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Test: Basic Nodal & Mesh Analysis - 1 - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Basic Nodal & Mesh Analysis - 1

Test: Basic Nodal & Mesh Analysis - 1 for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Test: Basic Nodal & Mesh Analysis - 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Basic Nodal & Mesh Analysis - 1 MCQs are made for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Nodal & Mesh Analysis - 1 below.
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Test: Basic Nodal & Mesh Analysis - 1 - Question 1

 Of the two methods of loop and node variable analysis

Test: Basic Nodal & Mesh Analysis - 1 - Question 2

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

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 2

Number of junction point for the given circuit is 6.

Test: Basic Nodal & Mesh Analysis - 1 - Question 3

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

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 3

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

Applying nodal analysis at node (x), we get 

Test: Basic Nodal & Mesh Analysis - 1 - Question 4

An electric circuit with 8 branches and 4 nodes will have

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 4

Number of loop equation
= b - ( n - 1)
= 8 - ( 4 - 1 ) 
= 8 - 3 = 5

Test: Basic Nodal & Mesh Analysis - 1 - Question 5

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

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 5

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

i1 = 2.36 A and i2 = 4.91 A 

Hence, the current through the battery of 10 V is 

i2 = 4.91 A

Test: Basic Nodal & Mesh Analysis - 1 - Question 6

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

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 6

The two mesh equations are: 
5I1 - 3I2 = 10 
-3I1 + 7I2 = -15 
Solving the equations simultaneously, we get I1 = 0.96A and I2 = -1.73A.

Test: Basic Nodal & Mesh Analysis - 1 - Question 7

In the network shown in Figure find the value of V2 using nodal analysis

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 7

 

Test: Basic Nodal & Mesh Analysis - 1 - Question 8

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

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 8

To determine the number of Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) equations for the given circuit, we need to analyze its structure. The circuit is a Wheatstone bridge with a voltage source E E E and six resistors (R1 R_1 R1​ to R6 R_6 R6​).

Circuit Analysis:

  • Nodes: Identify the distinct nodes where currents split or combine. In a Wheatstone bridge:
    • There is one input node (top) connected to R1 R_1 R1​, R2 R_2 R2​, and R3 R_3 R3​.
    • One output node (bottom) connected to R4 R_4 R4​, R5 R_5 R5​, and R6 R_6 R6​.
    • Two middle nodes where R2 R_2 R2​ and R3 R_3 R3​ meet, and R5 R_5 R5​ and R6 R_6 R6​ meet.
    • Total nodes = 4 (assuming the positive and negative terminals of E E E are separate nodes).
  • Loops: A loop is a closed path. The circuit has:
    • One outer loop (through E E E, R1 R_1 R1​, R4 R_4 R4​, and back).
    • Two inner loops (one through R2 R_2 R2​, R6 R_6 R6​, R4 R_4 R4​, and the other through R3 R_3 R3​, R5 R_5 R5​, R4 R_4 R4​).
    • Total independent loops = 3 (using the formula for a planar graph with n n n nodes and b b b branches: L=b−n+1 L = b - n + 1 L=b−n+1, where b=6 b = 6 b=6 branches, n=4 n = 4 n=4 nodes, L=6−4+1=3 L = 6 - 4 + 1 = 3 L=6−4+1=3).

KCL Equations:

  • KCL applies at each node, stating that the sum of currents entering equals the sum leaving.
  • With 4 nodes, we can write KCL at each. However, one equation is dependent (e.g., the sum of currents at all nodes equals zero due to the circuit being a single connected component). Thus, the number of independent KCL equations is n−1=4−1=3 n - 1 = 4 - 1 = 3 n−1=4−1=3.

KVL Equations:

  • KVL applies around each independent loop, stating that the sum of voltage drops equals the source voltage.
  • With 3 independent loops, we need 3 KVL equations.

Total Equations:

  • Total number of independent equations = KCL + KVL = 3 (KCL) + 3 (KVL) = 6.

However, in circuit analysis, the number of KCL and KVL equations is typically reported as the count of each type needed to solve the circuit, considering independence. For a Wheatstone bridge with one voltage source, the standard approach confirms:

  • 3 independent KCL equations (at 3 nodes, excluding one dependent equation).
  • 3 independent KVL equations (for 3 loops).

Thus, the number of KVL and KCL equations is 3 each.

Test: Basic Nodal & Mesh Analysis - 1 - Question 9

The current l flowing in the given figure is

Test: Basic Nodal & Mesh Analysis - 1 - Question 10

The value of dependent source for the circuit shown below is

Detailed Solution for Test: Basic Nodal & Mesh Analysis - 1 - Question 10

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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