Net Electronics (Physics)

8/12/2019 Net Electronics (Physics)

1/23

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1

CHAPTER 1: NETWORK THEORY

Basic Circuit Analysis and KVL/KCL

Resistance: It is the property of a resistor to oppose current.

Case-I:Ri

V+ Conventional current flow from high to low potential.

Case-II: RI

+Electron current flow from low to high potential

In both case I and case II: V iR

Where resistance of Resistor:l

RA

where l= length of conductor, A = Area of conductor, = resistivity of conductor..

So

RA

l and conductivity =1

Resistivity

1

Current:

The phenomenon of transferring charge from one point in a circuit to another is described by the termelectric current.

An electric current may be defined as the time rate of net motion of electric charge across a cross-

sectional boundarydq

idt

q= Coulombs, t= sec, i= Amp

Voltage:

To move the electrons from one point to other point in particular direction external force is required.In analytical circuit external force is provided by emf and it is given by

dWV

dq

where a differential amount of charge dq is given a differential increase in energy dW.

Voltage = Energy per unit charge = Work per unit charge

A voltage can exist between a pair of electrical terminal whether a current is flowing or not. Anautomobile battery, for example, has a voltage of 12V across its terminals even if nothing whatsoever

is connected to the terminal.Ohms Law:

At constant temperature, the potential difference V across the terminals of a resistor R as show below,is directly proportional to the current i flowing through it. That is

V i V iR R+ V

i

Ohms law can also be expressed in terms of conductance G as

1i GV G

R

G= Conductance [mho or Siemens]


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2

Field Interpretation of Ohms Law:

At constant temperature current density is directly proportional to electric field intensity.

aI

lJ E

J E J ........... Current density (A/m2), ........... Conductivity 1[( ) ]m E ........... Electric field intensity (V/m)

Linearly test of Resistor:

When resistive element obeys ohms law then the element is called as linear resistor otherwise it isa non-linear resistor.

Power Dissipation in a Resistor:

2 22 2( ) V I

P Vi iR i i R V GR G

Practical Voltage Sources: It delivers energy at specified (V) which depends on current delivers bysources.

Rs

Vs

i

V(t)

DC

Rs

Vs(t)

I

AC

i

V

( )V t Vs IRs

Current Sources:Ideal current source delivers energy at a specified (I), which is independent on voltage

across the source. Internal resistance of ideal current source = .

Is

I

V

DC

I (t)s

i(t)

AC

ideal

I

V

Practical Current Sources: Practical current sources delivers energy at specified current I, which is dependent

on voltage across the source. In real time system current source does not exist.

Is

I

V

DC

I (t)s

i(t)

AC

V

I

R1

+


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3

Capacitance: It is the property of capacitor to hold charge qto confined electric field.

EAC

d

+ V

V

d

I

1V idt

C

d= distance between plates, A = Area of cross section of each conducting plates

Note: When the capacitance of capacitor is independent of current is called as linear capacitor.

|Constant

QC

V

Q

V

I

dVdt

dV

i cdt

So under steady state i.e. at t 0i

Capacitor acts as open circuit:

Inductance:

If the current i flowing in an element of figure below changes with time, the magnetic flux

produced by the current also changed which causes a voltage to be induced in the circuit, equal to

the rate of flux linkage that is

dV

dt

dtV

dt

i.e.di

V L

dt

where, L is constant of proportionality and is called self inductance.

i(t)+

(t)v(t)

(a)

GS

AB

i(t)

(b)

Inductive circuit Mutually coupled circuit

Therefore, there is an induced emf in coil B, which is equal to

de N

dt

ordi

e Ldt

As (di/dt) is directly proportional to ( / )d dt . Comparison of these two equation gives

N Li


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4

From above it may noted that an inductor is a device, while inductance is the quantityL.

NL

i

Asdi

V Ldt

, so in steady state i.e. t 0V

And circuit behaves like shots circuit.

Active Element: When an element is capable of delivering energy for infinite period of time is called as active

element.

Example: Voltage, source, current source.

Passive Element: When an element is not able to deliver energy for infinite time period is called as passiveelement.

Example: Resistor, inductor and capacitor

Active Elements:

Voltage Source: Ideal voltage source delivers energy at a specified V. Internal resistance of ideal voltagesource is zero.

Voltage Divider Rule:

11

1 2

s

RV V

R R

,2

2

1 2

s

RV V

R R

~Vs

V1 R1

V2 R2

Current Division Rule:

~

I I1

R1

I2

R2eq 1 2

1 1 1

R R R

2 11 2

1 2 1 2

,R R

I I I IR R R R

Same for Inductor:

~

I1I I2

L2L1 1 2

1 1 1eq

L L L 2 11 2

1 2 1 2

;L LI I I IL L L L


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5

For capacitor:

~

I I1

C1

I2

C2 1 2eqC C C 1 2

1 2

1 2 1 2

,C C

I I I IC C C C

Kirchoffs Voltage Law (KVL):

It states that algebraic sum of all voltages in a closed loop is equal to zero. It is based on conservationof energy.

+

+Vs1 Vs2

V2

R2 +

R1

V1+

I

Applying KVL:

1 2 1 2 0s sV V V V 1 2 1 2s sV V V V

Q. Calculate values of V1and V

2across 4 and 5 resistor by KVL.

6V

1V 5V3V

2V5V2+

I

V1+

4

2

Soln. Applying KVL

6 + 3 + 1 5 4I 2I 2 5I = 0

3 = 11I,3

I11

Amp

V1

= 4I = 4 3

11

12V

11 volt

V2= 5I = 5

3

11=

15

11 volt

Q. Calculate VAB

in the given circuit

5V

2 A

4

X

2V

Y 10V

B

2 5I2I1


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6

Soln. KVL at loop 1: (Left loop)

4I1 2I

1+ 5 = 0, I

1= 5/6 Amp

KVL at loop 2: (Right loop)

10 2I2 5I

2 = 0

I2=

10

7 Amp, VVAB= VAX+ VXY+ VYB

=5 10

4 2 26 7

=20 20

26 7

=32

21 1.53VoltABV

Q. The value of I1in the circuit is

+

+

3A

8V

7

I1

2V

95A5

Soln. The circuit can be redrawn in the form of voltage source as [Current voltage source]

+

15V

45V

2V

7 + 5

+

9

+

I1

8V

Applying KVL

8 7I 15 5I + 45 9I 2V = 0

53 17 21 I = 0, I =36

21 I 1.71

Kirchoff Current Law (KCL)

(i) KCL states that algebraic sum of all current meeting at a point is equal to zero.

0ci (ii) KCL worked on principle of law of conservation of charge.

I1 I5

I4I3

I2 I1+ I2+ I3 I4 I5 = 0 1 2 3 4 5I +I +I =I +I


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7

Q. Calculate the values of V1and V

2by KCL:

6A 1 3 3V

5 V1 2 V2 1

Soln. Applying KCL at node 1: 6 =1 1 2

1 2

V V V ... (1)

KCL at node (2) : 2 2 2 13

1 3 2

V V V V ... (2)

By solving equation (1) and (2);

We get V1= 5V and V

2= 3V

Limitations of KVL and KCL:

KCL and KVL will fail for the high frequency circuit.

KVL and KCL will fail in distributed elements since in distributed element it is not possible to separateeffect of R, L, C.

Source Transformation:

RV I V/R R

Series Combination of Batteries:

R RV2

V V1 2

2R

V1

Parallel combination of current sources:

I1 R1 I2 R2 I +I1 2 R || R1 2

Q. Obtain single current source for network shown

+

+

56

10V18V

A

Soln. 2.733A 6 2A 5 5A


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8

Q. Convert given circuit into a single voltage source

12A

8A 4

6

X

Y

Soln.+

+

6

4

32V

72V

X

Y

+ 104V

10

Linear and non-linear Elements:A linear network shows linear characteristics of voltage versus current. For a non-linear element thecurrent passing through it does not change linearly with the linear change in applied voltage at a

particular frequency. Semiconductor devices are usually examples of non-linear element.

In V.I. relation if output is zero for zero input and relation is linear then it is called linear network or

element obeys ohms law.

V = ki linear V = ki2 Non-linearV = 2i + 3 Non-linear V = in Non-linear

I

V

Lineari

V

Non-Linear

Simple resistors inductors and print at end capacitors are linear elements and their resistance inductanceand capacitances do not change with a change in applied voltage on the circuit current .

Active and Passive Elements:

If a circuit element has the capability of enhancing the energy level of a signal passing through it then

it is called an active element. Vacuum tubes and semiconductor devices are active elements on the

other hand resistors, inductors, capacitors, thermistors etc. are passive elements as they do not haveany intrinsic mean of signal boosting.

In V-I relation if any portion has V-I value as negative then it is active network.

Bilateral and Unilateral Element:

If the magnitude of the current passing through an element is affected due to change in polarity ofpolarity of the applied voltage then element is called unilateral element. On the other hand if current

magnitude remains the same even if the applied voltages polarity is changed then it is called a bilateralelements.

If by changing variable relation between both dependent variable and independent variable remains

same then it is called bilateral network otherwise unilateral network.


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9

Mesh Analysis:

Mesh is a property of a planner circuit and is undefined for a non-planner circuit.

We define a mesh as A loop that does not contain any other loops within it.

+

+

Fig. (a) Fig. (b)

[Planner] [Non-Planner]

No branch passes over or under any other branchNetwork Theorem:

Superposition Theorem:

The statement of superposition theorem follows as below:In any linear bilateral network having more than one source, response in any one of the branches

is equal to algebraic sum of the response caused by individual source while rest of the sources arereplaced by their internal impedances.

Note:

The principle of superposition is useful for linearly test of the system.

This is not valid for power relationship. Sources can be made inoperative by

(a) Shot circuiting the voltage sources

(b) Open circuiting the current sources

A linear network comprises independent sources, linear dependent source and linear passive elementslike resistor, inductor, capacitor and transformer. Moreover, the components may either be timevarying or time invariant.

Other Definition: The superposition principle states that the voltage across (or current through) on elementin a linear circuit is the algebraic sum of the voltage across (or current through) that element due to

each independent source acting alone.Q. Calculate value of V for the given circuit by using superposition theorem.

+6V 3

4 V+

8

Soln. At one time only effect of one source is considered. Voltage source is short circuited and currentsource is open circuited.

Since there are two sources, So, V = V1+ V

2


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10

Let V1is the voltage across 4 ohm due to 6V voltage source alone (in this case current source is

open circuited).

+6V 4 V1

+

8

16

48 4V , 16

4 2V12V

Let V2is the voltage due to 3A current source alone (in this case voltage source is short circuited).

3A4 V2+

8

3 = 2 24 8

V V , 3 = 2 {2 1}

8

V 2 8 voltsV

So total voltage V is V = V1+ V

2= 2 + 8 = 10 volt

Q. Calculate the value of V0by use of superposition theorem.

3

+

5

20V8A2V0+

Soln. Since there are two sources so total voltage due to both sources will be0 1 2

V V V

Let V1is the voltage only due to 8A current source, (In this case 20V source is short circuited)

3 5

8A2V1+

3 5

2V0+

+ 20V

1

8 54

3 2 5i A

, V

1= 2 4 = 8 volt

Let V2is voltage due to 20V source alone

2

202

10V , V2 = 4 volt, V0= V1 + V2= 8 + 4 0 12voltV

Thevenins Theorem:

Any two terminal linear bilateral network can be replaced by a voltage source in series with impedance.

N/W

A

B +

A

B

Vth

Zth


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11

Calculation of Thevenins equivalent for independent source:

For Calculation of Rth

: All independent current sources are open circuited and all independent voltagesources are short circuited.

For Calculation of Vth

:Voltage across open circuit terminal.

Calculation of Thevenins equivalent for dependent source:

For calculation of Vth: Voltage is calculated across open circuited terminal by assuming current zero in thatterminal.

For calculation of Rth

: First Ise is calculated by assuming short-circuited terminal and then Rth

is calculated

by /Th Th ScR V I

Q. For the given circuit calculate the power loss in the 1 ohm resistor by use of thevenins theorem.

+

1510

10V

X

Y

2A

Soln. For calculation of Vth

branch XY terminal is assumed as open circuited and let voltage is Vth

-

+

10

10V

X

Y

2A

I1 I2

5

+

+

Vy

Vx

2 = I1+ I

2, 2 =

10

10 5x xV V , 2 =

1 1 110 5

xV

,

3 310

xV

10xV Volts , 5 2 10x yV V 10ThV Volt

For Calculation of RTh

:(If dependent source is present) 2A current source is open circuited and 10Vsource is short circuited and let resistance is R

Th:

RTh510

10 5

10 5ThR

,10

3.33ThR ,

102.31A

3.33 ITh

Th

Th L

VI

R R

Q. Find the current through the 5 resistor in the circuit by use of Thevenins theorem.

2A 1 5 5V

2

+


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12

Soln. For calculation of VTh

= 5 ohm resistor is open circuited and so current in 5 ohm will be zero.

5Th abV V V

2A 1 5V

2

+a

b

For Calculating RTh

: 2A current source is open circuited and 5V source is short circuited.

1

2

a

b R

Th=

0 30

3 0

,

51

5 10Th

L

Th L

VI A

R R

Nortons Theorem:Any two terminal linear bilateral network containing active and passive element can bereplaced by an equivalent current source in parallel to an equivalent impedance current source.

N/WA

B

A

B

IN ZN N ThZ Z

Note:

+ VTh

RTh

(a) Thevenin's equivalent circuit

RN

(b) Norton's equivalent circuit

IN

Both of the above circuits are convertible to each other with the relations given as below:

,Th N Th N N N ThR R V I R I R

Q. Find the Nortons equivalent circuit across a-b for the network shown in figure:

5A5

2V

2

a

b

Soln. 5A

5

2V

Isc


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13

2 + 5(5 + Isc

) = 0, Isc

=2

55

= 0.4 5 4.6scI A

2 5 101.43

2 5 7Th Th

R R

Q. Find Nortons equivalent to the right of a-b terminal (across 3V source)

+ 5

10a

3V

b

i = 1A0

Soln. The equivalent circuit after shorting ab terminal.

+5

10

Isc

V

i1

i = 1A0

(a)

5

10

(b)[To calculate R ]Th

i0= i

1+ I

sc, I = i

1+

10

V

I = 5 10

V V

, I =

3

10

V

, V =

10

3 and Isc=

1

3 = 0.33A 10 5 15ThR Reciprocity Theorem:

A linear network is said to be reciprocal or bilateral if it remains invariant due to the interchange ofposition of cause and effect in the network.

For verification of the reciprocity theorem following conditions must be satisfied.

Circuit should consist of linear, time-invariant bilateral element.

Circuit should consist of only a single independent sources.

When circuit consist dependent source, reciprocity theorem can not be verified.Maximum Power Transfer Theorem: Maximum power transferred from source to load is only possible

when(i) Source impedance = Load impedance

(ii) Thevenin impedance = Load impedance

For DC Circuits:

DCSource

Network

X

Y

RL

I

+ VTh

I

RL

X

Y

RTh

(Load connected to the do source network) (Equivalent source network and load)


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14

Th

Th L

VI

R R

while the power delivered to the resistance load is

2

2 0

L L LTh L

V

P I R RR R

.... (i)

Differentiating equation (i) with respect to RLand equating to zero. We get L ThR R

Hence, it has been provided that power transfer from a dc source network to a resistance networkis maximum when the internal resistance of the dc source network is equal to the load resistance.

Again value of that maximum power is,

2 2

max 2( ) 4Th Th Th

Th Th Th

V R VP

R R R

or2

max4

Th

L

VP

R 2[ ]ThR R

The total power supplied is thus:

2

24

Th

Th

VP

R

2

2Th

Th

VP

R

During maximum power transfer the efficiency becomes,

max 100%P

P

2

2

4100%

2

Th

Th

Th

Th

V

R

V

R

50%

So that efficiency in this case is 50% i.e. half of the total power is transferred to the load RL.

[Maximum Power Transfer Theorem for ac circuits]

Consider the Thevenins equivalent circuit for an ac network as shown below:

~

ZTh

ZLVTh

Here, ,Th Th Th L L LZ R JX Z R JX

Now let us consider different cases for maximum power transfer.Case-1: Both R

Land X

Lare variable

When both RLand XLare variable then maximum power from source to load will be transferred ifload impedances is complex conjugate of internal impedance of the network.

i.e. *L ThZ Z

In this case maximum power (active) will be calculated as2

max4

Th

L

VPR


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15

Also during maximum power transfer efficiency will be 50%.

Case-2: RLis variable but X

Lis constant.

In this case maximum power will be transferred when, | |L Th LR Z JX

or 2 2( )L Th Th LR R X X

Efficiency can be calculated as 100 %L

Th L

R

R R

Efficiency comes out to be greater than 50% 50%

Case-3: Load impedance is purely resistive.

In this case maximum power will be transferred when, | |L ThR Z

Efficiency comes out to be greater than 50%.Case-4:RLand X

Lare variable but the impedance angle is constant i.e.

L L LZ R JX and

1tan ConstantLX

QR

In this case maximum power will be transferred when, L ThZ Z

Q. Calculate value of R in circuit such that maximum power transfer takes place and also calculate

amount of this power.

+

+6V

4V

1 5

2 1 R

Soln. Above circuit can be solved by Thevenins theorem: VTh RRTh

To find RTh

1 5

2 1 RTh (2 ||1) 5 ||1ThR 0.85ThR

To find VThopen load R

1 5

2 1

VTh+

4V 6.4ThV V


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16

Maximum Power =

2 2(6.4)12W

4 4 0.85Th

Th

V

R

Q. Assuming maximum power transfer from source to load R calculate the value of R and maximumvalue of power transferred.

+1 50V 1 5

10 R

3 Y

X

Soln. VTh

across XY = 100

3V, R

Thacross XY =

25

3

Pmax =

2 2(100 / 3) 100

254 343

Th

Th

V

R Watt


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17

Assignment - Network Theory

1. A segment of a circuit is shown in figure VR= 5V, V

C= 4 sin 2t. The voltage V

Lis given by

Q

2AP

51A +

1F

+ VcR

VL2H

+

S

VR

(a) 3 8 cos 2t (b) 32 sin 2t (c) 16 sin 2t (d) 16 cos 2t

2. In the circuit of figure, the magnitudes of VL

and VCare twice that of V

R. Given thatf= 50 Hz, the

inductance of coil is

VCC

VR

5

~5


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18

6. The RMS value of the voltage u(t) = 3 + 4 cos (3t)

(a) 17 V (b) 5V (c) 7V (d) (3 2 2)V

7. Assuming ideal element in the circuit shown below, the voltage Vab

will be

+ 5V1A Vab

2

i

a

b

(a) 3V (b) 0V (c) 3V (d) 5V

8. The current through the 2k resistance in the circuit shown is

D

C

A B

1k 1k

1k1k

2k

6V

(a) 0mA (b) 1mA (c) 2ma (d) 6mA

9. The time constant for the given circuit will be

1F1F 3A

3

31F

(a) 1/gs (b) 1/4s (c) 4s (d) gs

10. In the circuit given below, the value of R required for the transfer of maximum power to the load

having a resistance of 3 is

R

6 3 Load

+

10V

(a) 0 (b) 3 (c) 6 (d) 11. The r.m.s. value of the current i(t) in the circuit shown below is

7~

i(t)

(1 sin t) v

1H

1

1F

(a)1

2A (b)

1

2A (c) 1A (d) 2 A


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19

12. In the given figure, the Thevenins equivalent pair (voltage, impedance), as seen at the terminals P

Q, is given by

4V2010

10 UnknownNetwork

(a) (2V,5 ) (b) (3V,5 ) (c) (4V,5 ) (d) (2V, 7.5 )

13. In the circuit shown, the power supplied by the voltage source is

1A

+ 10V

1

2A

1

1

11

(a) 0W (b) 5W (c) 10W (d) 100W14. The voltage e

0in the figure is:

8A 10 6 12 e0

+

2

16V

(a) 48V (b) 24V (c) 36V (d) 28V15. In the circuit of figure, the value of the voltage source E is

+1V

0VV2

10VV1 4V

+ +

2V+

E

(a) 16V (b) 4V (c) 6V (d) 16V

16. The voltage V in figure is equal to:

++

+ 4V

5V

V+ 3

2 4V

(a) 3V (b) 3V (c) 5V (d) None of these

17. The voltage in figure is

2A+ 10VV

3+

(a) 10V (b) 15V (c) 5V (d) None of these


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20

18. What is the value of R required for maximum power transfer in network shown above?

25V

5

20 3A

4

R

(a) 2 (b) 4 (c) 8 (d) 1619. The voltage across the terminal aand bin figure is:

3A1V

22

1a

b

(a) 0.5V (b) 3.0V (c) 3.5V (d) 4.0V

20. The nodal method of circuit analysis is based on

(a) KVL and ohms law (b) KCL and Ohms law(c) KCL and KVL (d) KCL, KVL and Ohms law

21. The current i4in the current of figure is equal to:

5A 3A

7A

i = 4A3i4

(a) 12A (b) 12A (c) 4A (d) None of these

22. Thevenin equivalent voltage VABand resistance RThacross the terminals AB in the above circuit are

A B10V

2

3

3

2

(a) 6V,5 (b) 4V,5 (c) 2V,2.4 (d) 2V,2.5

23. The differential equation for the current i(t) in the circuit of the figure is

sin t+

2 2H1F

i(+)

(a)2

22 2 ( ) sin

d i dii t t

dt dt (b)

2

22 2 ( ) cos

d i dii t t

dt dt

(c)2

22 2 ( ) cos

d i dii t t

dt dt (d)

2

22 2 ( ) sin

d i dii t t

dt dt


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21

24. The RC circuit shown in the figure is

R C

CR+

Vi V0

+

(a) a low-pass filter (b) a high-pass filter

(c) a band-pass filter (d) a band-regeat filter

25. In the circuit shown below, the value of RLsuch that the power transferred to R

Lis maximum.

+

+5V 2V 1A

10 10

10RL

(a) 5 (b) 10 (c) 15 (d) 20

26. Find the value of C to deliver the maximum power to load.

+ 2 sin 2t2

2 1H

C 4Load

(a) 0.125F (b) 0.5F (c) 2F (d) 4

27. Find ZLsuch that maximum power is transferred to it.

~ ZLJ2+

Vs

2 J2 J2

(a) (2 2)J (b) (2 2)J (c) ( 2 )J (d) 2

28. The voltage across a capacitor is triangular in waveform. The waveform of current is

(a) triangular (b) trapezoidal (c) Sinusoidal (d) Rectangular

29. The value of Rin

of the network

Rin

1 1

1 1

1

1

1

(a) 1.62 (b) 2 (c)1

3 (d)

1

2

30. The maximum power that can be distributed in the load in the circuit shown is

9V RL

13

6

10

(a) 0.396 W (b) 6W (c) 6.75W (d) 13.5W


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22

31. Find Vxfrom the given circuit.

50

5

Vx 10+

10A

(a) 42.2 V (b) 83.3 V (c) 97.3V (d) 103V

32. If Vc(f) = 4 cos (105 t)V in the circuit, find Vs.

+

2mH

80nF +VcVs

(a) 6.4 cos 105 t V (b) 2.4 cos 105 t V

(c) 6.4 cos 105

t V (d) 2.4 cos 105

t V33. Determine the current through the branch AB of network shown below.

+ 10V

J5

J10

J5A

5

B

5


23/23

23

ANSWER KEY

1. (b) 2. (c) 3. (d) 4. (c) 5. (b)

6. (a) 7. (a) 8. (a) 9. (c) 10. (a)

11. (b) 12. (a) 13. (a) 14. (d) 15. (a)

16. (a) 17. (a) 18. (c) 19. (c) 20. (b)

21. (b) 22. (b) 23. (c) 24. (c) 25. (c)

26. (a) 27. (d) 28. (d) 29. (a) 30. (a)

31. (b) 32. (d) 33. (a) 34. (c) 35. (a)

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