.l lel c:irc:uit I 3· While the voltage across the two ends of each resistances R1 , R2 and R3 is the same and equals the supply voltage V. +

v -

Now let us study current distribution. Apply Ohm's Jaw to each resistance. V = 11 R 1 , V = 12 R2 , V = 13R 3 11 =

v R;"'

v

v

12 = Rz ' 13 = R3

v

v

v

= 11 +12 +13 = - + - + Rt Rz R3 ... (1)

For overall circuit if Ohm's Jaw is applied, V = I Req

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Circuit Theory

1 - 17

and

Basic Circuit Analysis & Networlt Reduction Techniques

I=Rv

... (2)

eq

Rcq = Total or

Where

equiv~lent

rHistance of the circuli

Comparing the two equations, 1

1

1

1

- - = - +- + Rt R2 R3 Rcq Where R is the equivn.lcnt

resista~

of the pa.rallel combination.

In general if 'n' resistances are connected in p arallel, 1

R

1 I 1 1 = Rl+ R2 + R3 +......+ Rn

Conductance (G) : It is known that,

I

1 R =

G

(conductance) hence,

G = G t +G2 +G3+......+Gn

... For parallel drcult.

Important result : Now if n = 2, two resistances are in parallel then,

1

R

1 Rt

1 R2

= - +-

R = This formula is directly used herea fter, for two resistances in parallel. 1.8.1.1 Characteristics of Parallel Circuits I) The same potential difference gets across aU the resistances in parallel. 2) The total current gets divided into the number of paths equal to the number of

resistances in parallel. The total current is always sum of all the individual current~.

I = It +l2 + I J +.. ....+ln 3) The redprocal of the equivalent resistance of a parallel drcuit is. equal to the sum of the redprocal of the individual resistances.

4) The equivalent resistance is the smallest of all the resistances.

R < Rt , R
Circuit Theory

1 · 18


5) The equivalent ::onductancc is the aritlunctic addition of the individual conductances. Key Point : Th~ equiwltnt resistance is sma//er than lite smallest of a// the rtsistances connt'cled in para/lei.

1.8.2 Inductors In Parallel Consider the Fig. 1.22 (a). Two inductors L 1 and L 2 are connected in parallel. The currents flowing through L1 and L 2 are i 1 and i 2 respectively. The voltage developed across L1 and L2 are VL 1 and VL2 respectively. The equivalent circuit is shown in Fig. 1.22 (b). ;, +

o-o--<-~---oo

Leq

lz

{b)

(a)

Fig. 1.22 Inductors In parallel

For inductor we have,

r.I JVLI d1,

t

t

i1 ;

L2

- ~

JV1.2 dt,

while

.

I

I = -

-~

Leq

t

--

jVLd t

For parallel combination, VL = VLI = VL2

= il + L,-q

J' V..dt

--

Lcq

and

jl I

= =

I

I JvLdt - I J VLdt + l.o L2

--

(I

' = L.+i2I )_J_vL dt

-w

I I -+ -

L,

t2

That means, reciprocal of L'<juivalent inductance of the parallel combination of reciprocals of the individual inductances.

i~

the sum

For n inductances in parallel, I

Lcq =

1

I

I

I;'+ l2 +....L;

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Circuit Theory

1 ·19


1.8.3 Capacitors In Parallel Consider the Fig. 1.23 (a). Two capacitors C1 and C1 are connected in parallel. The currents flowing through C1 and C2 are i 1 and i 2 respectively and voltages developed across c,, C2 are Vel and Vc2 respectively. The equivalent circuit is shown in the Fig. 1.23 (b).

c,

Vc

....-~·u c..,

•

Cz (a)

(b)

Fig. 1.23 Capacitors in parallel

1=

For capacitor w e have,

ccqdt dVc

For parallel combination, Vc 1 "

Vc2 " Vc

and

= i, + i2

c

..

..

dVc

dVc, c dVc2 = c 'd!+ ld(

dVc

dt = ('c, + c2 )dYe

"'ld!

c

"'ld!

C.:q =

c, + c2

That means, equivalent capacitance of the parallel combination of the capacitances is the sum of the individual capacitances connected in series. For n capacitors in parallel,

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Ci rcuit Theory

1 - 20

Basic Circuit Analysis & Networll Reduction Techniques

The Table 1.2 gives the equivalent at 'n' basic elements is series. Element

Equival ent.

·n• Resistances in series

R, R , R,

R •

· ·· - ~

....

Req = R1 + R2 • R3 + ... + R,

'n' lnductOts 1n series L,

L,

L,

· · ·-~ l•

....

Leq = L1 + L2 + L3 + ... + ln

'n' Capacitors in series

o-ll c,

0

II

0

c,

11--.... o-ll--

c,

c.

1

I

t

t

= - + - + ···+Ceq Ct C2 Cn

Table 1.2 Series combinations of elements

The Table 1.3 gives the equivalent of 'n' basic elements in parallel. Element

Equivalent

'n' Resistances in parallel

~:~r~r~:JR.

1 -

Req

Q

1 1 1 - + - + ·-·+ R1 R2 Rn

'n' Inductors In parallel

~=r~~r~=J·

t

1

1

1

-l cq = -l 1 + -L2 +·-·+Ln-

'n' Capacitors in parallel

c.I __ , 1 c., c,f I __ . J c.

0

Ceq

=C1 + C2 + ...+ Cn

0

Table 1.3 Parallel combinations of elements

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Circuit Theory

1 -21

Basic Circuit Analysis & Network Reduction Technlquas

Key Point : Tht currtnt through strits combination remains same and voltage gets dillidtd

I..

while in parallel combitwlion voltage across combination remains same and current gets divided. Example 1.3 : Find the equivaltnt resistance between the two paints A and B shown in

the Fig. 1.24. 30

....A. A. AA

40

.?.R

4{l

10 A

6!1

50

B

Hl

Fig. 1.24 Solution : Identify combinations of series and parallel resistances.

The resistances 5 n and 6 n are in series, as going to So equivalent resistance is 5

carry same current.

+ 6 = 11 n

While the resistances 3 n , 4 n, and 4 n are in parallel, as voltage across them same but current divid es. :. Equivalent resistance is,

R = 3 +4+4 =i2

1

1

1 1

10

..

R =

12 iii = 1.2n

Replacing these combination.< redraw the figure as shown in the Fig. 1.25 (a). 20

Series 1.2 0

ParaUel

10 A

Fig. 1.25 (a)

Fig. 1.25 (b)

Now again 2.2 n and 2 n are in series so equivalent resistance is 2 + 1.2 = 3.2 n while 11 n and 7 n are in parallel. 77 Using formula ~t+Ri equivalent resistance is ~~:; = 18 =4.2770.

2

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Circuit Theory

1 - 22


Replacing the respective combinations redraw the circuit as shown in the Fig. 1.25 (b). Now 3.2 and 4.277 arc in parallel. 3 . 2x 4.277 3 .2+4 .277

Replacing them by

~

1.8304

n

RAB. ~ 1+ 1.8304 ~ 2.8304 0

1.9 Short and Open Circuits In the n etwork simplification, short circuit or open circuit existing in the network plays an important role.

1.9.1 Short Circuit

.------

'•a

When any two points in a network are ed directly to e:ach other with a thick metalic conductin,g wire, the two points are said to be short circuited. The resistance of such short drcuil is zero.

,

A ~----~~.---1

Thiel< conducting -

Netwod<

wire a~---....1.--<

·'------

The part of the network, which is short circuited is shown in the Fig. 1.26. The points A and B are short circuited.

Fig. 1.26 Short circuli

The resistance of th e branch AB is R.., = 0 0 . The curent JAB is flowing through the short circuited path. According to Ohm's law, VAB

=

R.., X lAB

= 0 X lAB = 0 V

Key Point : 'flrus, voltage across slrort circuit is always zero though cu.rrent f/17WS througlr

the short circuited patlr.

1.9.2 Open Circuit When there is no connection between the two points of a network, having some voltage across the two points then the two points are said to be open circuited. A+

t

Open

circuit

R,

T.o

l ~___ B-

: - - - - --

:

R t 2 :

.__.....:

Netwol1!

' L------

Fig. 1.27 Open c ircuit

As there is n o direct <:onneclion in an open circuit, the resistan ce o f the op en circuit is oo.

The part of the network which is open circuited is shown in the Fig. 1.27. The points A and B arc said to be open circuited. The resistance of the branch A B is Roc ="" 0.

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Circuit Theory

1 -23


There exists a voltage across the points AB called open circuit voltage, VAB but

Roc=- 0. According to Ohm's law, loc

= VAs =VAs =0 A Roc oo

Key Point : Thus, currtnt through open circuit is always uro thuugh there exists a voltage

across

optn

circuited terminals.

1.9.3 Redundant Branches and Combinations The redundant means excessive and unwanted.

Key Point : If in a circuit there are bra11dres or combinations of tlements which do not carry any curre11t thtn such bra11ches and combinations an! called redundant from circuit point of view. The redundant branches and combinations can be removed and these branches do not affect the performance of the circuit The two important situations of redundancy which may exist in practical circuits are, Situation 1 : Any branch or combination across which there exists. a short circuit, becomes redundant as it does not carry any current.

U in a network, there exists a direct short circuit across a resistance or the combination of resistances then that resistance or the entire combination of resistances becomes inactive from the circuit point of view. Such a combination is redundant from circuit point of view. To understand this, consider the combination of resistances and a short circuit as shown in the Fig. 1.28 (a) and (b).

·~ ~ (a)

-.

R2

I

Short

•

vc ~

-

A

~ 1=0

R;, R,

R.. ;

No current through R3

*I

:\

B

\

Short

No current

through R3 and R4

(b)

Fig. 1.28 Redundant branches

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Circuit Theory


1 - 24

In Fig. 1.28 (a), there is short circuit across R3. The current always prefers low resistance path hence entire curr~'llt I es through short circuit and hence resistance R3 becomes redundant from the circuit point of view. In Fig. 1.28 (b), there is short circuit across combination of R3 and R4. The entire current flows through short circuit across R3 and and no current can flow through combination of ~ and ~· Thus that combination becomes meaningless from the circuit point of view. Such combinatioJU can be eliminated while analysing t:he circuit.

a.

Situation 2 : U there is open circuit in a branch or combination, it can not carry any current and becomes redundant. In Fig. 1.28(c) as there exists open circuit in branch BC, the branch llC and CD can not carry any current and are become redundant from circuit point of view. ~Redundant

R,

A

B

~

c

!

I%0

Rz

I)

R4

1=0 E

F

ranches

D

Fig. 1.28(c) Redundant branches due to open circuit

1.10 Voltage Division In Series Circuit of Resistors Consider a series circuit of two resistors R 1 and R2 connected to SO\liCe of V volts. As two resistors are connected in series, the

current flowing through both the resistors is same, i.e. L Then applying KVL, we get, V = I Rt +I R2

v

Fig. 1.29

:. I= Rr + R2

Total voltage applied is equal to the

sum

of voltage drops

VRr

and

VR2

across Rt and

R2 respectively.

..

..

VRI

= I. Rt

VRI =

Similarly,

VRl

V

Rt + R2

·Rt = [

RI ] V Rt + R2

= I· R2

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Basic Cin:uit Analysis & Network Red uc tion Techniques

1 - 25

Circ:uit Theory

So this circuit is a voltage divider drroil Key Point : So in general, voltage drop llCI'OSS any resistor, or combination

of resistors, in a series circuit is equal to the ratio of that resistance value to the total resistance, multiplied by the source voltage.

1.11 Current Division in Parallel Circuit of Resistors Consider a parallel circuit of two resistors R1 and Rz connected across a source of V volts. Current through R 1 is l1 and Rz is l2, while total current drawn from source is h.

.. But i.e.

Fig. 1.30

'• = lz

.,. =

'• + lz

v

'• =

~,

12 =

v

Ri

V= IIRI = lz Rz

(~~ )

Substituting value of 11 in I,., IT = lz (

lz =

Now

•• ••

~~ ) + l2

= 12

[~~ +1]

= 12 [ R• : R2] 1

[R.~RJ ,,.

= ,,. - 12 =

,,. - [ R1R1+R2 ] ,,.

1 2 1 = [R +R -R ] IT

'• = [

Rt +Rz

R,- ] .,. R1 +R2

Key Point : lr1 general, the current in any branch is equal to tloe ratio of opposite branch resista11ce to the total resistance VQ/ue, multiplied by the total current in tl1e circuit.

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Circuit Theory

••

1 • 26

Basic: Circ:ult Analysis & Network Reduction Techniques

Example 1.4 : Find the voltage across tht three resistances shown in lltt Fig. !.3!.

100

200

300

(' + -

60V

Fig. 1.31 Solution :

=

v

... Series circuit

Rt +R2 + R3

60 = 10+20+ 30 = lA VRI = !RI =

VxRt = l xlO = 10 V Rt +Rz +R3

VxR 2 = lx20=20V VR2 = I R, = • R t +Rz +R J

and

VRJ = IR J =

VxR3 =l x30 = 30 V Rt +Rz + R3

Key Point : II can be seen that voltage across any resistanc~ of series circuit is ratio of that

resistance to the total resistance, multiplied by the source voltage. ••

Example 1.5 : Find the magnitudes of total CIITrtnt, current through R 1 and R 2

Rt= 10 Q, Rz= 20

Q,

if.

and V= 50 V.

t

•• R, : Fig. 1.32

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Circuit Theory

1 • 27

Basic Circuit Analysis & Network Reduc tion Techniques

Solution : The equivalent resistance of two is,

v lr = -Rcq

50 = -- =

6.67

7.5 A

As per the current distribution in parallel circuit,

Il = = and

12 =

lr( RI~R2 )= 7

sx( 10~20)

SA

lr( RI~IR2 )= 7 .5x( 10~20)

= 2.SA It can be verified that

1.12 Source Transfonnation Con.
if Ill~ supply equal load current to

tire load, with sa= Wad connectd llCTOSS its terminals The current delivered in above case by voltage source is,

Rsc and R L in series

.. . (1)

If it is to be replaced by a current so= then load

current must be (

v

. )

Rsc +RL

Consider an equivalent current sowce shown in the Fig. 1.33 (b). Fig. 1.33 (b) Current source

The total current is · I ·.

Both the resistances will take current proportional to their values.

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Cln:ult Theory

Bask: Cln:ult Analysis & Network Reduction Teehn.Jqua

1 • 28

From the current division in paraUel circuit we can wrile,

1L = !>< Now this IL and R

.,R

.:..R~sb~"7

must be same, so equating (1) and (2),

sc + L

v

= R.., = R,h = R say.

Let internal resistance be,

v =

Then,

1

or

..

=

1=

Key Point :

.•.(2)

(Rob +RL)

I>< Rsb

= I>< R

v Rsh

v R

=

v Rse

If wltAge source is converted to current source,

thnt CUI'Tent source I = - v

R,..

with parallel irrtmu1l resistAnce el(UIII to R". Key Point :

with series

If currtnt source i.< converted to voltage source, tllm voltage source V = I R.I.

intmu~l

resistance equal to R.h.

The direction of current of equivalent current source is always from Internal to the source. While converting current source to voltage source, voltage is always as +ve terminal at top of arrow and -ve temtinal at bottom direction of current is from -ve to +ve, internal to the source. This ensures flows from r
ve to + ve, polarities of of arrow, as that current

Note the directions of transformed sources, shown in the Fig. 1.34 (a), (b), (c) and (d) .

• :::: I

(a)

I-~ Rae

(b) I • -

v

R..

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Clreult l'Mory

Basic: Cireuit Analysis & Network Reduc:tlon Tec:hnlqun

1 • 29

R,.

-

v

(c:) V • l x R ah

(d) V • I XRst>

Fig. 1.34 Sourc:e transfonnatlon ,_,. Example 1.6 :

Transform a voltage sour~

of 20 volts

with an inttnUJi rtsist;Jnct of 5 0

to 11 current source. Solution: Refer to the Fig. 1.35 (a). 50

50

(b)

(a)

Fig. 1.35 Then current of curren t source is, I

=

R: = ;o =

4 A with internal p arallel resistance

same as R"' . :. Equivalent current source is as shown in the Pig. 1.35 (b).

'' *

Example 1.7 :

1April!May·20041

Convert into a w/tage source.

10A

t

50

Fig. 1.36

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Circuit Theory

1 • 30

Bailie Circuit Analysis & Network Reduction Techniques

so

Solution :

I = 10 A,

v

= IX R

R = 5 !l

~

10 X 5 = 50

v

The polarities are positive at top and negative at bottom of the anow. Thus the voltage source is as shown in the Fig. 1.37.

Fig. 1.37

1.13 Combinations of Sources In a network consisting of many sources, series and parallel combinations of sources exist. II such combinations Me replaced by the equivalent source then the· network simplification becomes much more easy. Let IL~ consid er such series and parallel combinations of energy sources.

1.13.1 Voltage Sources In Series If two voltage sources are in series then the equivalent is dependent on the polarities

of the two sources. Consider the two sources as shown in the Fig. 1.38.

Fig. 1.38 (1.1) Thus ;J the polarities of the two sources arc same then the equivalent single source is the ~ddition of the two sources with polarities same as that of the two sources. (a)

Consider the two sources as shown in the Fig. 1.39.

(a)

Fig. 1.39

(b)

Thus if the polarities of the two sources are different then the equivalent single source is the difference between the two voltage sources. The polarities of such source is same as that of the greater of the two sources.

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Circuit Theory

1 • 31


Key Point : The voltage sources to be connected in series must have s.ame current ratings

though thrir voltage ratings may be same or different. The technique can be used to reduce the series combination of more than two voltage sources connected in series.

1.13.2 Voltage Sources in Parallel Consider the two voltage .sources in parallel as shown in the Fig. 1.40.

v, •

The equivalent single source has a value same as V 1 and V 2. It must be noted that at the terminals open circuit voltage provided by each source must be equal as the sources are in parall2l.

Fig. 1.40

Key Point : Heuce the voltage sources to be connected in paratltl rrurst IUW
1.13.3 Current Sources in Series Consider the two current sources in series as shown in the Fig. 1.41. The equivalent single source has a value same as 11 and 12.

Key Point : The current through series circuit is always same hence it must be noted that the current sources to be connected in series must have same current ratings though 1/reir voltage ratings may be same or different.

Fig. 1.41

1.13.4 Current Sources In Parallel Consider the two current sources in parallel as shown in the Fig. 1.42.

=::;

(a)

Fig. 1.42

(b)

Thus if the directions of the currents of the sources connected in parallel are same then the equivalent single source is the addition of the tWo sources with direction same as that of the two sources.

Circuit Theory

1 • 32


~----------------~------------------~ Consider the two current $0Urces with opposite directions connected in parallel as

shown in the Fig. 1.43.

(a)

(b)

Fig. 1.43

Thus if the directions of the two $0urces are different then the equivalent single source has a direction same as greater of the two soun:es with a value equal to the difference between the two sources. Key Point : Tire cu"o't S()urces to be conntcled in parallel must l!lzw same ooiU!ge ratings though thtir cumnt ratings may be same or diff~ent.

1.14 Kirchhoff's Laws In 1847, a German physicist, Kirchhoff, formulated two fundamental laws of electricity. These laws are of tremendous importance from network simplification point of view.

1.14.1 Kirchhoff's Current Law (KCL) Consider a junction point in a oomplex network as shown in the Fig. 1.44. At this junction point if It= 2 A, l 2 = 4 A and !3 = 1 A then to determine 14 we write, total current entering is 2 + 4 = 6 A while total current leaving is 1+ 14 A Fig. 1.44 Junction point

And hence,

14 = 5 A.

This analysis of currents entering and leaving is nothing but the application of J
Another way to state the law is,

Tlu algebraic 61lm of all the currmt meeting 11t a junction point Is always zero. The word algebraic ~)

meaN

considering the signs of various cuJTents.

at junction point = 0

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Circuit Theory

1 - 33


Sign convention Currmts flowing towards a jlmction point .are .auum~ to be positive while currm ts flowing away from a jumtion point ass~ to be negative. e.g. Refer to Fig. 1.44, currents !1 and !2 arc positive while l 3 and l4 arc negative. Applying KCL,

L I at junction 0

= 0

l1 +l 2 -J3 - 14 = 0 i.e. l1 +l2 = l3 +l4

The law is very helpful in network simplification.

1.14.2 Kirchhoff's Voltage Law (KVL) "In IUf'!/ network, the algebraic sum of the voltage drops across the circuit elemmts of any clo~ p.ath (or loop or mesh) Is equal to thl! algebraic sum of the e.mfs in the path" In other words, ''the algebraic sum of all the branch voltages, around any closed path or closed loop is always zero."

I

Around a closed path

R

R

Ao---J\II/IIIr--8 -

L V= 0

A o---J\II/IIIr--8

+

-I

+ 1-

The law states that if one starts at

a

goes on tracing and noting aU the potential changes

certain point of a closed path and

(either drops or rises), in any one particular direction, till the starting point is reached again, he must be at the same potential with which he started tracing a closed path. (a)

Fig. 1.45

(b)

Sum of aU the potential rises must be equal to sum of all the potential drops while tracing any closed path of the circuit. The total change in potential along a closed path is always zero.

nus law

is very useful in the loop analysis of the network.

1.14.3 Sign Conventions to be Followed while Applying KVL When current flows through a resistance, the voltage drop occurs across the resistance. The polarity of this voltage drop always depends on direction of the cu:rrcnt The current

always flows from higher potential to lower potential. In the Fig. 1.45 (a), current I is flowing from right to left hence point B is at higher potential than point A, as shown. In the Fig. 1.45 (b), current I is flowing from left to right, hence point A is at higher potential than point B, as shown.

Once all such polarities are marked in the given circuit, we can apply KYL to any closed path in the circuit. Now while tracing a closed path, if we go from - vc marked tcrmirull to + ve marked terminal, that voltage must be taken as positive. This is called potential rise.

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Ci~~ Theo~ ~------1_-_34 __-=B~as~i~c~C~Ir~c~ui~t~Ana~l~y~al~a~&_N_e_hNG ___~__ Red -=u~c~ti~ ·o~n_T~ech -=n_l~q-uea __ For example, If the bronch AB is traced from A to B then the drop across it must be considered as rise and must be taken as + I R while writing the equations. While tracing a closed path, if we go from +ve marked tertnin:al to - ve marked terminal, that voltage must be taken as negative. This is called potential drop. For example, in the Fig. 1.45 (a) only, if the branch is traced from B to A then it should be taken as negative, as - I R while writing the equations. Similarly in the Fig. 1.45 (b), if branch is traced from A to B then there is a voltage drop and term must be written negative as - I R while writing the equation. If the branch is traced from ll ! tl A, it becomes a rise in voltage and term must be written positive as + I R while writing the equation.

Key Point : 1) Potential rist i.e. travelling from negative to positively marked ltrmina/, must ~ considaed as Positivt. 2) Potential drop i.e. travelling from positiV< to negatively marked termitull, must ~ considaed as Negative. 3) While tracing a closed path, select a11y 011c dircctilm clockwise or anticlockwiu. This selection is totally indq~mdent of the directions of currents and' voltages of various branches of that closed path. 1.14.4 Application of KVL to a Closed Path Consider a closed path of a complex network with various branch currents assumed as shown in the Fig. 1.46 (a). As the loop is assumed to be a part of complex network, the branch currents are assumed to be different from each other. Due to these currents the various voltage drops taken place across various resistances are marked as shown in the Fig. 1.46 (b). B

I

R1

E1

- n;l I l

~-----

+

1'2111B

Rise

e,

Rise

•

1z -

R, ~ Drop

Fig. 1.46 (a), (b) Closed loop of a complex nehNork Copyngh!cd materia

Circuit Theory

1 • 35


The polarity of voltage drop along the current di.rection is to be !fllarlced a.s positive (+) to negative (- ). Let us trace this clo.sed path in clockwise direction I.e. A-B.C-0-A. Across R1 there l< voltage drop 11 R1 and as getting traced from +Ve to -ve, it is drop and must be taken as negative while a pplying KVL Battery E 1 is getting traced from negative to positive i.e. it is a rise hence must be considered as positive.

Across Rz there is a voltage drop 12 R 2 and as getting traced from ·+ve to -ve, it is drop and must be taken negative. Across R 3 there is a drop !3 R3 and as getting traced from +ve to -ve, it is drop and must be taken as negative. Across R4 there is drop I4 R4 and as getting traced from +ve to - ve, it is drop must be ta.ken as negative. Battery Ez is getting traced from - ve to +ve, it is rise and must be taken as positive. :. We can write an equation by using KVL around this closed path as, -I t R1 + Et -lz Rz -13 R 3 - !4 R4 + Ez = 0

i.e.

... Required I
Et + Ez =It Rt +lzRz +I3R3 +I4R4

If we trace the closed loop in opposite direction i.e. along A·D-C·B-A and follow the same sign convention, the re;ulting equation will be same as what we have obtained above. Key Point : So while applying KVL, direction in which loop is to be traced important but following the sign con~1tion is most important.

is not

The same sign convention is followed in this book to solve the problems.

1.14.5 Steps to Apply Kirchhoffs Laws to Get Networtc Equations The steps are stated based on the branch current method. Step 1 : Draw the circuit diagram from the given information and insert all the values of sources with appropriate polarities and all the resistances. Mark aU the branch currents with some assumed directions using KCL at various nodes and junction points. Kept the number of unknown currents minimum as far as possible to limit the mathematical calculations required to solve them later on.

Step 2 :

Assumed directions may be wrong, in such case answer of such current will be mathematically negative which indicates the correct direction of the current A particular current leaving a particular source has some magnitude, then same magnitude of current should enter that source after travelling through various branches of the network

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Circuit Theory

1 -36

Basic Circuit Analysis & Networtl Reduction Techniques

Step 3 : Mark all the polarities of voltage drops and rises as per directions of the assumed branch cu.r rents flowing through various branch resistances of the network. 1lUs is necessary for application of KVL to various closed loops. Step 4 : Apply KVL to different closed paths in the network and obtain the corresponding equations. Each equation must contain some element which is not considered i~ any previous equation. Key Point : KVL must II<' applied to sufficient number of loops such tlult tllch dement of

tire network is inclutkd atltJISI once in any of tire equations. Step 5 : Solve the simultaneous equations for the unknown currents. From these currents unknown voltages and power consumption in different resistances can be calculated. What to do if current source exists ? Key Point : If tlrert is current source in tire ndwork thtn compltlt tht currtnt distribution considering tire current source. But whik applying KVl. the loops should not II<' considered inooltring current source. The loop tqiUitions must II<' written to ~ loops whicll do not include any current source. This is because drop llCTOSS current source is unknown.

For example, consider the circuit shown in the Fig. 1.47. The C\lrrel\t clistribution is completed in of current source value. Then KVL must be applied to the loop bcdcb, which does not include current source. The loop abefa should not be used for KVL application, as it includes current source. Its effect is already considered at the time of current distribution.

/lAII!!I+.SS!il:ll'-,lilt!".,. HI ciS 1 ) 1 0 This loop 8 1 6/Jr.tllllll!::,.. ,---'>IIM--i'"'-.....:..IIJI.'Jv--lc can not be (5-1,) considered lor application ol KVL • •

SA

e

(5 - J,)

!'.: ~ ~r=

considered in of current disllibulion.

10V

d

Fig. 1.47

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Circuit ThiKifY

1 • 37

Basic Circuit Analysis & Network Reduction Technlq~,~ea

1.15 Cramer's Rule If the network is complex, the number of equations i.e. unknowns increases. In such case, the solution of simultaneous equations can be obtained by Cnmer's Rule for determinants. Let us assume that set of simultaneous equations obtained is, as follows : 8J1Xt +a12X2+ ·····••• +8tnXn :

Ct

•2t"t +aux2+ ........ +a2n xn = ~

Where C1 ,C2 , .••.. •. ••• ... •. c. are constants. Then Cramer's rule says that form a system determinant 6 or D as,

=

D

Then obtain the subdeterminants Djby replacing jtb column of 6 by the column of constants existing on right hand side of equations i.e. C1,C2 , •• • Cn;

01

~

c, •tz

"'"

•11

c,

Cz

•22

a2n

• zt 02 =

C2

c.

•n2

3nn

•t2

c,

• 21 •22

Cz

•••

Cn

•11

and

Dn

=

'n2

.., c.

•tn •zn 8nn

The unknowns of the equations arc given by Cramer's rule as,

D, x, = 1)' Where

D 1,

0 2 , ... ,

Dn

Xz

Dz = D ' ..............,

x.

On

=D

and D are values of the respective determinants,

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Cln:ult Theory

1 - 38

1-+

Apply Kirchhoffs current law and !"'llage law to lk circuit shown in the

Example 1.8 : Fig. 1.48.

..

Basic Cln:ult Analysis & Network Reduction Techniques

lndialte the various brRnch currents. Write down lk equations refilling lk various lmmch curretlls.

300

150

,.

••

•

So/VI! these equations to find IN values of these currmts.

SOV ': ="

Is the sign negati!N! ?

Flg. 1.48

If yes,

of

any of the calcu/altd currents

explain the signifiCJlTict of the negaliVI! sign.

Solution : Application of Kirchhoff's law : Step 1 and 2 : Draw the d rcuit with all the values which are same as the given network. Mark all the branch currents starting from +ve of any of the source, say +ve of 50 V source. Step 3 : Mark all the polarities for different voltages across the resistances. This is combined with step 2 shown in the network below in Fig. 1.48 (a). t, - lz 300 8 ' • • ··- lz • ··150

A

'•

• sov-::-

F

Step 4 :

•

200

E

'•

Fig. 1.48 (a) Apply KVL to different loops.

c t, - 12

•

··-·2

100V

D

Loop 1 : A-8-E-P-A.,

- 15 11 - 20

lz + 50

• 0

... (I)

lv - 100 + 20 1z = o

... (2)

Loop 2 : B-C-D-E-0, - 30 (II -

Rewriting all the equations, taking constants on one side.

15 11 + 20

lz =

50

...(1)

and

- 30 11 + 50

lz

= 100

...(3)

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Circuit Theory

1 • 39


Apply uamer's rule,

Calculating 0 1,

115 201

0 = - 30 50 = 1350

o,

,, Calcula ling 0 2,

=

II: 50201

= 500

= 500 = 03 7 A = Dt 0 1350

02 = 12 =

15

so

I

- 30 100 = 1 02

D

3000

3000 = 1350 = 2.22 A

For 11 and 12, as answer is positive, assumed direction is correct. :. For I1 answer is 0.37 A. For I2 answer is 2.22 A. 11 - I2

=

0.37 - 2.22

= -1.85 A Negative sign Indicates assumed direction Is wrong.

i.e. I1 - I 2 = 1.85 A flowing i.n opposite direction to that of the assumed direction.

1.16 A.C. Fundamentals Let us discuss in brief, the fundamentals of altemabng circuits consisting of various alternating current and voltage sources, resistances alongwith inductive and capacitive

reactances. An alternating quantity is the one which changes periodically both in magnitude and direction.

Generated o.mJ.

Amplitude

Practically a purely sinusoidal waveform is accepted as a standard alternating waveform for alternating voltages and currents, d ue to its advantages. Let us consider the waveform of an alternating quantity which is purely sinusoidal as shown in the Fig. 1.49.

lime period T seconds

onecyde

Fig. 1.49 Waveform of an alternating e.m.f.

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Circuit Theoty

1 • 40

Bale Circuit Analyals & Network Reduction Techniques

1.16.1 Instantaneous Value The value of an alternating quantity at a particular instant is known as its instantaneous value. e.g. e1 and e2 are the instantaneous values of an alternating e.m.f. at the instants t 1 and ~ respectively shown in the Fig. 1.49.

1.16.2 Waveform The graph of instantaneous values of an alternating quantity plotted against time in called its waveform.

1.16.3 Cycle Each repetition of a set of positive and negative instantaneous values of the alternating quantity is called a cycle. Such repetition occurs at regular interval of time. Such a waveform which exhibits variations that reoccur after a regular time interval is called periodic waveform. A cycle can also be defined as that interval of time during which a complete set of non-repeating events or wave form variations occur (containing positive as wcll as negative loops). One such cycle of the alternating quantity is shown in the Fig. 1.49. Key Point: One cycle corresponds to 2n radians or 360°.

1.16.4 Time Period (T) The time taken by an alternating quantity to complete its one cycle is known as Its time period denoted by T seconds. After every T seconds, the cycle of an alternating quantity repeats. This is shown in the Fig. 1.49.

1.16.5 Frequency (f) The number of cycles completed by an alternating quantity per second is known as its frequency. It is denoted by f and it is measured in cycles I second which is known as Hertz, denoted as Hz. As time period T is time for one cycle i.e. seconds I cycle and frequency is cycles/second, we can say that frequency is reciprocal of the time period. V.i

v.l

..

., flme

~----H~hT----~

Less number of cycles per second I.e. low frequency

Less T

-More number of cyCles per second i.e

high frequency

Fig. 1.50 Relatio n between T and f Copyrgh!cd malcri~

Cil'(;uit Theory

Basic Cil'(;uit ~alysis & Network Reduction Techniques

1 • 41

A~

time period increases, frequency decreases while as time period decreases, frequency increases. This Is shown in the Fig. 1.50. ; I

In our nation, standard frequency of alternating voltages and currents is 50 H7_

I

1.16.6 Amplitude The maximum value attained by an alternating quantity during positive or negative hall cycle is called its amplitude. It is denotL'
1.16.7 Angular

Frequency(~

It is the frequency expressed in electrical radians per second. As one cycle of an alternating quantity corresponds to 2 It radians, the angular frequency can be expressed as (2 n >< cycles/sec.) It is denoted by • ro' and its unit is radians/second. Now, cycles/ sec. means frequency. Hence the relation between frequency 'f' and angular frequency · ro' is,

I

2 nf

Ill =

radians/sec.

or

oo =

:.;.

radians/sec.

I

1.16.8 Equation of an Alternating Quantity As alternating quantity is sinusoidal in nature, its quation is expressed u~ing sin where 6 is angle expressed in radians. Hence alternating voltage is expressed as,

I

c= fmsinO

a

I

While alternating current is expressed as,

I

i

= lmsinO

I

This equation gives instantaneous values of an •ltem•ting quantity, •t any time t. Now

a

=

rot

in radians

Thus various forms of the equation of an alternating quantity are, e

and

t)

=

2 fmsin (rot)= Em sin (2n f I) = Em sin ( ;

=

2 lm sin (Cill) = 1, sin (2rt f I) = I, sin ( ; t)

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Circuit Theory

1 - 42


Important Note : In all the above equations, tl1e angle 6 is expressed in radians. Hence,

while calculating the instatJtaneous value of tht t.mf, it i.~ necessary to CtJiculatt the sine of the expressed in radians. Key Point: Made of the calculator should be converted to radillns, teo calculate tht sine of 1/u: angle expressed in radians, before substiluti11g in any of the above equations.

at~gle

In practice the alternating quantities are expressed in of their r.m.s. values. The relation between r.m.s value and the maximum value is,

I

Yr.ms

and

= V.n .J2

Ir.m.s -72, - 1,.

I

The r.m.s values are denoted by the capital letters as V or I.

1.17 Phasor Representation of an Alternating Quantity In the analysis of a.c. circuits, it is very difficult to deal with alt.c:mating quantities in of their waveforms and mathematical equations. The job of adding, subtracting, etc. of the two altematin.g quantities is tedious and time consuming: in of their mathematical equations. Hence, it is necessary to study a method which gives an easier way of representing an alternating quantity. Such a representation is called phasor representation of an alternating quantity.

The sinusoidally varying alternating quantity can be represented graphically by a straight line with an arrow in this method. The length of the line rep.resents the magnitude of the quantity and arrow indicates its direction. This is similar to a vector representation. · Such a line is called a phuor. Key Point: The phosors art assumed to be rolared in antic/ockwise direction. One complete cycle of a sine wave is represented by one complete rotation of a phasor. The anticlockwise direction of rotation is purely a conventional direction which has been universally adopted. Consider a phasor, rotating in anticlockwise direction, with uniforrn angular velocity, with its starting position 'a' as shown in the Fig. 1.51. If the projectio.n s of this phasor on y •

CUtr~t

c Antidockwlse \ rotation

9

9

Fig. 1.51 Phasor representation of an alternating quantity Co yrgh!cd malcri~

Circuit Theory

1 - 43

Basic Circuit Analysts & Network Reduction Techniques

Y-axis arc plotted against the angle turned through'

e ',(or time as e =

w t), we get a sine

wavcfonn.

Consider the va.rious positions shown in the Fig. 1.51. 1. At point 'a', the Y-axis projection is zero. The ins tan taneous value of the current is aJso zero.

2. At point 'b', the Y-axis projection is [ I (ob) sin 0) ]. The length of the phasor is equal to the maximum value of an alternating quantity. So, in~tantaneous value of the cu.r rent at this position is I = Im sin 0, represented in the waveform. 3. At point 'c', the Y-axis projection ' oc' represents entire length of the phasor i.e. instantaneous value equal to the maximum value of current Im·

4. Similarly, at point d, the Y-axis projection becomes Im sin 0 which is the instantaneous value of the current at that instant. 5. At point 'e', the Y-axis p rojection is zero and instantaneous value of the current is z.ero at this instant.

6. Similarly, at points f, g, h the Y· axis p rojections give us instantaneous values of the current at the respective instants and when p lotted, give us n egative half cycle of the alternating quantity. Thus, if the length of the phasor is taken equal to the maximum value of the alternating quantity, then its rotation in sp ace at any instant is s uch that the length of its projection on the Y-axis gives the instantaneous value of the alternating quantity at th at particular instant. The angular velocity ' w ' in an anticlockwise direction of the phasor should be such that it completes one revolution in the same time as taken by the alternating quantity to complete one cycle i.e. 9 = wt, Where

w = 2 n f rad/sec

Points to :

In practice, the alternating quantities are represented by their r.m.s. ·v alues. Hence, the length of the phasor represents r.m.s. value of the alternating quantity. In such case, projection on Y-axis does not give directly the instantaneous value but as lm = ..fi 1, .• ,..1 the projection on Y-axis must be multiplied by ..fi to get an instantaneous value of that alternating quanl!ty. Phasors are always assumed to be rotated in antlcloc:k wlse direction.

Two allemating quantities of same frequencies can be represented on same phasor diagram.

If frequmcies of the two qwmlities are differmt, then sudr IJ!Ialltities camrot be represented on the same plulsor diagram.

Key Point:

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, Cin:uit Theory

1 • 44

Baaic: Cin:uit Analysis & Network Reduction Techniques

1.18 Concept of Phase of an Alternating Quantity In the analysis of alternating quantities, it L• necessary to know the position of the phasor representing that alternating quantity at a particular instant. It is represented in tenns of angle e in radians or degrees, measured fr.1m certain reference. Thus, phase can be defined as, Phas~ : The phase of an alternating quantity at any instant is the angle ~ (in radians or degrees) travelled by the phasor representing that alternating quantity upto the instant of con~deration, measured from the reference.

2•

O=mt

Fig. 1.52 Concept of phase Let X-axis be the reference axis. So, phase of the alternating current shown in the Fig. 1.53 at the instant A is ¢ = 00. While the phase of the current at the instant B is the angle q> through which the phasor has travelled, measured from the reference axis i.e. X-axls.

In genera I, the phase ~ of an alternating quantity varies from ~ = 0 to 2 lt radians or ~ = 00 to 3600.

A

Current

~~~~~~~~--~~ 9=~

-- o Reference

T

2:t

4

lnstant

B

1 - - - T --=-----1 llme period

Fig. 1.53

Another way of defining the phase is in tenns of a time period T. The phase of an alternating quantity at any particula:r instant is the fraction of the time period (T) through which the quantity is advanced from the reference instant. Consider alternating quantity represented in the Fig. 1.53. As per above definition,. the phase of

quantity at instant A is {, while phase at instant B is

3 T. Generally, the phase is 4

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Circuit Theory

a..ic Circuit Analysis & Network Reduction Techniques

1 • 45

expressed in tenns of angle ¢ which varies from 0 to 2 n radians and measured with respect to positive x-ruc.is direction.

In of phase the equation of alternating quantity can be modified as,

I

Where

e = Em sln(wt±$)

I

9 = Phase of the alteiThlting quantity. Let us consider three cases; Case 1: $ =0°

When phase of an alternating quantity is zero, it is standard pure sinusoidal quantity having in.~tantaneous value zero at t = 0. This i~ shown in the Fig. 1.54 (a). Dse 2 : Positive phase ¢ When phase of an alternating quantity is positive it means that quantity has some positive inst.tntaneous value at t = 0. This is shown in the Pig. 1.54 (b). Dse 3 : Negative phase Q When phase of an a!tem.lting quantity is negative it means that quantity has some negative inst.tnt.tneous value at t = 0. This is shown in the Fig. 1.54 (c). Voltage

or

Vottage

vonage

current

or

or

current

current +Ve

t =0

~~~(@Jl}ml! Phase ~=

o•

:'

-o:

!

. / - Phasor

------~L -

t

0 Phasor

Negative phase - o

Positive phase $

diagram

''

Phasor •

diagram 't-

~

o

diagram

(ajZero phase

' _____ _o~· - ----

(bl Poaltiva phaao

(cl NogaUvo phase

Fig. 1.54 Concept of phase

Copyrgh!cd malcri~

Circuit Theory

1 • 46


1. The phase is measured with respect to reference direction i.e. positive x axis direction. 2. The phase measured in anliclockwise direction is positive while the phase measun.>d in clockwise direction is neg~tive.

1.18.1 Phase Difference Consider the two alternating quantities having same frequency f Hz having different maximum values. e

Em sin (w I) ~

and

Im sin (w t)

Em > Im

where

The phasor representation and waveforms of both the quantities a.re shown in the Fig. 1.55.

'"

!-l---:=-1--t~--.\···-······E,II 0

0 =WI

Fig. 1.55 The phasors

OA

and After 0 ~

7t

2

radians, the OA p hasor achieves its maximum E, while at the same

instant, the 08 phasor achieves its maximum ~· As the frequency of both is same, the angular velocity w of both is also the same. So, they rotate together in synchronism. So, at any instant, w e can say that the phase of voltage e will be same as phase of i. Thus, the angle traveUcd by both within a particular time is always ·t he same. So, the difference between the p hases of the two quantities is zero at any instant. The difference between the phases of the two alternating quantities is caiied the phase difference

Copyrghtcd malcri~

Circuit Theory


1 - 47

which is nothing but the angle difference between the two phasors representing the two alternating quantities. Kay Point : When s11ch phase diflnence bet-.uern the two altern11ting quantities is ::.eru,

the hvo quantities are said to be in p/rase. The two alternating quantities having same frequency, reaching ma ximum positive and negative values and zero values a t the same time are said to !be in phase. Their amplitudes may be different In the a .c. analysis, it is not necessary that aU the alternating quantities must be always

in phase. It is possible that if one is achicv;ng its zero value, at the same instant, the other is having some negative value or positive value. Such two quantities arc said to have phase difference between them. If there is difference between the phases (angles) of the two quantities, expressed in degrees or radians at any particular Instant, then as both rotate with same speed, this difference remains same at all the instants. Consider an e.m.f. hav;ng maximum value Em and cu.rrcnt hav;ng maximum value lm. Now, when c.m.f. 'e' is at its zero value, the current 'i' has some negative value as shown in the Fig. 1.56. e,i

... ,.....

••

/•

...

..... .f .......

. ..

. , __j__ __f.-....- : l

/.

!

\

',

\

Of

•

'.. •.,.

0

I

....

........ .

....

~

__________

' :A

'• (I) ' ~~~-i·--···

I / :

'

~•

\

\.

:

;

E,

........

-··+--::.,-_-------''1--~-.:,._

I

•

•

.. .. B·------~, ...- - - - · · - - - ·· ·· ,• -lm

/l'l ..

.

..

f)

..............,_....:.;·_·______::-1---------"-- - - -Em

Fig. 1.56 Concept of phase difference (Lag)

Thus, thef" eJrists a phase difference 9 between the two phasors. Now, as the two are rotating in anticlockwisc direction, we can say that current is falling back with respect to

voltage, at aU the times by angle \) . This is called lagging phase difference. The current i is said to lag the voltage e by angle <1> • The current i achieves its m.axi.ma, zero values 4> angle later than the corresponding maximum, zero values of voltage. The equations of th.e two quantities arc written as,

e = Em sin rot

and

i = lm sin (rot-~

'i' is said to lag 'e' by nngle $.

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Circuit Theory

1 • 41

Basic Circuit Analysis & Networtl R.eduction Techniques

It is possible in practico: that the current 'i' may have some positive value when voltage 'e' is zero. 1lUs is shown in the Fig. 1.57.

Fig. 1.57 Concept of phase difference (Lead)

It can be seen that there exists a phase difference of oangle between the two. But in this case, current 'i' is ahead of voltage 'e', as both are rotating in anticlockwise direction with same speed. Thus, current is said to be leading with respect to vo ltage and the pha~ difference is called luding phue difference. The current I achieves its maximum, zero values 9 angle before than the corresponding maximum. zero values of the voltage. At all instants, current i is going to remain ahead of voltage 'e' by angle 'fl. The equations of such two quantities are written as

e

~

Emsincot

and

i~Imsin(cot+~

' i' is said to lead 'e' by angle $. Key Point : Thus, rtlaJed to the plulse difftrenee, it can be remnnbered that a plus(+) sign of angle indicates lead where as a minus (-) sign of angle indicates lag with respect to the rtfertJtee.

1.18.2 Phasor Diagram The diagram in which different alternating quantities of the same frequency, sinusoidal in nature are represented by individual phasors indicating exact phase interrelationships is known as phasor diagram. The phasors are rotating in anliclockwise direction with an angular velocity of co= 2 1t f rad/sec. H ence, all phasors have a particular fixed position wi th respect to each other. Key Point : Hence, p}lflS<)r dingram can be considered as a still picture of tlrese phasors at a

particular instant. To clear this point, consider two alternating quantities in phase with each other. e

~

E,., sin co t

and

i

~ 1m

sin cot Copyrgh!ed malcri~

Circuit Theory

1 • 49


At any instant, phase difference between them is zero i.e. angle diffe rence between the two phasors is zero. Hence, the phasor diagram for such case drawn at different instants will be alike giving us the same information that two quantities are in :phase. The phasor diagram drawn a t different instants are shown in the Fig. 1.58. E~

!-- - -:, - --

0

E~

~

0

0

Fig. 1.58 Same phasor diagram at different Instants

Consider another example where current i is lagging vollage c by angle ¢. So, difference between the angles of the pha>Ors representing the two quantities is angle 9 . e = E, sin ro t

= lm sin (rot -

and

¢)

The phasor diagram for such case, at various instants will be same, as shown in the Fig . 1.59 (a), (b) and (c).

eJ

0

--

Em

\

9

1m Lagging

/

Lagging

/ 1m Lagging

9

[Em

lm

0

0

Fig. 1.59

The phasor diagram drawn at any instant gives the same information. Key Point : tluit the liigging and ltadlng W<>fll l; fl!latlvt to lilt refernru. In tlze above CJZS<, if we take current as rcfererrce, we luwe to say tltot tire voltage ICDds current by angle • '[he directwn of rotation of pltosors is always a11ticlockwist. Important Points Regarding Phasor Diagram : 1) As phasor diagram can be drawn at any instant, X and Y rucis are not included in

it. But, gcneraUy, the refcr~'flce phasor chosen is shown along the positive X axis direction and at th.1t instant other phasors are shown. This is just from convenience point of view. Th e individu~l phase of an alternating q uantity is always referred with respect to th e positive x-axis direction. Copyrghicd malcri~

Circuit Theory

1 -50


2) There may be more than two quantities represented in phasor diagram. Some of them may be current and some may be voltages or any other illtcmating quantities like flux, etc. lbe frequency of aU of them mus t be the same. 3)

Generally, length of phasor is drawn L'CJUal to r.m.s. value of an alternating quantity, rather than max.imum value.

4) The phasors which are ahead, in anticlor.kwise direction, with. respect to reference

phasor a.r e said to be leading with respect to reference and phasors behind arc said to be lagging.

l•

5) Different arrow heads may be used to differentiate phasors drawn for different altem.tting'quantities like current, voltage, flux, etc. Example 1.9 : Two sinusoidal cu"cnts arc given by, i1 = 10 sin (rot + W3J and

i 2 = 15 sin (rot - 'lr/4)

''

'/

Calculate the phase difference between them in degrees. radians i.e.

6()0

''

n/3

Solution : The phase of current i 1 is

'

while the phase of the current

i 2 is - n/4 radians i.e. - 45°. This is shown in the Fig. 1.60.

Hence the phase difference between the two is, o = 01

-

The individual phases 60' and - 45° 0

_ ------- - - -

arewHhrespectto

•ve x-axis direction

02 = 60' - (- 45") = 105•

And i 2 lags i 1•

Fig. 1.60

1.19 Mathematical Representation of Phasor Any phasor can be represented mathematically in two ways, 1) Polar co-ordinate system Let

i =

and

I., sin (wt

2) Rectangular co-ordinate system + ¢l

The phase of current i is 9. The phase is always with respect the x-axis as shown in the Fig. 1.61.

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Circuit Theory


1 ·51

I I

c ··------------

I(OA) = I, = r

A

polar syst~m, the phasor is represented as. r L ± 9. So current i above, is represented as I, L + i> in polar system. In rectangular system, the phasor is divided into x and y components i.e. real and imaginary components as x ± j y. The current i above is represented as, lm cos i> + j im sin 9 in rectangular system.

Fig. 1.61

r

phase = + 9

In

r

Polar system,

and

L

± ¢ while

= r cos

and

X

while

r = Jx2+y2,

Rectangular system, x ± j y y=rsin$ ...(1)

9= tan· I(~)

... (2)

The equations (1) and (2) can be used to convert rectangular form to polar or vice versa. Such rectangular to polar and polar to rectangular conversion is often required in phasor mathematical operations like addition, subtraction, multiplication and division. Key Point : Instead of using above rtlations, sh1dmts can use the polar to rectangular (P-+ Rl and rectangular to polar (R -+ P) functions available on calculator for lht required con~JUSions.

As the gt'aphical method is time consuming which includes plottin,g the phasors to the scale, generally, analytical method is used. Also, graphical method may give certain error which will vary from person to person depending upon the skills of plotting the phasors. The answer by analytical method is a.lways accurate. Very Important : Tire polar fonll of eqomtion or pllasc as,

1111

e = Em

altmrnting quantity ca11 be easily obtniucd frou< its sin(Cllt±~

then

E i11 polar = E L±~ whtre E = r.nu. v alue ,...

Example 1.10 : Write the polAr fonn of tire voltage given by, V = 100 sin (100 n I+ n/ 6) V

Obtain its rectangular fqnn. Solution : Vm = 100 V and¢ =

+~ rad = +30°, v,,m_<. = ~

= 70.7106 V

CoPY

lied 1a "'

Circuit Theory

1 • 52

In polar form =

Baalc Circuit Analyala & Network Reduction Technlquea

70.7106 L + 30° V

:. Rectangular form = 61.2371 + j 35.3553 V

of an altmrating qwmtity aists in its polAr form and tlbt in rectangu/IIT form. Tlrtu to find r.m.s. valru of an alternJJting qumdity e:qnus it in polar form. , . Example 1.11 : find r.m.s. t1alut and phRse of the current l = 25 + j 4Q A. Key Point: 1k r.m.s. u/Que

Solution : The r.nu. value is not 25 or 4Q as it exists in polar form. Converting it to polar form, I = 47.1699L57.99° A "' Ir.m._ L 'A

r.m.s. value of current = 47.1699 A Phase = 57.99° Key Point: To obtain phRse, express 1M equation in sint form

if givm in cosine as,

u

e = Em cos (rut)

then

e = Em sin (rut + 90) as sin(90 + e) = cos 9

Thus the phase is 90• and not zero. In general,

e = Em e<,.;( rot± qt

then

e = Em The phase =

,.

sin(rot +90±~

90±~

Example 1.12 : A wltagt is defined as - Em cos Oll. Express it in polAr form.

Solution : To express a voltage in polar form expres.• it in the form, e = E,. sin rut Now

e = -Emroswt = -Emsin(rut+¥) 3

= Em sin ( Clll+ ; )

assin(CJlt+¥)=coswt as sin (n+ll) = - sln 0

Now it can be expressed in polar form as, 3n e = Em L + ·rad = Em L + 2.700 V

2

But + 210• phase is nothing but - 90° e = EmL-90"V

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1.20 Multiplication and Division of Phasors In th~ last section, th~ addition and subtraction of phasors is discussed, which is to be carried out using rectangular form of phasors. But the rectangular form is not suitable to perform multiplication and division of phasors. Hence multiplication and division must be performed using polar form of the phasors. Let P and Q be the two phasors such that, and

p = XJ +jyl

To obtain the multiplication Px Q both must be expressed in polar Corm P

and

q L¢1

Q = r2 L¢2

Then Key Point : Thus itl multiplication of complex 1111mbers

i11

polar fonn, the magnitudes get

multiplied while their angles get added. th~

The result then can be expressed back to rectangular form, if required. Now consider division of the phasors P and Q .

..!: Q

= fJ LQJ

r2L¢2

=I'rJ! 21 L¢1 - Qz

Key Point : Tlws in division of compltr 11umbers in polar form, the trrAgllitudts get divided

wlrile their angles get subtracted. Note : For converting polar to rectangular form, the students can use the function PH R on calculators without using basic conversion expressions. Similarly for rectangular to polar conversion, the students can use the function R H P on calculators without using basic conversion expressions. :

While <~di1ition and subtraction, use reclal•gular form. While mu1tiplication and division, usc polar form.

1.21 Impedance In the alternating circuits alonb•with the resistances, inductances and capacitances also play an important role.

The inductances are represented by inductive reactances in a.c. circults. An inductive reactance is the ohmic representation of an inductance denoted as XL and given by,

I

XL =

(I)

L=2

1t

f L !l

I Copynghl o"'

,;,,

Circuit Theory

1 • 54


The capacitances arc represented by capacitive reactances in a.c. circtrits. A capacitive reactance is the ohmic representation of a capacitance denoted as Xc and given by, 1

1

Xc = roe = 21tfC n The combination of R, XL and Xc present in the circuit is called an Impedance of the circuit. The Impedance is denoted by letter Z. But the behaviour of R, L and C is different from each other in a.c. circtrits hence R, XI. and Xc cannot be algebraically added to find total impedance of the circuit. Let us summarize the behaviour of R L and C in the tabular form. Parameter

Punt rctaistance R Pure Inductance l

Cha..cterlstlcs

v and inphase

I

•••

I logo V by 90"

Impedance In noctangular form

Impedance In polar form

Z • R+jO

Z • R <: O'

Z=O+JXL

Z = Xt_ L +90'

v rv Phasor dlag..m

I

I Pure capacitance c

I leads V by 90•

Z=O - JJ

Z=Xc <:- 90'

LV

Table 1.4 Inductive reactances are represented by positive sign + XL in the impedance while capacitive reactances are represented by neg.otive s ign - Xc in the impedance. Thus for R • L series circuit, the impedance is represented as,

z where

IZ I

= R + J x._ = =

IZI ..::o•n

J R2 +(Xd 2

and 8 = tan-1

xi

In such circuit, current lags voltage by angle 0.

For R..C series circuit, the impedance is represented as,

Z = R -J Xc =IZI.<:O" where

IZI = ~ R2 +(Xc)2

In this case 0 is negative and current leads voltage by angle 0.

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Circuit Theory

1 ·55


Find out tilt resistance a11d inductance or CDparitaPict of tire impeda11ces if frequency is 50 Hz. i) 25 L 45" C! iii 6 + j 8 !l iiil 8 - j 10 Cl

,..,. Example 1.13 :

give~~

Solution : i) Z = 25 L 45" !l

Converting to rectangular fonn,

z = 17.68 + J 17.68 n Comparing with z = R + j Xt n n

R = 17.68 Now

Xt.

=

n

2nfl 68 =2n17·xSO =0.0562 H

:. L ii)

XL = 17.68

and

z=6 +i8n

Comparing with Z = R + j XL Q .. R = 6

n

8

Now iii)

Xt = 8 n

and

= 2n xSO = 0.0254 H

z = s - i 10 n

Comparing with. Z = R - j X c :. R =

Now

sn

and

Xc

=

1 2nfC

:. c

=

I 2nfXc

Xc = 10

I

2rtx50xl0

n

= 3.18 X 10-4 F = 318 IJ.F

1.22 Power Factor The numerical value of cosine of the phase angle between the applied voltage and the current drawn from the s~pply voltage gives the power factor.

It is also defined as the ratio of resistance to the impedance. It is denoted as cos<>·

1cos q, = p .f. = For pure L and C,

~=

~

1

90" hence the pJ. is zero.

For other combinations, the p.f. is de fined as lagging or leading i.e. whether the resultant current Jags or leads the supply voltage.

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1.23 Power The power in a.c. circuit is given by, I P=V I cos¢1 where cos 0 = Power factor of the circuit. Now pure inductance and capacitance does not consume any power as cos ~ = 0 for such circuits. Only resistance consumes power. Hence power also can be obtained as, p = [ZR

where R = Resistive part of the equivalen t impedance of the circuit. Key Point: While calculating the equivalent impedance, it should be noted that while

adding tltt impedances in Sfries, the impedances must be expressed in rectangular form while multiplication and division of impedances mu..
1.24 Series R-L-C Circuit The series R·L.C circuit is shown in the Fig. 1.62. R

L

c

The impedance is given by,

z = R + i x~. - i Xc = R + j (XL - Xc)

v Fig. 1.62

If XL > Xc then resultant impedance is inductive and current I lags voltage V. If XL < Xc then resultant impedance is capacitive and current I leads voltage V. If XL = Xc then circuit becomes purely resistive

and I is in phase with V. The voltage d rops across the various clements arc denoted as V R• VL and VC' The m~gnitudes of these drops are given by, and Note that algebraic sum of VR, VL and Vc is not V as these three voltages are not in phase. But the veclor addition of VR, VL and Vc is the supply voltage V.

The network shown in the Fig. 1.63 is operatirtg in a sinusoidal steady state. Find voltage across capacitor, resistor and inductor.

,. . Example 1.14 :

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Circuit Theory

1 -57


sn

20mH

1 00~F

I:::::J

10 cos SOOt rv

Fig. 1.63 Solution : The applied voltage is 10 cos (500 t). Expressing it as,

v

= 10 sin (500 t + 90') . •. as sin (90 +

e) = cos e

Compare with, V = '!Im sin (
:. w = 500 rad/scc,

)i = 7.071

V

= 7.071 L 90"

v

: . V(RMS) =

:. v Now

Vm = 10 V, 0 = 90"

R = 5 0 ,L = 20mH, C = 100 IJF :. XL = roL = 10 Q

and

Xc :. Zr

=

I we= 20 n

= R + j (XL - Xc) = 5 + j (10- 20) = 5- i 10

n=

11.1803 L-63.43"0

7.071 L90" • • v :. i = - = ll.180 L - . • = 0.6324 L b3.43 A 3 63 43 ZT Hence voltages across the various elements are,

vR VL

I i I X R = 0.6324 X 5 = 3.162 v = Iil )(XL =0.6324 X 10 =6.324 v

Vc

= Iil x Xc = 0.6324 x 20 =12.648 v

=

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Circuit Theory Sr. No.

1 • 58

Basic Circuit Analysis & Network Reduction Techniqun 0

p.f . coa ~

o·

1

Unity p.f.

90'

0

Zero lagging

- 90'

0

Zero leading

D' L 9 L 90'

cos 9

Lagging

cos 0

Leading

Impedance (Zl

Circuit

Polar

Rectangular

;a n

1.

Pute R

R tt. O' Q

R•

2.

Pure L

XL L 90' !l

O+ j XL Q

3.

Pure C

Xc L -

0 -j

4.

Series RL

IZI L + 9'CI

R+ JXL f!

5.

Series RC

IZI

L - ~· n

R-jJ

90' (!

Remark

Xc!l

- 90' L

•

~

L 0

Xt > Xc Lagging 6.

Series RLC

IZI /. :t ~ n

R + j X 11 COSQ

0

X= XL - Xc

XL <

Xc Leading

Xt = Xc Unity Table 1.5 Summary of R, L and C series circuits

1.25 A.C. Parallel Circuit A parallel circuit is one in which two or more impedances are connected in parallel across the supply voltage. Each impedance may be a separate series circuit. Each impedance is called branch of the parallel circuit. The Fig. 1.64 shows a parallel circuit consisting of three impedances connected in parallel across an a.c. supply of V volts. Key Point: The voltage across all tile impedar~ces is same as supply t-ollage of V volts.

-

~------~~~------~ VvO:Is

Fig. 1.64 A.C. parallel circuit

The current taken by each impedance is different. Applying Kirchhoff's law,

... (J>hasor addition)

It + l2+l3

v

v

v

v

z.

z2

Z3

1

I

1

=

=-+~+=-

z=

~ z. · ~·~ z2 Z3

z I

Where Z is called equivalent impedance. This result is applicable for 'n' such impedances connected in parallel.

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Basic: Circuit Analysis & Network Reduction Techniquaa

1 · 59

1.25.1 Current Division for Parallel Impedances If there are two impedances connected in pa.rallcl and if IT is the total cuncnt, then current division rule can be applied to find individual branch currents.

-'• ; -IT

X ,__

-" = -IT

X ,--....:.,-

•

z2 ..::;"'= z. + Zz

:z,

Z1 + Zz

1.25.2 Concept of ittance ittance is defined as the ~iprocal of the impedance. It is denoted by Y and is measured in unit siemens or mho. Now, current equation for the circuit shown in the Fig. 1.65 is,

I

;

11 +12 + 13

I

=

V

VY

;

x(·d )+ V x(iz ) + V x(~3 )

VY1

1

+ VYz + VY3

y = Y1 + Yz +·Y3

..

where Y is the' ittance of the total circuit. The three impedances connected in parallel can be replaced by an equivalent circuit, where three ittances are connected in series, as shown in the Fig. 1.65.

~----~~}-----------J

v

L---------~~¥---------~

v

Fig. 1.65 Equivalent parallel circuit using ittances

1.25.3 Components of ittance Consider an impedance given as, Z = R±jX

Positive sign for inductive and negative for capacitive circuit. ittance

Y

= ~ =R;jX Copyrgh!cd malcri~

Circuit Theory

1 - 60


Rationalising the above expression, Y=

R'l'jX R'l'jX = (R ± j X)(R + j X) R2 + X2

= (R2:x2)+

i (R2:x2)=~

'!''X

I

z2

Y=G+j8

In the above expression,

G = Conductance = -

and

8 = Susceptance =

R

z2

~

z2

1.25.4 Conductance (G) It is defined as the ratio of the resistance to the square of the impedance. It is measured in the unit siemens. 1.25.5 Susceptance (B) II is defined as the ratio of the reactance to the square of the impedance. It is measured in the unit siemens.

Note :

The sign convention for the reactance and the susceptance are opposite to each

other. , Y = G + j 8 =

IY I

=

IY I

L 9 siemens or mho

,Jc 2 + B2 • o= tan-1 ~

Key Point: Impedances in parallel get 1xmoerled to illa1JCts i11 series while impedances

in series gel converted to illar1ces in parallel.

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1 - 61

01/cu/ate tire equivalent impedance of the 11e1tvork as vitWed tl1rnugh the temtiuats A-B showu iu the Fig. !.66. If atr altenwting voltage of 150 L {)"Vis connected across A-B, calculate tire curreut drawn from the source. Hmce carculate the pawn

,,.. Example 1.15 :

consumed. 4 + j31l

5L. Jo•n 6-jSO B

A

Fig. 1.66

Z2

6-

z3 = 5 L

Now

n = 5 L. 36.86• !l j 8 n = 10 L- 53.13• n 3o• n " 4.33 + i 2.5 n

4 + 3i

Solution : Let

z, and Z2 arc in pa.rallel, . .. : . (Z, II Zz) =

z--1Z1+Zzz--2

Z 1 Z2 must be in polar form while (2 1 :. (Z,

II Zz) "

"

Z2 ) in redangular form

5L.36.86°x lOL.-53. !3° SOL.-16.27" ( = --,.,.., ,........,.,....4+i3+6- i 8) (I 0 -j5) SOL.-16.27° 6 56• = 4.4721 L - 16.27°+26.56°

= il.lSL.-Z .

= 4.4721 L ... 10.29• " 4.4 ... ;

o.s n

4.4 + i 0.8 + 4.33 + ; 2.5 = 8.73 + i 3.3 n

= 9.332 L. + 20.706• n

This is the equivalent impedance.

150L o•

Hence the circuit becomes as shown in the Fig. 1.67 (a). V 150 L0° I = ZAO = 9 .332 L.20 .706• = 16.07 L.- 20.706• A

Figure 1.67 (a)

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Thus the current is 16.07 A, lagging voltage by angle 20.706°. The phasor diagram is shown in the Fig. 1.67 (b). The power consumed can be calculated as, P = VI cos

I

9 = 150 x 16.07 x cos( -

20706• )

= 2254.798 w

Fig. 1.67 (b)

Alternatively the power can be obtained as, p = 12 X (R.~a) Now

ZAB =

.. RAo :. p

8.73 + i 3.3 n

= 8.730

=

(16.07) 2 X 873 = 2254.4

w

This is because inductive part j 3.3 of an equivalent impedance does not consume any power.

Two impeda11ct Z1 = 5 - j 13.1 Q and Z 1 = 8.57 + j 6.42 n are connected in parallel across a voltage of (100 + j200) t'()/ts. Estimate:i) Branch cu"mts in complex form ii) Total power corrsumed, Draw a neat phasor ditlgram showing voltage, lmmcl1 cummts ~ud all phase angles.

II~ Example 1.16 :

Solutlon : The circuit is shown in the Fig. 1.68.

V = 100 + j 200 = 223.607 L 63.43° V

••

z,

z, =s- i 13.1 =14.021 ~

i)

L - 69 .t09• n

= s.s7 + i 6.42 = 10.71 L + 36.83• n

11 =

V

l;"

223. 607 L63..43• = 14.021 L69.109°

= 15.948 L 132.539° A Fig . 1.68 I _ 2-

= - 10.782 + I 11.75 A

V _ 223.607 L 63.43• 20.878 L 26. 6• A = 18.668 + j 9.3483 A - 10.71L+36.83• =

z2

ly = f 1 + f2 = - 10.782 + j 11.75 + 18.668 + j 9.3483

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1 • 63

Basic Circuit Analysis & Networit Reduction Techniques

= 7.886 + j 21.0983 A = 22.5239 L 69. s• A Q T = Angle between V and IT

= 69.5 - 63.43 = 6.075° leading Fig. 1.69

PT

= V IT cos h = 223.607 X 22.5239 X

COS (6.075)

= 5008.211 W

The phasor diagram is shown in the Fig. 1.69.

1.26 Star and Delta Conneetlon of Resistances In the complicated networks involving large number of resistances,, Kirchhoffs laws give us complex set of simultaneous equations. It is time consuming to solve such set of simultaneous equations involving large number of unknowns. In s uch a case application of Star-Delta or Delta-Star transformation, considerably reduces the complexity of the network and brings the network into a very simple form. This reduces the number of unknowns and hence network can be analysed very quickly for the required result. These transformations allow us to replace three star connected resistances of the network, by equivalent delta connected resistances, without affecting currents in other branches and vice-versa.

Let us see what is star connection ?

U the three resistances are connected in such a manner that one end of each is connected together to form a junction point called St.u point, the resistances are said to be connected in Star. The Fig. 1.70 (a) and (b) show s tar connec_!!!d-resistances. The star point is indicated a.~ S. Both the connections Fig. 1.70 (a) and (b) are exactly identical. The Flig. 1.70 (b) can be

redrawn as Fig. 1.70 (a) or vice-versa, in the circuit from simplification point of view.

s (a)

(b)

s (c)

Fig. 1.70 Star connection of thrwe resistances

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Clrwlt Theo ry

1 • 64 Basic Circuit Analyals & Network Reductio n Techniques ~-----------------------------------------------

Let ua see what Ia

d"a" connection

?

If the three resistances are connected in such a manner that one end of the first is connected to first end of second, the second end of second to fi~t end of third and so on to complete a loop then the resistances are said to be connected in Delta.

Key Point: Delta connection alW
00

00

00

Fig. 1.71 Delta connection of three resistances

1.26.1 Delta..Star Transformation Consider the three resistances Rt2 . R23 . R31 connect
1

terminals between w!tich these are connected in Delta arc named as 1. 2 and 3.

R,

s

Now it is always possible to replace these Delta COMectcd resistances by three equivalent Given Delta Fig. 1.72 connected resistances Equivalent Star Star Rt. Rz, R3 between the same terminals 1, 2. and 3. Such a Star is shown inside the Delta in the Fig. 1.72 which is called ~qulval~nt Star of Delta connected resistances. Key Point : Now to call tkse two arrangtma•ts as equivalent, the resistance belwttn any twa tenniuals must be S
2

Let us analyse Delta connection first, shown in the Fig. 1.72 (a).

Copyrighted materia

Circuit Theory

1 • 65

Basic Circuit Analysis & Networlo: Reduction Techniques ParaHel

.-------o1

~

1

2

.__ _R23 ..;__ _ _ _ _ 3

2 o---.......L----'

(a) Given Delta

Fig. 1.72

(b) Equivalent between 1 and 2

Now consjder the terminals (1) and (2). Let us find equivalent resistance between (1) and (2). We can red.raw the network as viewed from the terminals (1) and (2), without considering terminal (3). This is shown in the Fig. 1.72(b). Now terminal '3' we are not considering, so between terminals (1) and (2) we get the combination as, R12 parallel with (R31 +R23) a.~ R31 and R23 are in series.

:. Between (1) and (2) the resistance is, =

R 1z (R31 +Rz.l) R12 +(R31 +R23)

[using : •

~2

I+

.. . (a)

for parallel combination)

Now consider the same two terminals of equivalent Star connection shown in the Fig. 1.73.

R, R,

S·

s 2 , __ __ ___,

3

Fig. 1.73 Star connection

2

Flg. 1.74 Equivalent between 1 and 2

Now as viewed from tennina.Is (1) a.n d (2) we can see that terminal (3) is not getting connected anywhere and hence is not p laying any role in deciding the resi~tance as viewed from terminals (1) and (2). And hence we can red.r aw the network as viewed through the tcrm.ina.Is (I) and (2) as shown in the Fig. 1.74. :. Between (!)and (2) the resistance is= Rt + R2

... (b)

Copynghtcd materia

Cln:ult Theory

1 • 68

Basic: Cln:ult Analysis & Networ1c Reduction Tec:hnlqUM

Rz

R2., = Rtz R12 + R23 +R31

... (h)

R3

R31 = R12 R23 +R23 +R31

...(i)

Now multiply (g) and (h), (h) and (i), {I) and (g) to get foUowlng three equations. Rtz

2

R31 R23

(Rtz +R23 +RJt)

···0>

2

2 R23 Rt2 R31 (Rtz +R23 +RJt)

...(k)

2

2 R 31 R tz R23 (Rtz +R23 +RJt)

••.(!)

2

Now add O> ,{k) and (I)

=

R12 2 R3tR23 +R23 2 RtzR31 +RJt 2 Rtz R23 2 (Rtz +R23 +RJt)

=

RtzRJtR23(Ru +R2.1 +RJt) 2 (R12 +R23 +R3J) R u R 31 R23 Rtz + R23 +R31

But

Prom equation (g)

Substituting in above in RH.S. we get, RtRz +RzR3 +R3R1

I

=

Rt R23

R23 = Rz +R 3 +

R~~ 3

Similarly substituting in RH.S., remaining values, we can write relations for remaining two resistances.

and

R12

= Rt +Rz + R~:l

R31

= RJ+Rt +Rk~3

Copyrghtcd matcri~

Circuit Theory

1 • 69

Basic Circuit Analysis & Networtc Reduction Techniques

Equivalent dena of given star

3

Fig.

1.n

Star and equivalent Delta

Easy way of ing the result :

Tltt tquivalf.llt delta !;{JIIIti!Cttd ~TI'Sislana to be Ct>mm:teil belwi'en any two ttmti11als is >11m of the fu,'Q rcsisf4ttci'S comlt'Cied belttle!'n the same tuJO ttnnlna/s. an4 star point respectively iu stqr, plus lire product of tlte samt two star resistances divided bY tire third star rt'liMmru. So if we want equivalent delta resistance between terminals (3) and (1), then take sum of the two resistances coru\ected between same two terminals (3) and (1) and star point respectively i.e. terminal (3) to star point R 3 and terminal (1) to .star point i.e. R1 • Then to this sum of Rt and R3, add the term which is the product of the same two resistances i.e. R 1 and R 3 divided by the third star resistance which is R2 . .

RtRJ

.. We can wr1te, R31 = Rt + R 3 +~ which is same as derived .above. Result for equal resistances In star and delta : U all resistances in a Delta connection have same magnitude say R, then its equivalent Star will contain,

i.e. equivalent Star contains three equal resistances, each of magnitude one third the magnitude of the resistances connected In Delta. U all three resistances in a Star connection are of same magnitude say R, then its

equivalent Delta contains all resistances of same magnitude of , Rx R R12 = R31 = R23 = R+R +-y = 3 R

i.e. equivalent de.lta contains three resistances each of magnitude thrice the magnitude of resistances connected in Star.

Circuit Theory

1 • 70


Delta.Star

Star-Delta

Equ.va!ent delta

EQUiviMent star

Table 1.6 Star-Delta and Delta-Star Transformations ,. . Example 1.17 :

Convert the given Dtlta in the Fig. 1.78 into equivalent Star.

3

150

Fig. 1.78 Solution : Its equivalent star is as shown in the Fig. 1.79. Where R1 =

10x5 1.67 0 5 + 10+15 =

R2 =

15x 10 5+10+15

= 50

=

5x15 5+10+15

= 2.5 0

R3

R,

s 3

2

Fig . 1.79

Copyrgh!cd malcri~

Circuit Theory

1 - 71


,,... Example 1.18 : Cm1vertthe given star into an equivala1t delta. 1

3

2

Fig. 1.80 Solution : For star to delta conversion,

(4+ .3)(3 -j4)

= 4 + j 3 + 3 - j4 + ~..,'~:,.,...:....:.. (5+j0)

. 5L36.86 °X5L - 53.13 o

= 7 - J+

SLO O

= 7- j + 4.8 -; 1.4 Similarly

~

= 7 -j+SL- 16.27 °

= 11.s - j 2.4 n

and 2 31 can be obtained. The equivalent delta is shown in the Fig. 1.81.

Fig. 1.81 Key Point : Use rectangular system for addition or subtraction while polar for

multiplication or division.

Copyrghtcd malcri~

Circuit Theory

1 • 72

Beale Cln:ult Analyala & Networtt Reduction Technlquea

Examples with Solutions ,_. Example 1.19 : Find

resi.tllna between points A-8.

~iuaktu

60

40

Fig. 1.e2 Solution : Redraw the circuit, A o---------r------,

i'~i'='-

150

150 <

100 •

c

0 40 ~

60

""' 60 '

<

,

."~'

,, ..

4Q

~..

B

B

Fig. 1.83 (a) A

A 15 x10 15 •10

=60

"'

"' 6x4 6+4

B

a

150

=

RA·B s 6+2.4 8.40

2.40

B

Fig. 1.83 (b)

..

·. RAB = 8.40

Copyrgh!cd malcri~

Circuit Theory

1 • 73

Basic Cin:ult Analysis & Network Reduction Techniques

n. . Example 1.20 : 01/culate the tfftdivt resisllmce betwem points A and 8 in the gi~~m circuit in Fig. 1.84. 40

30

2 A

Sll

20

60

Sll

20

20

50 B

3Cl

Fig. 1.84 Solution : The resistances 2, 2 and 3 are in series while the resistances ·4, 2, and 5 are in

series. 2+2+3 = 70 and

4+2+5 = 110

The circuit becomes as shown In Fig. 1.85 (a).

6

5 ............~................~--oB

~

Fig. 1.85 (a) Converting 6 PQR to equivalent star~

6x 3 RPN = 6+3+6 = 1.20 6x6 RPN = 6 + 3 +6 = 2.40 6x3

~N ~ 6 +3+ 6 = 1.2 o

Hence the circuit becomes as shown in the Fig. 1.85 (b).

R 5

~------~--------~~8

Fig. 1.85 (b)

Copyngh!cd malcria

1 • 74

Circuit Theory

Basic Circuit Analysis & Network Reduction Techniques 3.2

The resistances 2 and 1.2 are in series.

A~.---~~---r--------~

1.2 and 11 are in series.

5 and 2.4 are in series.

7

:. Circuit becomes after simplification as shown in the Fig. 1.85 (c).

7.4 L--------L--------~~ 8

The resistances 7.4 and 12.2 are in parallel. 7.4x 122 .. 7.4 11 12.2 4.6061 n . + 7 4 122

=

Fig. 1.85 (c)

=

So circuit becomes,

3.2 A

A

7

Series

"U.

-

4.6061

B

(d)

B

(e)

Fig. 1.85

Now the two resistances arc in parallel. 7 x 7.8061 RAB = 7 +7.8061 = 3.69

Q

,._ Example 1.21 : Find the currmt in the branch A · B in the d.<:. circuit shown in the Fig. 1.86, using Kirchlroffs laws. 16A A

Hl

SA

4A

Hl B 7A

Fig. 1.86

Copynghl<'d rna .ria

Circuit Theory

1 • 75

Basic Circuit Analysis & Networl\ Reduction Techniques

Solution : The various branch currents are shown in the Fig. 1.86 (a). Applying KVL to loop ADBA App¥ng KCL al various nodes

16A A

... (I) Applying KVL to the loop ACBA,

,

- (16 - 11 - l:z) - (12 - It -l:z) + It = 0

16 - ·,-~

0

c

SA

:.- 16 + 11 +

4

:.311 + 212

lz- 12 + 11 +

= 28

12 + 11 = 0 ... (2)

Add (1) and (2), B

411 = 23

7A

11 = 5.75 A

Fig. 1.86 (a)

... This is the current through branch

AB. ,_,. Example 1.22 : Determine the equiwlent resistance the circuit si10W11 in Fig. 1.87.

0

A

~tween

G

1 Cl

C

50

D

80

B

A

40

E

70

F

100

F

the terminals A and B for

B

"CEEi". Fig. 1.87

Solution : In Fig. 1.87, the resistances 2 0 and 1 0 are in series and resistances 10 0 and 9 0 are in series.

Fig. 1.87 (a)

Copyngh!cd materia

Circuit Theoty

Bulc: Circuit Allalysl8 & Networtl Reduction TIChniqua

1 • 76

Converting delta AEC to star

c

c

~ 30

s

<

• 40

E

Fig. 1.87 (b) 3

X

3

"' 3 + 4 + 3 "' 0·9 (} 4x3

"' 3 + 4 + 3 = 1•2 (}

3X4

Res "' 3 + 4 + 3 = 1.2

..

a

Converting Delta DB.F to star 80

0

0

B

::::::

B

s RFs

F

F

Fig. 1.87 (c) 6x8 8 + 19

..

Ros = 6 +

..

19 = 4.6()6 (} Rss = 6 +8 X8 + 19 . . .

.. ..

RPS =

= 1.4S4S (}

6 X 19 6 + 8 + 19"' 3·4545 (}

RAB "' 1.2 + (7.3545

c

50

11 11.6545)

+ 4.606 B

0

B

3.45450

A

A

E

70

f

Fig. 1.87 (d) Copyrgh!cd malcri~

13asic Circuit Analysla & Network Reduction Techniques

1 -77

Circuit Theory

= 12. + 7.3545 X I 1.6545 606 =10.31S O 7.3545 + I 1.6545 + 4.

4.6060 11.65450

Fig. 1.87 (e)

1-+

For the circuit shown in the Fig. 1.88, ruritt KCL al!d KVL equalion and so!N to find cumnts I1 and 12. Example 1.23 :

2Cl

Fig. 1.88

as s hown in the

Solution : Consider the various branch currents and node voltages Fig. 1..88 (a).

20

Applying KCL at node A,

A V

0

tI,

'•

12

5 1, 13

t

41,

5 11 - f3 + 12 = 0

tl2

10

13 =

Now

v; and

20

20 -Va 1

20V

Note the direction of 12 which is coming towards the node hence the 20 V source is at

B

Fig. 1.88 (a)

v.

higher potential than V• forcing 12 towards node A from the base.

20-V. = 0 1 511 = 1.5 11 = 0.3 Applying KVL to the loop ABCOA, - 2 13 + 12 - 211 = 0 511 - - + 2

v. -20 v. - 4

. .. (1)

Copyrghlcd malcri~

Circuit Theory


1 • 78

-2xv•+12-2I 1 2

..

V• + 2 !1

~

0

~

12

... (2)

Pu tting 11 from (I) in to (2),

v. + 0.6 v. - 8 12 1.6 v, = 20 v. = 12.5 v ~

.. .. ..

II

and ••

=-

12 =

0.25 A i.e. 0.25 A

20-V, = 7.5 A 1

.J.

r

Using successive source tran.<Jommtion, find the voltage across 1 Cl resistor behveen 'a" and 'b" shown in the Fig. 1.89.

Example 1.24 :

10

20

Hl L---~---------------4b

Fig . 1.89 Solution : Converting all the voltage sources to the current sources, 1fl

10

2Cl

211

10

L----L----~----L------------4 b

Fig . 1.89 (a)

2111 •0.6670

211 1 = 0.667 0

Hl

L---~--------------~ b

Fig . 1.89 (b) Copynghtcd materia

Circuit Theory

1 - 79

Basic Circuit Analysis & Network Reduction Techniques 4V

0.667Q

0.66Hl

10

~------------------~b Fig. 1.89 (c) 1.3340

,----.------<jla

'•b 1.3340

Hl L----Ob

Hl

L---.l.------Ob

Fig. 1.89 (d) & (e)

l,.b

2xt.334

v•b ,. . Example 1.25:

... Current division rule.

= (1. 334+1) = 1.143 A

= I,b X 1 n = 1.143

v

with 'a' positive.

Using source transformation, find V 1 and V2 s1wum in the circuit.

2Q

2A

t

t

2(l

1A

Fig. 1.90

Solution :

Cunv~rting

all voltage sources to current sources, )Q

v, 2A

t

2

t

0.5A

2

t

1A

2

1A

Fig. 1.90 (a) Copyrghlcd malcri~

Circuit Theory

1 • 80

2.5A

Basic Circuit Analysis & Netwolil Reduction Technlqun

2112 =1 0

t

t

20

2A

Fig. 1.90 (b) 10

Converting sources,

current

sources

to

voltage

+

10

+

20

Applying KYL, - 21 - 1 - 1 - 2.5 + 4~0

+ 4V

00

-

4I = - 1.5

I = 0.375 A Drop across 2 n = 2 x 0.375 = 0.75 y

Fig. 1.90 (c) and Drop across Hl = 1 x 0.375 = 0.375 y

v,

= 0.375 + 2.5 = 2.875 V +ve with respect to base.

and

v2

= -0.75 + 4 = 3.25 V +ve with respect to bue.

,.

Example 1.26 : For tire circuit shown in tM Fig. 1.91, use source tran!;formations to detmnine currmt I. Hl

20 10 40 10

Fig. 1.91 Solution : Redrawing the given network we get,

Copyrgh!cd malcri~

Circuit Theol)'

1 • 81

Basic Circuit Analysis & Network Reduction Techniques 21l

Hl

40

20

40

Fig. 1.91 (a)

10

20

Hl

j

10--:1

1A

100

1114 =0.80

(b)

(c)

10 1.80

0.80 100

2V +

-+

100

1 •0.8 =0.8V

(d)

(e)

I

2V •

0.811.8 =0.444A

2V +

1.52540

j 0.444A

(g)

(f) 1.52540 2V

• -

+

0.444. 1.5254 =0.678 v

(h) Fig. 1.91

Copyrigh!ed malcria

Circuit Theory

1 • 82


Applying KVL, -1.5254 X I - 0.678 ... 2 = 0

2-0.678

l.S = 0.8667 A 254

,,_. Example 1,27 : A Wheatstone bridge consists of the folluwing resistanas. Branclles AB = 3 0, BC = 6 Q, CD = 12 Q and DA = 10 n.

A 2 V cell is connected btlwten points 8 and D and ga/VQI!omder of resistance 20 Q is connected betwa:n A and C. Find the cu"ent through glllvanomder by Kircluaoffs laws. So lution : Step 1 : Circuit diagram from given inJormation. Step 2 and 3 : Mark all the branch currents applying KCL at various codes and then mark the polarities of various voltages due to these assumed branch cu.r rcnts. Step 4 : Apply KVL to various loops. 0

Loop 1 : Loop A-D-B-1\ (through cell) - 10 (!1 - 12 + IJ) + 2 - 3 (11 - lz) = 0 .. - 13 II - 13 l2 - 10 l3 = - 2

... (I)

Loop 2 : Loop A·D·C·A

.-

- 10 (!, - 12 + 13) + 12 (12 - !3) - 20 13= 0 - 10 I I- 22 '2 - 42 !3 = 0

c

2V

.. . (2)

Fig. 1.gz

A

+

-

2V

Fig. 1.92 (a) Copyrgh!cd malcri~

Circuit Theory

1 • 83


Loop 3 : Loop A-8-C-A

+ 3 (11 - 12 ) - 6 ~ - 20 I3 ~ 0 ... (3)

We want current through galvanometer i.e. ~- Applying Cramer's rul!c 0 =

-13 - 10

13 - 10 22 -42

3 -9

..

-20

0 = 6156,

13 -2 - 10 22 0 3 -9 0

-13 03

;

0 3 = -48

13

~

03

D

- 48

~ 6156 = - 0.00779 A

..

Cu rrent through galvanometer is 7.79 mA but negative sign indicates its d i.reclion is from A to C.

n•• Example 1.28 : Two batttries A and 8 luzving e.m.fs of 209 V and 211 V having intenull resistance 0.3 n a11d 0.8 n respectively are to be cluzrged from a d.c. source of 225 V. If for th.lt purpose they wert connected in parallel and resistance of 4 n was connected betwem the supply and batteries to limit charging curmrt, find i) MP.grritude and direction of current through each battery. ii) Power delivered lly source. Solution : The circuit diagram is as shown in Fig. 1.93.

I,

A

We can use branch current method. Show the branch currents and polarities.

40

•

Apply KVL to different loops.

11 - 12 +

o.sn

209V

'•

... (1)

Loop 2 : Loop A·B-C·O·E-F-A, i.e. - 0.8 (I1 - Ill - 211 + 225 - 4 11 • - 4.8 11 - 0.8

12 +

Source

- 0.3 12 - 209 + 225 - 4 II = 0

i.e.

c

0.30

Loop 1 : Loop A·B-E·F·A

i.e. 4 I 1 + 0.3 I2 • 16

t, -12

B

Iz + 14

E

211

v

D

Fig. 1.93

0

• 0

... (2)

Copyrghtcd matcri~

Circuit Theory

1-U

i.e.

Basic Circuit Analysis & Networtc Reduction Tec:hnlques · 4.8 It - 0.8 12 = 14

Solving equations (1) and (2) simultaneously, It = 3.663 A 12 = 4 .482 A

It - 12 = - 0.8183 A i.e. it is in opposite direction to what is assured. i) Magnitude of current through source

= 3.663 A l

Magnitude of current through battery A = U82 A

J.

Magnitude of current through battery B = o.8183 A l ii) Power delivered by,

Source = 225 x 3.663 = 82.4.175 watts ,_. Example 1.29 :

Find tire tquilllllent resislllnct betwten 8 Rrrd C.

(.ApriVMay - 2004)

80

Fig. 1.94 Solution : Converting internal star to delta, RAB

= 3.84 + 2.4 +

Roc = 3.84 + 1.6 + RCA = 2.4 + 1.6 +

3

·~~ 2.4 = 12

3

·~~ 1.6 = s o

2.~:'s!· 6 = 5

0

Q

The circuit reduces as shown in the Fig. 1.94.

Copyngh!cd malcria

Circuit Theory

1 • 85

Basic Circuit Analysis & Network Reduction Techniques A

RAB 1211

80 B '----'WVIr----> C 8118

• 40 Shaded : Parallel combinations

(a)

{b)

Fig. 1.94 Rearranging to calculate Roc, :. Roc= (4) 11 (6 + 2.5) 4x85 = 4+8.5 = 2.72 n

Fig. 1.95

1m+

Example 1.30 :

Determine the value of R irr tire circuit tuloc• the Cllrretlt is uro in the

branch CD.

(Aprii/May-2004)

R

..

A

sn ~

? 20!l ~

~

v -:.;:-

c

A

10!:! ~

\ A.

0

::-$ R

B

Fig. 1.96 Solution : Use Kirchhoff's laws. The various branch currents are shown in the Fig. 1.96(a).

Copynghted rna .ria

Circuit Theory R

I, E


1 • 86 _A

A

Current through branch CD is uro.

"

+ •

·· -

12 +

5C!_ : v ~=-

Apply KVL to EACBFE,

+

;

<

.

-

1=0

c 12

•

~

~ _<

~

F

Apply KVL to EADBFE,

+

10Cl :

- ltR-2013 -R13 + V = 0

R

:.l tR+(20+R) !3 = V

B

i.e.lt - 12 -! 3

=

... (2)

... KCL at node A.

Fig. 1.96 (a)

D

... (1)

:. Rit+l512 =V

•••

R 15 R 0 1 - I

0 20+R - 1

=0

... (3)

= 15 (20 + R) + 15 R + R (20 -+ R)

= R2 + 60R +300 R 02

03

v v

0 20+R -1

=

R 1 0

=

V (20 + R)

=

=- RV + V (20 + R) + RV

R 15

v

R 0 1 -1

v =15 V- RV + RV 0

= 15 v V (20+ R)

02 o= R2 +50R+300

..

!2

=

and

13

= PJ D

=

lSV R 2 +50R+300

To have current through CD uro, the points C and D are equipotential i.e. drop across 5 n d uc to I 2 must be same as drop across 20 n due to I 3 . .. 5 12 = 2013

.. .. ..

5x

V (20+ R) 20x15 V = 2 R2 +50R+300 R +50R+ 300 100 + 5 R = 300

R = 40 0

Copyrgh!cd malcri~

Circuit Theory

1 ••

1 • 87

Example 1.31 :


Using squra lransfomtlltion, find tilL power deliutnd by !he 50 V

voltage SOilrctl.

(ApriVMay 2004)

511

30

20 50V

10V

Fig. 1.97

Solution : Converting 10 V voltage source to c:urent source and drawing a network, 10

A

I'= J = 3.333 A

50

Adding two current sources, we get single current sow-ce of value (10 + 3.333) A i.e. 13.333 A.

3.333A 10A

20 r

.

30

Find equivalent resisbl.nce for parallel connection of 2 n and 3 n i.e.

Fig. 1.97 (a)

R' = 2 X 3 = -6 = 1.2 n and redrawing a network as shown in the Fig. 1.97 (b), (2 +3) 5 A

50

I"

13.333A

1.20

Flg. 1.97 (b) Now, converting current source back to voltage source,

V' = f.

R = (13.333) (1.2) = 15.996 volts A

50

•

•

1.20

50V

Fig. 1.97 (c) Copynghlcd materia

Circuit Theory

1 - 88

Basic Circuit Analysis & Network Reduction Technlqun

Applying KVL, - 1.2 I - 15.996 + 50 - 5 I

;

0 0

..

0

I (1.2 + 5) = (50 - 15.996)·

..

I =

(50 - 15.996) (5+1. 2)

= 5.484 A

: . Power supplied by 50 V source,

p = (50) (5.484) = 274.2 watts CalculJlt~ tlu CWTtnt through 6 n usisiJlna appliallion of KircJrlwffs laws.

,_ . Example 1.32 :

~

20

givm nttumk by

[April/May-2004)

40

Hl

10V ., :;.

of the

0.

40

-:::- 20V

60

Fig. 1.98

Solution : Mark all the branch currents starting from positive of 10 V source. Mark all the polarities for different voltages across the resistances.

...

20

A

I,

B u,-12) •

+ ••

12

1

10V

+

~=H

10

40

c
"

+

13 +

40 : ~· ~

0:

60 ;

G

-

-

D

-:,; 20V -;; _

F

Fig. 1.98 (a) Now applying I
Loop 1 : A·B-G-H·A Loop 2 : 8-C-F..C.B

- 2 ·~- 4 12 + 10 = 0

... (1)

- 1 ( II - 12) - 6 ll + 4 12 = 0

. .. (2)

Copyrgh!cd malcri~

Circuit Theory

1 • 89


... (3)

Loop 3 : C-0-E-F-G Rewriting above equations by taking constants on one side, 2 11+ 4 12 = 10 - 11+ 5 12 - 6 13

=0

- 4 11 + 4 12 + 10 13 = 20 D

Applying Cramer's Rule,

=

2 4 0 - 1 5 -6 =284 -4 4 10

Now we are interested in finding current through 6 n i.e. 13.

03 =

13 =

2 4 10 - 1 5 0 =440 -t 4 20 03

0 =

440 284 = 1.5492 A

As the answer is positive, assumed direction of 13 is correct. So current through 6 resistance is 1.5492 A flowing from C to F.

n

,_. Example 1.33 : Calcuh!te the resista11ce Ru when all the resistana values art equal to <April/May-2005) 1 n for the circuit in Fig. 1.99.

b

d

Fig. 1.99 Solution : Converting deltas aeb and ced to equivalent star,

1

Fig. 1.99(a) Copyrgh!cd malcri~

Circuit Theory

R

1 -90

R "~ 2 = •"3

I=

~


1x1 1 1 +1 +1 - 3

n ,

The circuit reduces as shown in the Fig. 1.99 (b) Series

77

a

a

Req

e

H

•

113

113

113

~

(b)

Series

c

1/3

113

ROQ

1 +(1/3)= 4130

!•

•

1/3 + 113

2130

•

1/3 ~

b , . 113 = 4130

d

(c)

Fig. 1.99

Converting sbds delta into star, we get

• •

213

,.

• s,d

1/3 ;.

b

4/3

d

b

Fig. 1.99 (d)

Copyrgh!cd malcri~

Clrc:ult Theory

1 • 91

Basic Circ:ult Analysis & Network Reduction Techniques

The circuit reduces as shown in the Fig. 1.99 (c) a

113

-

Req

413

e

'j s' 0.38090

Patallel

~

s,d

0.19040 b (f)

Fig. 1.99

I..

~q = (!. 7!42J0.4285)+0.1904: 0.533211

Example 1.34 : An impedance consumes a PQU1" of 60 walls ami lllkes a cu"enl 10- j 8 A when conru:cted to a source of 'a + j 5' V. Find 'a' and cirCIUI tlemenls.

of

Solution : The given circuit is shown in the Fig. 1.100.

V

2 = T R + jx =

10-f8 A

a+j5 = t0-j8

t.

a +j5 10-j8

atjSV

!

Rationallsing right hand side,

Fig. 1.100

(a +j5)(10+j8) R + jx = (10-j8)(10 +j8)

=

R + jx =

(a +j5)(10+j8)

(100) -(j2 64)

lOa -40 164

= .:..(1-0a '-_ -_ 4_0;..,)+,..;i.:.;.<SO.c....+S.:..a~) 164

. 50 +8a

+1

164

... (1)

Equating real and imaginary parts,

and

R =

l Oa-40 164

... (2)

X=

50+8a 164

.. . (3)

The power consumed by impedance Z is,

Copyrigh!ed malclia

Circuit Theory

· 1 • 92 p

..

.. ..

Basic; Cln:ult Analysis & Network Reduction Tec:hnlqun

=
60

= 164 R = 0.3660

R

2

(J10 2 +8 2 )

60

xR

•.• (4)

Substituting in (2), 0.366

.. ..

=

lOa -40 164

= 40 + 0.366 X 164 a = 10 50+80 Substituting in (3), X = = o.7926 n 164

I..

10 a

•.. (5)

Hence the voltage applied is 10 +j 5 volts while the impedance is (0.366 + j0.7926) n.

Example 1.35 :

The network s!wrun in II~ Fig. 1.101 crmsists tJj two star connected

circuits in parallel. Obtain tht single tklla connected equivalent. 10

10

""""" j10

""""" 'j",'Q

5

~ap

•

I

Fig . 1.1 01

Solution : Converting both the stars to delta, .

.

Zt = JIO + JIO +

''

'

$>,, ,,<1: 5 ''

' ,,

,,

,

j!Ox jlO

5

.

= - 20 + ]20

~

= jlO + 5 +

jl~~ 5 = 10 + jlO

Z:l

= jlO + 5 +

jl~~ 5 = 10 + jlO

z; = 10 + 10 +

10

~10 = 20 -

;2o

... .,.1 =- J. I

Fig. 1.101 (a)

Copyrghtcd malcri~

Cln:ult Theory

Basic Circuit Analysis & Network Reduction Techniqun

1 • 93

z',

,

~---------c::t --------: :

10

''

'

10

~' '

:

,, ' ' ,,

,,

,

.

z 2 = 10 + 1s +

lOx jS . -w = to + 11o

, . lOx jS z 3 = 10 + 1s +

--ro- ~ 10 + 1.1o

Connecting both the deltas we get the circuit as shown in the Fig. 1.101 (c).

Fig. 1.101 (b)

z

z,

z• = (- 20+ J20) (20 - j20) 1

-

20+ j20 + 20- j20

= .. open circuit

z, Zi

l 11

Z:! II Z2 = (10 + j10) II (10 + j10)

Z:J

= S + jS ~ II Z'3 = S+jS

Fig. 1.101 (c) So effective single delta connected circuit is as shown in the Fig. 1.101 (d).

Fig 1.101 (d)

,,... Example 1.36 :

Obt4in lht thiiA connected equivalent of tht ndwork show11 in Fig.1.102. -J2n

---1:

i~!l-

-~

Fig. 1.102

Copyrgh!cd malcri~

Circuit Theory

1 • 94

Basic Clreult Analysis & Network Reduction Techniques

Solution ; Consider a star network formed with - j2 Fig. 1.102 (a).

n,

j5

n

and 5

n

as shown in

Ze

,·-- -t=J- --, ' -j2Cl

·• : I'

o '

z..

Parallel

jSQ

"""""

II

; 50

•

'

OZc .

-7.5+j5)Q

100

(a)

10Cl

(b)

Fig. 1.102 Converting star network into its equivalent d elta network as follows. zA = - j2 + 5 +

H~~ <5 rel="nofollow"> = - 12 +5-2 = (3- j2) n

z 8 = - j2 + j5 +

<-i 2~ (jS) =- 12 + J5- 2 = (2- j3) n

Zc

= j5 + 5 +
12.5

= (- 7.s

+ j5) n

Replacing sta.r network by its equivalent delta network as shown in the Fig.1.102(b). It is dear that Zc and 10 n in parallel. H ence we can write, Z(: = Zc ll l 0={- 7.5+j5)(10) = (- 7.5 +j5)(10) - 7.5+j5 + 10 2.5 + j5

= (9 . 0138 Ll46 . 3" )(10 L O") : 16 .1245 L S:!.ll7' (5 .5901 L63 .43" ) = (2+j16)Q Hence delta connected equivnlent of the ootwork is as shown in the Fig. 1.102 (c).

(2+]3)0

(3-]2)0

z,.

Zc

(2+j16)fl

Fig. 1.102 (c)

Copyrghlcd matcM~

Cln:ult Theory

1 • 98 Basic Cln:uit Analysis & Network Reduction Techniques ----------------------~------------------------

!1. Find tht equill
iii A and N

lAns. : u1.33 n

m0.11n 1

A

8

6U

Flg. 1.107 12. Detmnint ammt In bran
150

200

150

A

300

300

B

Fig. 1.108 13. In tht following circuit, dtlmnine I) It, h and 1), iiJ Valou of R and

iii) Volou of Ex

(Ant.: l1 • 0.3 A. h •1.5 A. E, • 174 V,l3 • 0.7 A, R •11.4!l) A

Fig. 1.109

Copyrghlcd malcri~

Circuit Theory

1 • 99


14. Rtdu« lilt mtwork,using souru trrmsfomulfiQn IU105S tM

A-B.

lnmi71Als

A

211 ~ 10V

+ 3A

211

2ll

~

8

Fig. 1.110 15. Deriw tM rel41iol15hip to express thm stor tonnttUd resistJin«> into "'"iOJiimt deltJI.

16. Deriw the reWiDmlrip to express three th/14 .,._ted r<>istiUices into equiOJiimt slllr. 17. Two vo/lmeltrs A and 8, having rtsisllmt:ts of 5.2 1<0 and 15 1<0 resp«tiody are "'"ntdtd in serils IUTOSS 240 V supply. Wh.ot is lilt reAding on =h voltmeter ? (Ana. : 61.78 V, 118.21 V) and 20 n art conntdtd in pamllel. A r<Sistana of 12 n is =ntdtd in serils wifh the combi7Ultiott . A voltagt of 120 V is 11pplitd IUTOSS the entirt tirtuit. Find lilt

18. Two resisllm«> 15

n

turrtnl in tatlt resistance. vollngt across 12 n T<Sisliln« and pow
two resistunces of 12 and 8 n. Tht total pow
19. A mistance R is ~ in suits with a parlllkl circuit comprising

20. In the series parallt l circuit shown in lilt Fig. 1.111.find the ii) lilt supply voltage V i) voltage drop tM 4 0 resistanct

=

(Ani.: 45 v, 140 V) A

l

v

1

8

.•~n

E

:

• 100

80:~

0 240~

•

120~ -~

I I

sov

' so<~

c Fig. 1.111 Copyrghlcd

nalcri~

Circuit Theory

1 -100 Bask; Circuit Analysis & Network Reduction Technlqun

21. Find tl!. cum:nt in

alii~!.

lml1u:hts

of II!. nd1vorl: shoum in II!. Fig. 1.112. tAns.: 39 A. n A. 39 A. 81 A. 11 A. 41 Al 80A

A

30A

80A

60A

70A

0

120A

Fig. 1.112 22.

If tht toW powu dissipaltd in II!. circuil sJrown in tht Fig.

1.113 is 18 Willis, foul tht zralue of R

and curmtl through it. 80

....

R

160

(ADs. : 12 a , 0.6 Al

40

,.

,.

+I , I'

12V

Fig. 1.113 23. The Comtnl in tJ1t 6

a rtsistmct qj IM nttwor/c sJrown in IM fig. 1.114 is 2 A. Ddnmint fht

curmds in all '"' other resulllnas and tht slq.ply t10114gt V. (Ana. : 1.5 A, l.5 A, 1 A. 46 V) 2A

HI

..••••'. 80

60

,..

AA AA.

.... 20h

y

8h' +

v Fig. 1.114

bllttay w1om w.uw by • resatance of so o gives tJre turni1W mu.ge of 48.6 v. If tht load resisl4n« is mcr-d to 100 a' tht ttrmhtal t10114ge is obsnwd to "" 49.2 v.

24. A parli&u!Rr

Copyrghtcd na

.ri~

Ci rcuit Theory

1 - 101 Basic Circuit Analysis & Network Reduction Techniques i) E.M.F. of ba!ttty

ii) lnttmal resistanct of batttty 25. A r..Utmlcr: of 10 0 is
26. Two storage ba!ltrits A and 8 an
27. A ntlwork ABCD is mode up as follaws :

A.B has • « t terminal connected to II; BC is a r..Utor of 25 n ; CD is n resistor of 100 n ; DA is n batltty of 4 V and negligible r..UtDnce with po
S.j80

8-tj60

L-------L-----------~8

Fig. 1.115 (Ana.: 757 + j 157(1)

29. Find lht single vollllgt smiTe< equi'oaltnl across the ltrminals A·B. Hence obtain the single current $DUree equivalmt also. r------r---{:=}---~----~A

4+j3ll 3+j20

'

2~0V

2+j51l

~-----L----------~----~8

Fig. 1.116 (Ans.: V •10.78 Lts.06•V, Z • 1.66 + j 2.U!I;, I • 4 L-36.86' A)

000

Copyrgh!cd malcri~

Circuit Theory By U A Bakshi A V Bakshi 2n5i4y

Overview 6h3y3j

More details h6z72

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