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QUESTION: 1

The current I in the circuit shown below is

Solution:

Let the current flowing through 5 Ω resistor be I' Applying KCL, we get

5 + I = I'.

Applying KVL, we get

So,

= -4.4 A

QUESTION: 2

For the circuit shown below, match List - I with List-II and select the correct answer using the codes given below the lists:

Codes:

Solution:

Nodes = 10 (a, b, c, d, e, f, g, h, k, p)

Junction points = 3(b, e, h)

QUESTION: 3

The value of current I in the circuit shown below for V = 2 volt is

Solution:

Applying KCL at the node connecting 0.5 V current source and 1 A current source, we get

1 + I = 0.5 V

For v - 2 volt, 1 + I = 1

or, I = 0 A

QUESTION: 4

For the circuit shown below, the value of (V_{1}+ V_{2}) is

Solution:

Using KVL in the loop, we have

V_{1 }= 10 + 15 = 25 volts

Also, V_{2} = -25 volts

∴ V_{1} + V_{2} = 25 - 25 = 0 volts

QUESTION: 5

The value of voltage source to be connected across the terminals X and Y so that drop across the 10 Ω resistor is 45 V is

Solution:

Let the required voltage be V

Then, voltage across 10 Ω resistor

or,

Hence, required voltage is

V = 180 volts

QUESTION: 6

In a voltage divider circuit as shown below with two resistances R_{1} and R_{2}, the voltage drop V_{1} is twice the drop V_{2} across R_{2}. The value of R_{1}, and R_{2} will be

Solution:

Using Ohm’s law,

Given, V_{1} = 2V_{2}

As I is same therefore,

R_{1} = 2R_{2}

∴ R_{T} = R_{1} + R_{2} = 2R_{2} + R_{2} = 3R_{2}

So, R_{T} = 3R_{2} = 4Ω

or,

and

QUESTION: 7

A delta connected load is shown in figure below. The equivalent star connection has a value of R in Ω is

Solution:

We know that

R_{Δ }= 3 R_{Y}

or,

QUESTION: 8

The voltage drop across the 2 Ω resistor for the circuit shown below is

Solution:

On combining the current sources, the circuit is reduced as shown below.

Hence, voltage drop across the 2Ω resistor

= 2 x 2 = 4 volts

QUESTION: 9

For the circuit shown below, the sum of the unknown currents I_{1}, I_{2}, l_{3} and I_{4} is

Solution:

Using KCL,

I_{2} = 6 - 2 = 4 A

Also,

or,

and I_{4} = I_{2 }- I_{3} = 4 - (-1.5) = 5.5 A

Also, I_{1} = 6 A

∴ I_{1} + I_{2} + I_{3} + I_{4} = 6 + 4 - 1.5 + 5.5

= 15.5 - 1.5 = 14A

QUESTION: 10

In the delta equivalent of the given star connected circuit is equal to

Solution:

= (10 + j10 + j20)

= (10 + j30) Ω

QUESTION: 11

Four resistances 80 Ω, 50 Ω, 25 Ω and R are connected in parallel. Current through 25 Ω resistance is 4 A. The total current of the supply is 10 A. The value of R will be

Solution:

Given, I_{25Ω}=4 A

Here, R' = 80║50║R

or,

or,

or, 29R + 400 = 40R or 11R = 400

or, R = 36.36 Ω

QUESTION: 12

For the circuit shown below, the equivalent resistance will be

Solution:

Here, a and c are at equipotential.

Also, b and d are at equipotential

∴

QUESTION: 13

Three resistors of R Ω each are connected to form a triangle. The resistance between any two terminals will be

Solution:

QUESTION: 14

A network has 10 nodes and 17 branches. The number of different node pair voltage would be

Solution:

Number of different node-pair voltage = Number of KCL equations = n - 1 =10 - 1 = 9

QUESTION: 15

Kirchhoff's laws are not applicable to circuits with

Solution:

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