network theorems part ii

8/10/2019 Network Theorems Part II

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Network Theorems (Part II)-MCQs

1. If an impedance ZLis connected across a voltage source V with source impedance ZSthen

for maximum power transfer, the load impedance must be equal to(a)source impedance ZS

(b)complex conjugate of ZS

(c) real part of ZS

(d)imaginary part of ZS

[GATE 1988: 2 Marks]

Ans. (b)

According to maximum power transfer = 2. A load, ZL=RL+jXLis to be matched, using an ideal transformer, to a generator of internal

impedance, ZS=RS+jXS. The turns ration of the transformer required is

(a)|/|(b)|/|

(c)|/|(d)|/|


Ans. (a)

= (

)

or

=

3. The value of the resistance, R, connected across the terminals, A and B, (ref. Fig.) which

will absorb the maximum power, is


2/7

AC

A B

4 K

4 K6 K

3 K

R

(a)4.00k

(b)4.11k

(c)8.00k

(d)9.00k

[GATE 1995: 1 Mark]

Ans. (a)

For maximum power transfer = finding Thevenins equivalent betweenpoint A, B replace voltage source by its internal impedance

= = =

4. The Thevenin equivalent voltage VTHappearing between the terminals A and B of the

network shown in the figure is given by

AC

3

VTH-j6j2 j4

10000V

A

B

a

(a)j 16(3-j4)

(b)j 16(3+j4)

(c)16(3+j4)

(d)16(3-j4)


Ans. (a)

The voltage at the node a is 100

= = =


3/7

5. Use the data of the figure (a). The current I in the circuits of the figure (b)

R2R2

R1

R1

R3

R3

R4R410V 2A 20Vi=?

a b

A A BB

(a) -2 A

(b)2 A

(c) -4 A

(d)+4 A


Ans. (c)

This is a reciprocal and linear network. According to reciprocity theorem, an ideal

voltage source in loop A, produces current I in loop B. By interchanging the

positions of voltage source and ammeter produces the same current in loop A. When

voltage source is doubled and is negative current also doubles with opposite

direction =

6. In the network of the figure, the maximum power is delivered to RLif its value is

20

50V

RL

40

i1

0.5 i1

(a)

16 (b)40/3

(c)

60 (d)20


Ans. (a)

For maximum power delivered, RLmust be equal to Rth.


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20

50V

40

I1

0.5 I1

P

Writing KCL at node P, let Vthbe the open circuit voltage

. =

=

Solving equation (i) and (ii) =, = = ,is short circuit current when RLis shorted = . = . . = .

20

50V

40

I1

0.5 I1

= = . =

7. For the circuit shown in the figure, Thevenins voltage and Thevenins equivalent resistance

at terminals a-b is


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5 10VRL

5

i1

0.5 i1a

b

+

-

1A

(a)5 V and 2

(b)7.5 V and 2.5

(c)4 V and 2

(d)3 V and 2.5


Ans. (b)

= KCL at node a

= 2 Vab = 7.5

For Rthdeactivate independent sources (10V, voltage source by zero

impedance and 1A current source by open circuit)

= = .

8. In the circuit shown, what value of RLmaximizes the power delivered to RL?

+-

+-

Vt

VX 4

44

-VX+

100V RL

(a)2.4

(b)8/3

(c)4

(d)6



6/7

Ans. (c)

For maximum power delivered to RL, RL= Rthof the network. (deactivate

independent source Vi)

Removing RLand connecting Vtest= 1 volt supplying current I. then Rth=impedance looking into the network

=

DC

+-

I

I1

4

44

(I - I1)VX

Rth

VX

+-

1V

Using KVL in outer and inner 100p

= = , = =

= = =

9. In the circuit shown below, if the source voltage VS= 100 53.130V then the Theveninssequivalent voltage in Volts as seen by load resistance RLis


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3

+

-

RL= 10

10 VL1 i2

3

i1

+

-

AC

j4 j6

-+

j40 i2VS

VL1

(a)100900(b)80000 (c)80090

0

(d)100600[GATE 2013: 2 Marks]

Ans. (c)

To find Vthor Voc(open circuit voltage), detach RL

=, = = (dependent voltage source is zero), . =

= + = =

network theorems part ii

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