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S.1Electrical Circuits (November/December-2012, R09) JNTU-Hyderabad

B.Tech. II-Year I-Sem. ( JNTU-Hyderabad )

Code No.: A109210204

Jawaharlal Nehru Technological University Hyderabad

II B.Tech. I Semester ExaminationsNovember/December - 2012

ELECTRICAL CIRCUITS

( Common to EEE, ECE, ETM )

Time: 3 Hours Max. Marks: 75

Answer any FIVE Questions

All Questions carry equal marks

- - -

1. (a) Name three passive elements of electrical circuit and deduce the relationship between voltage and current for

each passive elements. (Unit-I, Topic No. 1.1)

(b) Distinguish between ideal and practical voltage sources and draw their V-I characteristics. (Unit-I, Topic No. 1.2)(c) Determine the power being absorbed by each of the circuit element shown in figure. [4+4+7]

(Unit-I, Topic No. 1.1)

4.5 mA+

13.7 V + 25 V

+

+ 62 V12 cos 1200t mA

12.5 mA 5 sin 1200t V,t = 4 ms

2ix,ix = 10 A

4.5 mA+

13.7 V + 25 V

+

+ 62 V12 cos 1200t mA

12.5 mA 5 sin 1200t V,t = 4 ms

2ix,ix = 10 A

Figure

2. (a) Using mesh analysis, find the magnitude of the current dependent source and current through the 2 resistor

as shown in figure. (Unit-II, Topic No. 2.3)2

1

3

1 2 A

5 ii

+

2

1

3

1 2 A

5 ii

2

1

3

1 2 A

5 ii

2

1

3

1 2 A

5 ii

+

Figure

(b) Find the power loss in the resistors of the network for the figure shown using nodal analysis, [8+7]

(Unit-II, Topic No. 2.3)

+

4 A 2 A

10 V 3

1

2 2

+

4 A 2 A

10 V 3

1

2 2

Figure

R09Solutions

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S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2013


3. (a) Define the following for alternating quantity,

(i) R.M.S value

(ii) Average value

(iii) Form factor

(iv) Peak factor. (Unit-III, Topic No. 3.1)

(b) For the circuit shown in figure, determine the total impedance, total current and phase angle. [8+7]

(Unit-III, Topic No. 3.3)

~110 V, 50 Hz

10

70

100 F

210 F~110 V, 50 Hz

10

70

100 F

210 F

Figure

4. (a) Define the bandwidth and derive the expressions for bandwidth of series resonating circuit and its relation with

Q-factor. (Unit-IV, Topic No. 4.2.1)

(b) Write the applications of locus diagrams. For the circuit shown in figure, plot the locus of current. [8+7]

(Unit-IV, Topic No. 4.1)

~ R

jXc

~ R

jXc

Figure

5. (a) State and explain Faradays laws of electromagnetic induction. (Unit-V, Topic No. 5.2)

(b) An iron ring of 8 cm dia. and 14 cm2 in cross section is wound with 250 turns of wire for a flux density 1.8 Wb/m2

and permeability 450. Find the exciting current, the inductance and stored energy. Find corresponding quanti-

ties when there is a 1.8 mm air gap. [5+10](Unit-V, Topic No. 5.1)

6. (a) Define and illustrate the following with an example,

(i) Branch

(ii) Tree

(iii) Cut-set

(iv) Tie-set. (Unit-VI, Topic No. 6.2)

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(b) Figure shown below represents a resistive circuit. Determine the number of branches, number of nodes and

number of links. Write down the incidence matrix for the given network. Also develop the network equilibrium

equations. [8+7](Unit-VI, Topic No. 6.3)

5 2

2 4

3

1

10 V

+

32

4

1

5 2

2 4

3

1

10 V

+

32

4

5 2

2 4

3

1

10 V

+

32

4

1

Figure

7. (a) Is Norton theorem dual of Thevenins theorem? Justify your answer. (Unit-VII, Topic No. 7.5)

(b) Find the current in the 10 resistor as shown in below figure using superposition theorem. [5+10](Unit-VII, Topic No. 7.2)

+

1

5

1

10

1 A

10 V

2 A

A B+

1

5

1

10

1 A

10 V

2 A

A B

Figure

8. (a) State and explain the compensation theorem. (Unit-VIII, Topic No. 8.8)

(b) Apply Thevenins theorem and obtain the current passing through 210 F capacitor of figure. [7+8]

(Unit-VIII, Topic No. 8.4)

110 V, 50 Hz

10

70

100 F

210 F

110 V, 50 Hz

10

70

100 F

210 F

Figure

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Q1. (a) Name three passive elements of electrical circuit and deduce the relationship between voltage

and current for each passive elements.

Answer : Nov./Dec.-12, (R09), Q1(a) M[4]

Passive Element

For answer refer Unit-I, Q2, Topic: Passive Elements.

Relationship between Voltage and Current of Passive Element

For answer refer Unit-I, Q14, Topic: Voltage-current Relationship of Passive Element.

(b) Distinguish between ideal and practical voltage sources and draw their V-I characteristics.

Answer : Nov./Dec.-12, (R09), Q1(b) M[4]

For answer refer Unit-I, Q4, Topic: Ideal Voltage Source, Practical Voltage Source.

(c) Determine the power being absorbed by each of the circuit element shown in figure.

4.5 mA+

13.7 V + 25 V

++

62 V12 cos 1200t mA

12.5 mA 5 sin 1200t V,t = 4 ms

2ix,ix = 10 A

4.5 mA+

13.7 V + 25 V

++

62 V12 cos 1200t mA

12.5 mA 5 sin 1200t V,t = 4 ms

2ix,ix = 10 A

Figure

Answer : Nov./Dec.-12, (R09), Q1(c) M[7]

The given circuit elements are shown in figure,

4.5 mA+

13.7 V + 25 V

++

62 V12 cos 1200t mA

12.5 mA 5 sin 1200t V,t = 4 ms

2ix:ix = 10 A

4.5 mA+

13.7 V + 25 V

++

62 V12 cos 1200t mA

12.5 mA 5 sin 1200t V,t = 4 ms

2ix:ix = 10 A

To determine,

The power absorbed by each element = ?

We know that,

The power absorbed by any element is given as,

P = VINow,

The power absorbed by element-1 is given as,

P = 13.7 4.5 103

= 0.06165 W


P = 25 ( 12.5 103)

= 25 1000

5.12

= 0.3125 W

SOLUTIONS TO NOV./DEC.-2012, R09, QP

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P= (12 cos 1200t) 103 (5 sin 1200t) 103

But, t= 4 ms

Substituting tvalue, we get,

P= [12 cos(1200 4 103) 103

[5 sin(1200 4 103)] 103

= (12 cos 4.8) 103 (5 sin 4.8) 103

= (12 0.9965) 103 (5 0.0837) 103

= (11.958 103) (0.4185 103)

= 5.004 106 Watts


P= 62 2 ix

But, ix= 10 A

Substituting ix

value, we get,

P = 62 2 10

= 620 2

= 1240 Watts

P = 1.2 kW

Q2. (a) Using mesh analysis, find the magnitudeof the current dependent source andcurrent through the 2 resistor asshown in figure,

2

1

3

1 2 A

i

+

2

1

3

1 2 A

i

+

Figure


Given circuit is shown in figure (1),

2

1

3

1 2 A

i

+

2

1

3

1 2 A

i

+

Figure (1)

To determine,

Using mesh analysis,

Current dependent source = ?

Current through 2 resistor = ?

By using mesh analysis, marking the mesh currents

as shown in figure (2),

2

1

3

1 2 A

i

+

I3

5 i

I1

I2

2

1

3

1 2 A

i

+

I3

5 i

I1

I2

Figure (2)

From mesh-1, we have,

I1= 2 A

Applying KVL to mesh-2, we get,

5i + 3(I2+I

1) + 1I

2= 0

5(I1+I

2) + 3(I

2+I

1) +I

2= 0 [Qi =I1 +I2]

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5I1+ 5I

2+ 3I

2+ 3I

1+I

2= 0

8I1+ 9I

2= 0

8(2) + 9I2 = 016 + 9I

2= 0

9I2= 16

I2=

9

16= 1.778 A ... (1)

Applying KVL to mesh-3, we get,

5i = 2I3+ 1(I

1+I

3)

5(I1 +I2) = 2I3 +I1 +I3 5I1 I1 + 5I2 = 3I3

4I1 + 5I2= 3I3 4(2) + 5(1.778) = 3I3 8 + 8.89 = 3I3

0.89 = 3I3

I3 = 389.0

I3= 0.297 A

I3 0.3 A

The value of current dependent source,

= 5 i

= 5(I1+I

2)

= 5(2 + 1.778)

= 1.11 Volts

And current through 2 resistor is given as,

I3= 0.3 A

(b) Find the power loss in the resistors ofthe network for the figure shown usingnodal analysis.

+

4 A 2 A

10 V 3

1

2 2

V1 V2 +

4 A 2 A

10 V 3

1

2 2

V1 V2

Figure


Given circuit is shown in figure,

+

4 A 2 A

10 V 3

1

2 2

V1 V2 +

4 A 2 A

10 V 3

1

2 2

V1 V2

Figure

Note: Node voltages V1

and V2

are assumed as shown in

figure,

To determine,

Using noda1 analysis, power loss in each resistor,

i.e., P2 = ?

P1 = ?

P3 = ?

P2 = ?Applying KCL at node-1, we get,

4 =13

10

2

21211 VVVVV ++

+

4 =3

101

3

1

1

1

3

1

2

121 +

+

++ VV

3

2

3

4

6

1121 VV = 0 ... (1)

Applying KVL at node-2, we get,

2 =23

10

1

21212 VVVVV +

+

2 =3

10

2

1

3

11

3

11 21

+++

VV

316

6

11

3

421 +

VV = 0 ... (2)

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On solving equations (1) and (2), we get,

V1= 5.26 Volts

V2= 6.74 Volts

Power loss across 2 resistor connected at node 1.

P2 = RI

22

But, I2= 2

1V

=2

26.5

= 2.63

P2 = (2.63)

2 2

= 13.83 Watts

Power loss across 1 resistor connected betweennode-1 and node-2.

P1= RI .

21

But, I1 = 1

21 VV

= 5.26 6.74= 1.48 A

Here negative sign indicates that the current is in

opposite direction.

P1= RI

21

= ( 1.48)2 1

= 2.19 Watts

Power loss across 2 resistor connected across node-2.

P2= (I2)2

R

But, I2 = 2

2V

=2

74.6

= 3.37 A

P2= (3.37)

2 2

= 22.71 Watts

Power loss in 3 resistor is given by,

P3 = (I3 )

2.R

But, I3 = 3

10 21 VV +

=3

74.61026.5 +

= 2.84

P3 = (2.84)

2 3

= 24.19 Watts

Q3. (a) Define the following for alternating

quantity,

(i) R.M.S value

(ii) Average value

(iii) Form factor

(iv) Peak factor.


For answer refer Unit-III, Q1.

(b) For the circuit shown in figure, determine

the total impedance, total current andphase angle.

~110 V, 50 Hz

10

70

100 F

210 F

~110 V, 50 Hz

10

70

100 F

210 F

Figure

Answer : Nov./Dec.-12, (R09), Q3(b) M[7]Given that,

Voltage, V= 110 V

Frequency,f= 50 Hz

Resistance,R1= 10

Capacitance, C= 100 F

Resistance,R2= 70

Capacitance, C'= 210 F

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To determine,

(i) Total impedance,Ztotal

= ?

(ii) Total current,Itotal = ?(iii) Phase angle, = ?The given circuit is shown in figure,

~110 V, 50 Hz

10

70

100 F

210 F

Z1

Z2

Z3

~110 V, 50 Hz

10

70

100 F

210 F

Z1

Z2

Z3

Figure

Now,

The reactance,

1C

X =fC21

= 610100502

1

= 410502

1

= 410100

1

= 210

1

=

100

= 31.831 And also,

The reactance,

2C

X =Cf 2

1

=6

10210502

1

=06597.0

1

= 15.1584

The impedance,

Z1

=R1j

1CX

= (10 j31.831)

The impedance,

Z2

=R2= 70

The impedance,

Z3= (j15.1584)

Now,

The total impedance is given as,

Ztotal

=Z1+ (Z

2||Z

3)

= (10 j31.831) + (70 || j15.158)

= (10 j31.831) +

158.1570

158.1570

j

j

= (10 j31.831) +1061.06

70 15.158

j

j

= (10 j31.831) + (3.135 j14.479)

= 13.135 j46.31

= 164.741367.48 ohmsWe know that,

The total current is given as,

Itotal

=totalZ

V

Itotal = 164.741367.480110

Itotal = 2.285 164.74 Amps

And also,

The phase angle is given as,

= 74.164o

Q4. (a) Define the bandwidth and derive theexpressions for bandwidth of seriesresonating circuit and its relation withQ-factor.


For answer refer Unit-IV, Q10(i), Q11.

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(b) Write the applications of locus diagrams.For the circuit shown in figure, plot thelocus of current.

~ R

jXc

~ R

jXc

Figure


The following are application of Locus diagram,

1. The locus diagrams are used to determine the

behaviour or response of the circuit networks.

2. These are also used to pre-determine the operating

characteristics of A.C circuits under various

conditions.

3. These are employed in determining the magnitude

and phase of a resistance, inductance and capacitance

parameters.

For remaining answer refer Unit- IV, Q3, Topic:

VariableXC.

Q5. (a) State and explain Faradays laws of

electromagnetic induction.


For answer refer Unit-V, Q11.

(b) An iron ring of 8 cm dia. and 14 cm2 incross-section is wound with 250 turns ofwire for a flux density 1.8 Wb/m2 andpermeability 450. Find the excitingcurrent, the inductance and stored

energy. Find corresponding quantitieswhen there is a 1.8 mm air gap.


Given that,

Diameter of ring,D = 8 cm

Cross-section of ring,A = 14 cm2

Number of turns,N= 250 turns

Flux density,B = 1.8 Wb/m2

Permeability = 450

To determine,

The exciting current = ?

Inductance,L = ?

Energy stored = ?

And also,

For an air gap of 2 mm,

Determine,

(i) Exciting current = ?

(ii) Inductance,L = ?

(iii) Energy stored = ?

Now,

The length of flux path in the ring is given as,

Length of flux path = D

= 8

= 3.142 8

= 25.136 cm

= 0.25136 m

The flux is given as,

Flux, = Flux density (B) Cross-section area

= 1.8 14 104

= 0.00252 Wb

= 25.2 104 Wb

And also,

The Ampere turns per meter of flux path length, is

given as,

H=r

B

0

H=450104

8.17

H= 3183.09

We know that,

Total Ampere turns required is given as,

Ampere turns required =H Length of flux path

= 3183.09 0.25136

= 800.10

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(i) Exciting Current

The exciting current is given as,

Exciting current,

I =Total ampere turns required

Number of turns on ring

=250

10.800

= 3.2004 Amps

(ii) Inductance

The inductance is given as,

Inductance,

L =I

N

=2004.3

102.252504

=2004.3

63.0

= 0.1968 H

(iii) Energy Stored

The energy stored is given as,

Energy stored=2

2

1LI

=2

)2004.3(1968.02

1

= 1.0079 J

For an Air Gap of 2 mm

Ampere turns per/meter of flux path length in iron

portion is given as,

Ampere turns/meter,

H= 3183.09

Length of flux path in iron portion = 25.136 0.18

= 24.956 cm

= 0.24956 m

~0.25 m

Ampere turns required by iron portion,

ATi= 3183.09 0.25

= 795.773

Ampere turns required by 2 mm air gap = 0.796Blg 106

= 0.796 1.8 1.8 103 106

= 2579.04

Total ampere turns required =ATi+H

= 795.773 + 3183.09

= 3978.863

(i) Exciting current,

I= 250

863.3978

= 15.915 Amps

(ii) Inductance,

L=I

N

=915.15

102.252504

=915.15

63.0

= 0.03958

~ 0.04

~ 40 103

~ 40 mH

(iii) Energy stored=2

2

1LI

=

23

)915.15(10402

1

= 5.066 J

Q6. (a) Define and illustrate the following withan example,

(i) Branch

(ii) Tree

(iii) Cut-set

(iv) Tie-set.

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(i) Branch

The elements of a tree connecting one node to the other are called as branches. These are denoted by b and are

represented by dark lines. The branches of a tree are also called as twigs. The relation between the number of nodes n and

number of branches b in a tree is given as,

b = n 1

Example

Let us consider the graph as shown figure (1),

12

3

4

12

3

4

Figure (1)

The various branches for figure (1) is represented in figure (2),

2

1 3

4

2

1 3

4

1 2 3

4

1 2 3

4

Figure (2)

(ii) Tree

From an oriented connected graph, a subgraph formed with all the nodes connected by it without forming a closed

path is called as tree. The following are the conditions to be satisfied in order to form a tree.

The tree must have (n 1) branches where n is the number of nodes.

The tree must have same number of nodes as that of a graph i.e., n nodes.

All the nodes must be connected.

The tree should not have any closed path.

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The tree for figure (1) is represented as shown in

figure (3).

12

3

4

12

3

4

Figure (3): Tree

(iii) Cut-set

It is defined as the set of branches which when

removed will divide the connected graph into two connectedsub-graphs.

Example

1 3

2

4

56 1 3

2

4

56

1 3

2

4

56 1 3

2

4

56

Figure

When the branches 2, 4 are removed two subgraphs

of elements 1, 6 and 3, 5 are formed 2, 4 are the cut-sets.

(iv) Tie-set

For answer refer Unit-VI, Q9.

(b) Figure shown below represents aresistive circuit. Determine the numberof branches, number of nodes andnumber of links. Write down theincidence matrix for the given network.Also develop the network equilibrium

equations.

5 2

2 4

3

1

10 V

+

32

4

1

5 2

2 4

3

1

10 V

+

32

4

5 2

2 4

3

1

10 V

+

32

4

1

Figure


The given resistive circuit is shown in figure (1),

5 2

2 4

3

1

10 V

+

5 2

2 4

3

1

10 V

+

Figure (1)

The graph for given resisitive circuit is shown in

figure (2),

1

1 2

3

4

4

2

5

6

31

1 2

3

4

4

2

5

6

3

Figure (2): Graph

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From figure (2), we have,

Number of nodes, n = 4

Number of branches, b = 6Number of links, l = b (n 1)

= 6 (4 1)

= 6 3

= 3

Incidence Matrix

The incidence matrix is given as,

Incidence matrix,

=

111000

100101

010110

001011

1 2 3 4 5 61

2

3

4

Nodes(n)

branches(b)

111000

100101

010110

001011

1 2 3 4 5 61

2

3

4

Nodes(n)

branches(b)

Network Equilibrium Equations

The fundamental circuit for the given network is

shown in figure (3),

1 2

3

1

4 5

3

L3

6

L1

L2

2

4

1 2

3

1

4 5

3

L3

6

L1

L2

2

4

Figure (3): Fundamental Circuits

LetI1,I

2,I

3be the loop currents,

Now,

The loop matrix is given as,

Loops\Branches

B =

101001

110100

011010

1 2 3 4 5 6L1L2

L3 101001

110100

011010

1 2 3 4 5 6L1L2

L3

From the above matrix, we have,

V2

V4

V5

= 0 ... (1)

V3 + V5 V6 = 0 ... (2)

V1

+ V4+ V

6= 0 ... (3)

And also,


R1

= 3 , R2= 1 , R

3= 2

R4= 4 , R

5= 2 , R

6= 5

Now,

V1= i

1R

1 v

1

V1= i

1(3) 10

V1= 3i

1 10

V2= i

2R

2

V2= i

2(1)

V2= i

2

V3= i

3.R

3

V3= i

3(2)

V3= 2i3V

4= i

4.R

4

V4= i

4(4)

V4= 4i

4

V5= i

5.R

5

V5= i

5(2)

V5= 2i

5

V6= i

6.R

6

V6= i6(5)

V6= 5i

6

Substituting V1, V

2, V

3, V

4, V

5, V

6values in equations

(1), (2) and (3), we get,

i2

4i4

2i5

= 0 ... (4)

2i3

+ 2i5

5i6

= 0 ... (5)

(3i1 10) + 4i

4+ 5i

6= 0

3i1+ 4i

4+ 5i

6= 10 ... (6)

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Now,

BT =

110

011

101

010

001

100

We know that,

The branch currents in matrix form is given as,

Ib=BT.I

L

6

5

4

3

2

1

i

i

i

i

i

i

=

3

2

1

110

011

101

010

001

100

I

I

I

From the above matrix, we have,

i1=I

3

i2

=I1

i3=I

2

i4= I

1+I

3=I

3I

1

i5= I

1+I

2=I

2I

1

i6= I

2+I

3=I

3I

2

Substituting i2, i

4, i

5values in equation (4), we get,

I1 4(I

3I

1) 2(I

2I

1) = 0

I1 4I

3+ 4I

1 2I

2+ 2I

1= 0

7I1 2I

2 4I

3= 0 ... (7)

Substituting i3, i

5, i


2(I2) + 2(I

2I

1) 5(I

3I

2) = 0

2I2+ 2I

2 2I

1 5I

3+ 5I

2= 0

2I1+ 9I

2 5I

3= 0 ... (8)

Substituting i1, i

4, i


3(I3) + 4(I

3I

1) + 5(I

2I

1) = 10

3I3+ 4I

3 4I

1+ 5I

2 5I

1= 10

9I1+ 5I

2+ 7I

3= 10 ... (9)

The equations (7), (8) and (9) represent the network

equilibrium equations.

Q7. (a) Is Norton theorem dual of Theveninstheorem? Justify your answer.

Answer :

Nov./Dec.-12, (R09), Q7(a) M[5]Considering a Thevenin circuit, a voltage source is

connected in series with a resistance. However, when a

Norton circuit is considered, the source will be a current

source, which is a dual of voltage source and also the series

resistance will be replaced with a parallel resistance, which

is also a dual representation from series connection to parallel

connection. Hence, from the above, it is clear that the Norton

circuit represent a dual of Thevenin circuit. Further, we can

conclude that the Norton theorem is a dual of Thevenin

theorem.

(b) Find the current in the 10 resistor asshown in below figure using superposi-tion theorem.

+

5

1 1

10

1A

2A

10 v A B

+

5

1 1

10

1A

2A

10 v A B

FigureAnswer : Nov./Dec.-12, (R09), Q7(b) M[10]

Given circuit is shown in figure (1),

+

5

1 1

10

1A

2A

10 v A B

+

5

1 1

10

1A

2A

10 v A B

Figure (1)

To determine the current in 10 resistor usingsuperposition theorem,

To find the current through 10 resistor, consideringeach source at a time.

Firstly, considering the 10 V source i.e., open

circuiting the current sources. The circuit diagram when

considering 10 V source is shown in figure (2),

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+

5

1 1

10 10 V+

5

1 1

10 10 V

Figure (2)

The step by step series parallel reduction of the

circuit of figure (2) is shown as,

+

5

1 1

10 10 V+

5

1 1

10 10 V

Figure (3)

+

5

1 + 1 = 2

10 10 V+

5

1 + 1 = 2

10 10 V

Figure (4)

+

10 10 V

=+

43.125

25

(I1)10

+

10 10 V

=+

43.125

25

(I1)10

Figure (5)

Current through 10 resistor is given by,

(I1)10 = 43.110

10

+

= 0.87 A

Now considering the 2 A current source and

neglecting the other sources. Therefore, the circuit of figure (1),

can now be redrawn as shown in figure (6),

5

1 1

10

2 A5

1 1

10

2 A

Figure (6)


circuit of figure (6) is shown below,

5

1 1

10

2 A

5

1 1

10

2 A

Figure (7)

1

2A1 10 5

(I2)10

1

2A1 10 5

(I2)10

Figure (8)

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1

2A1 =+

33.3

105

105

1

2A1 =+

33.3

105

105

Figure (9)

Current through (1 + 3.33) branch is given as,

33.41

12

+

= 33.5

2= 0.375 A

Now from figure (8), we have current flowing through

10 resistor is given as,

(I2)

10 = 105

5375.0

+

= 0.125

Now, considering the 1A current source and

neglecting other source. The circuit from figure can now be

drawn as,

5

1 1

10

1 A

5

1 1

10

1 A

Figure (10)


circuit of figure (10) is shown in figure (11),

5

1 1

10

1 A

5

1 1

10

1 A

Figure (11)

1

1A1 10 5

(I3)10

1

1A1 10 5

(I3)10

Figure (12)

1

1A1 =

+

33.3105

105

1

1A1 =

+

33.3105

105

Figure (13)

1A 4.33 1

1A 4.33 1

Figure (14)

Current through branch 4.33 is given as,

I4.33 = 33.41

11

+

=33.5

1= 0.188 A

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From figure (12), current through 10 resistorbranch is given as,

(I3)

10 = 1055188.0

+

= 0.063 A

According to the super position theorem, currentflowing through 10 resistor branch is given as,

I10 = (I1)10 + (I2)10 + (I3)10

= 0.87 0.125 0.063

= 0.682.

Q8. (a) State and explain the compensationtheorem.


For answer refer Unit-VIII, Q20.

(b) Apply Thevenins theorem and obtainthe current passing through 210 F

capacitor of figure.

~110 V, 50 Hz

10

70

100 F

210 F

~110 V, 50 Hz

10

70

100 F

210 F

Figure


The given circuit is shown in figure (1),

~

10 100 F

70 210 F~

10 100 F

70 210 F

Figure (1)


Voltage, V= 110 0 V

Frequency,f= 50 Hz

Resistance,R1

= 10

Capacitance, C1

= 100 F

Resistance,R2= 70

Capacitance, C3

= 210 F

And also,

By analyzing the circuit we have,

X2

= 0

R3

= 0

Now,

The figure (1) is modified as shown in figure (2),

~

A

B

10 100 F

110 V, 50 Hz

Z1= R

1+

Z2= R

2= 70

1cjX

~

A

B

10 100 F

110 V, 50 Hz

Z1= R

1+

Z2= R

2= 70~

A

B

10 100 F

110 V, 50 Hz

Z1= R

1+

Z2= R

2= 70

1cjX

Figure (2)

From figure (2),

The Thevenin voltage is given as,

VTh

=21

2.ZZ

ZV

+... (1)

Now,

The capacitive reactance of branch-1 is given as,

1C

X =fC21

= 610100502

1

=03142.0

1

= 31.827

Now,

Impedance,

Z1

=R1 1C

jX

= 10 j31.827

Impedance,

Z2

=R2

= 70

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Substituting V,Z1,Z


VTh = 110 0 70)827.3110(

70

+j

= 110 0 )827.3180(

70

j

= 110 0 (0.7554 +j0.3005)

= 110 0 0.813 21.692

= 89.43 21.692

Now,

The equivalent Thevenin impedance is given as,

ZTh

=Z1

+Z2

= 10 j31.827 + 70

= (80 j31.827)

The equivalent Thevenin circuit is represented in figure (3),

ZTh

A

B

C = 210 FVTh

ZTh

A

B

C = 210 FVTh

Figure (3)

Now,

The current passing through the 210 F capacitor is given as,

iC

=3Th

Th

ZZ

V

+... (2)

But,

Z3

=3C

jX =32

fC

j

= 610210502

j

=066.0

j

= j 15.1515

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Substituting VTh

,ZTh

,Z3

values in equation (2), we get,

iC

=)1515.15()827.3180(

692.2143.89

jj +

=1515.15827.3180

692.2143.89

jj

=89.43 21.692

80 46.9785j

=422.30773.92

692.2143.89

iC = 0.963952.114

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