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This mock test of Basic Laws - 1 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Basic Laws - 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The equivalent resistance of the circuit shown in figure below across the terminals a-b will be

Solution:

On combining the resitances in parallel, equivalent circuit will reduce as shown in below.

QUESTION: 2

For the circuit shown below, the equivalent resistance across terminals a-b will be

Solution:

Hence, equivalent resistance is

QUESTION: 3

What is the value of m so that the current* I* in the circuit shown below is maximum?

Solution:

The equivalent resistance of the given circuit is

Hence, for current* I* to be maximum , R should be minimum.

QUESTION: 4

The given circuit shown below is converted to an equivalent voltage source V' connected in series with an equivalent resistance R'.

The values of V' and R' are respectively

Solution:

Converting the current sources into equivalent voltage sources, the circuit is reduced as shown below.

QUESTION: 5

The equivalent current source for the circuit shown below will be represented as

Solution:

Equivalent voltage in the given circuit is

V= 6 + 2 0 - 1 0 = 16 volt

And equivalent resistance is R = 1 + 5 + 2 = 8 Ω

QUESTION: 6

Assuming the batteries to be ideal voltage sources in the figure shown below, the value of current i is

Solution:

Converting all the voltage sources into equivalent current sources using source transformation, the equivalent circuit will be reudced as shown below.

QUESTION: 7

Two 100 W, 220 V bulbs are required to be connected across a 400 volts supply. The value of resistance to be inserted in the line so that the voltage across the bulbs does not exceed 220 V is

Solution:

Total power drawn from the circuit = 2 x 100 = 200 watts

Hence, supply current is

Let R be the series resistance to be inserted in the circuit such that the voltage across the bulbs is 220 V.

Applying KVL , we get

QUESTION: 8

The equivalent resistance across the terminal X-Y for the circuit shown below is

Solution:

Converting the star-connected resistors into Δ equivalent, the given circuit is reduced as shown below.

Hence, equivalent resistance between terminals Xand Y is

QUESTION: 9

Assertion ( A ) : KCL states that at any node of a circuit, at any instant of time, the sum of incoming currents is equal to the sum of outgoing currents.

Reason (R): KCL indicates the law of conservation of energy.

Solution:

KCL is based on law of conservation of charge while KVL is based on law of conservation of energy.

QUESTION: 10

The total power absorbed in the circuit below is

Solution:

The current through the resistors will purely depend upon the voltage source of 2V connected a cross them . Let the currents i_{1} , i_{2} and i_{3} respectively flowing through the resistors of 1Ω, 2Ω and 5Ω.

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