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# Networks Theorem - MCQ Test

## 20 Questions MCQ Test GATE Electrical Engineering (EE) 2022 Mock Test Series | Networks Theorem - MCQ Test

Description
This mock test of Networks Theorem - MCQ Test for Railways helps you for every Railways entrance exam. This contains 20 Multiple Choice Questions for Railways Networks Theorem - MCQ Test (mcq) to study with solutions a complete question bank. The solved questions answers in this Networks Theorem - MCQ Test quiz give you a good mix of easy questions and tough questions. Railways students definitely take this Networks Theorem - MCQ Test exercise for a better result in the exam. You can find other Networks Theorem - MCQ Test extra questions, long questions & short questions for Railways on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

Solution:

QUESTION: 3

### Which theorem assists in replacement of an impedance branch over the network by the other network comprising different circuit components, without affecting the V-I relations throughout the entire network?

Solution:
QUESTION: 4

A simple equivalent circuit of the 2 terminal network shown in fig.  is

Solution:

After killing all source equivalent resistance is R
Open circuit voltage = v1

QUESTION: 5

Solution:

The short circuit current across the terminal is

QUESTION: 6

VTH, RTH  = ?

Solution:

For the calculation of RTH if we kill the sources
then 20Ω resistance is inactive because 5 A source will
be open circuit
RTH = 30+25 = 55 Ω,
vTH = 5+5X30 = 155V

QUESTION: 7

RTH = ?

Solution:

After killing the source, RTH = 6 Ω

QUESTION: 8

The Thevenin impedance across the terminals ab of the network shown in fig. is

Solution:

After killing all source,

QUESTION: 9

For In the the circuit shown in fig. a network and its Thevenin and Norton equivalent are given

The value of the parameter are

VTH  RTH  I RRN

Solution:

v∝ = 2 x 2 + 4 = 8 V = vTH
RTH = 2 + 3 = 5Ω = RN

QUESTION: 10

v1 = ?

Solution:

If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent

QUESTION: 11

i1 =  ?

Solution:

If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent

QUESTION: 12

A circuit is given in fig.  Find the Thevenin equivalent as given in question..

As viewed from terminal x and x' is

Solution:

We Thevenized the left side of xx'and source transformed right side of yy'

QUESTION: 13

A circuit is given in fig. Find theThevenin equivalent as given in question.

As viewed from terminal y and y' is

Solution:

Thevenin equivalent seen from terminal yy' is

QUESTION: 14

A practical DC current source provide 20 kW to a 50Ω load and 20 kW to a 200 Ω load. The maximum power, that can drawn from it, is

Solution:

(r + 200)2 = 4(r + 50)2
⇒ r = 100 Ω
i = 30A,

QUESTION: 15

In the circuit of fig. P.1.4.15–16 when R = 0 Ω , the current iR equals 10 A.

The value of R, for which it absorbs maximum power, is

Solution:

Thevenized the circuit across R,RTH = 2 Ω

QUESTION: 16

In the circuit of fig when R = 0 Ω , the current iR equals 10 A.

The maximum power will be

Solution:

QUESTION: 17

A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2Ω, the power dissipated by the bulb is

Solution:

rbattery VocIsc = 24/30 = 0.8 ohms

P = V2oc / ( r + 2)2   * 2

= 24/ ( .8 + 2)2  * 2

= 146.9 W

QUESTION: 18

The following results were obtained from measurements taken between the two terminal of a resistive network

The Thevenin resistance of the network is

Solution:

QUESTION: 19

A DC voltmeter with a sensitivity of 20 kΩ/V is used to find the Thevenin equivalent of a linear network. Reading on two scales are as follows

(a) 0 - 10 V scale  : 4 V
(b) 0 - 50 V scale : 5 V

The Thevenin voltage and the Thevenin resistance of the network is

Solution:

= 50 μ A

For 0 - 10 V scale Rm = 10 x 20k = 200 kΩ
For 0 - 50 V scale Rm = 50 x 20k = 1 MΩ
vTH = 20μRTH + 20µ x 200k = 4 + 20μRTH     ...(i)
UTH = 5μ x RTH + 5μ x 1M = 5 + 5μRTH        ...(ii)
Solving (i) and (ii)

QUESTION: 20

Consider the network shown in fig.

The power absorbed by load resistance RL is shown in table :

Find the Thevenin equivalent.

Solution: