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Test: Network Theorems - 2 - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Network Theory (Electric Circuits) - Test: Network Theorems - 2

Test: Network Theorems - 2 for Electrical Engineering (EE) 2024 is part of Network Theory (Electric Circuits) preparation. The Test: Network Theorems - 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Network Theorems - 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Network Theorems - 2 below.
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Test: Network Theorems - 2 - Question 1

i1 = ?

Detailed Solution for Test: Network Theorems - 2 - Question 1

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

Test: Network Theorems - 2 - Question 2

A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question..

As viewed from terminal y and y' is

Detailed Solution for Test: Network Theorems - 2 - Question 2

Thevenin equivalent seen from terminal yy' is

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Test: Network Theorems - 2 - Question 3

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

Detailed Solution for Test: Network Theorems - 2 - Question 3


Test: Network Theorems - 2 - Question 4

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

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

Detailed Solution for Test: Network Theorems - 2 - Question 4

Thevenized the circuit across R, RTH = 2 Ω

Test: Network Theorems - 2 - Question 5

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

Q. The maximum power will be 

Detailed Solution for Test: Network Theorems - 2 - Question 5


Test: Network Theorems - 2 - Question 6

If vs1 = 6 V and vs 2 = -6 V then the value of vα is

Detailed Solution for Test: Network Theorems - 2 - Question 6

Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.33

Test: Network Theorems - 2 - Question 7

A network N feeds a resistance R as shown in fig.P1.4.34. Let the power consumed by R be P.If an identical network is added as shown in figure, the power consumed by R will be​

Detailed Solution for Test: Network Theorems - 2 - Question 7

Let Thevenin equivalent of both network

Test: Network Theorems - 2 - Question 8

A certain network consists of a large number of ideal linear resistors, one of which is R and two constant ideal source. The power consumed by R is P1 when only the first source is active, and P2 when only the second source is active. If both sources are active simultaneously, then the power consumed by R is

Detailed Solution for Test: Network Theorems - 2 - Question 8


using superposition 

Test: Network Theorems - 2 - Question 9

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 2 Ω, the power dissipated by the bulb is

Detailed Solution for Test: Network Theorems - 2 - Question 9

 0.8
146.93

Test: Network Theorems - 2 - Question 10

The value of Thevenin's voltage across terminal a - b will be ______.

Detailed Solution for Test: Network Theorems - 2 - Question 10

Thevenin Theorem

Thevenin’s theorem states that it is possible to simplify any linear circuit, into an equivalent circuit with a single voltage source and a series resistance.


Steps to calculate:

  1. Thevenin Resistance (Rth): Open the circuit to the terminal from where the Thevenin resistance has to be found. While finding the Rth, short circuit the independent voltage source and open circuit the independent current source.
  2. Thevenin Voltage (Vth): This is the open-circuit voltage across the terminals where the load was connected.

Calculation

The voltage drop across 5Ω is: 

V = 5 × 10 = 50 volts

Applying KVL in loop:

-12 + V - 50 = 0

V = 62 volts

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