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# Test: Basic Laws- 2

## 15 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: Basic Laws- 2

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This mock test of Test: Basic Laws- 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 15 Multiple Choice Questions for Electrical Engineering (EE) Test: Basic Laws- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Basic Laws- 2 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Basic Laws- 2 exercise for a better result in the exam. You can find other Test: Basic Laws- 2 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
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 (V1+ V2) is Solution:

Using KVL in the loop, we have
V= 10 + 15 = 25 volts
Also, V2 = -25 volts
∴ V1 + V2 = 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 R1 and R2, the voltage drop V1 is twice the drop V2 across R2. The value of R1, and R2 will be Solution:

Using Ohm’s law, Given, V1 = 2V2
As I is same therefore,
R1 = 2R2
∴ RT = R1 + R2 = 2R2 + R2 = 3R2
So, RT = 3R2 = 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
RΔ = 3 RY
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 I1, I2, l3 and I4 is Solution:

Using KCL,
I2 = 6 - 2 = 4 A
Also, or, and I4 = I- I3 = 4 - (-1.5) = 5.5 A
Also, I1 = 6 A
∴ I1 + I2 + I3 + I4 = 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, I25Ω=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: