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This mock test of Basic Nodal & Mesh Analysis - 1 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
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QUESTION: 1

Of the two methods of loop and node variable analysis

Solution:

QUESTION: 2

Which of the following is not true about the circuit shown below?

Solution:

Number of junction point for the given circuit is 6.

QUESTION: 3

For the circuit shown below, the current through the 1 Ω resistor is

Solution:

Using source transformation, the current source is converted to a voltage source as shown below.

Applying nodal analysis at node (x), we get

QUESTION: 4

An electric circuit with 8 branches and 4 nodes will have

Solution:

Number of loop equation

= b - ( n - 1)

= 8 - ( 4 - 1 )

= 8 - 3 = 5

QUESTION: 5

For the circuit shown below, the current through the 10 V battery is

Solution:

Using source transformation, first we convert the current source into an equivalent voltage source

Applying KVL in loop-1, we get

Applying KVL in loop-2, we get

On solving equations (i) and (//). we get

i_{1} = 2.36 A and i_{2} = 4.91 A

Hence, the current through the battery of 10 V is

i_{2} = 4.91 A

QUESTION: 6

Find the value of the currents I1 and I2 flowing clockwise in the first and second mesh respectively.

Solution:

The two mesh equations are:

5I1 - 3I2 = 10

-3I1 + 7I2 = -15

Solving the equations simultaneously, we get I1 = 0.96A and I2 = -1.73A.

QUESTION: 7

The value of V in volts for the circuit shown below is

Solution:

The equivalent resistance across the 5 A current source is

QUESTION: 8

The number of KVL and KCL equations for the circuit shown below are

Solution:

Number of nodes, n = 5

Number of branchs, b = 6

Hence, number of KCL equation

= n - 1 = 5 - 1 = 4

and, number of KVL equations = b - ( n - 1)

= 6 - (5-1) = 6- 4 = 2

QUESTION: 9

The current *l* flowing in the given figure is

Solution:

QUESTION: 10

The value of dependent source for the circuit shown below is

Solution:

Applying KVL in the loop,

5 = 2i — 2i + i or i = 5 A

∴ Value of dependent source

= 2i = 2 x 5 = 10 volt

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