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# Electrical Circuits 2 MCQ

## 20 Questions MCQ Test Mock Test Series for SSC JE Electrical Engineering (Hindi) | Electrical Circuits 2 MCQ

Description
This mock test of Electrical Circuits 2 MCQ for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Electrical Circuits 2 MCQ (mcq) to study with solutions a complete question bank. The solved questions answers in this Electrical Circuits 2 MCQ quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Electrical Circuits 2 MCQ exercise for a better result in the exam. You can find other Electrical Circuits 2 MCQ extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### If the energy is supplied form a source, whose resistance is 1 Ohm, to a load of 100 Ohms the source will be

Solution:

The internal resistance of an ideal voltage source should be zero and practically it is very low.

Given that, source resistance is 1 ohm

Hence the source will be a voltage source

QUESTION: 2

### A lamp rated at 60V, 40W is to be connected across 230V. What is the value of resistance to be connected in series with lamp?

Solution:

Rating of lamp = 60V, 40 W When lamp is connected, maximum current that can be flow through the circuit is Let RS be the series resistance. QUESTION: 3

### The sum of currents entering a junction is 9 A. If the current leaves the junction form 3 different paths having the same resistance, the current leaving from any one of the path will be:

Solution:

By KCL,

The sum of current entering a junction is equal to the sum of current leaving at that junction.

⇒ Sum of currents leaving = 9 A

This 9 A current will flow through three resistances having same value.

The current will be some through each resistor QUESTION: 4

A voltage source and two resistors are connected in parallel. Suppose that Vs = 150V, R= 50Ω and R2 = 25Ω. Then each resistance contain current?

Solution:  QUESTION: 5

A current source and a voltage source are connected in series with a resistor as shown below. Suppose that Vs = 10V, is = 3A and R = 5Ω. What is the voltage across the resistor and the power absorbed by the resistor? Solution: QUESTION: 6

The voltage source in the given circuit supplies 24W of power. The current source supplies 6.0 W. Determine the value of the resistance R and R2: Solution: Given that, voltage source supplies 24 W By applying KVL,

- 24 + I1 R1 + I2 R2 = 0

⇒ I2 R2 = 24 – R1 → (1)

Given that, current source supplies 6 W By KVL in second loop

I2 R2 = 6 → (2)

From (1) and (2)

⇒ R1 = 18 Ω

By KCL at point ‘O’

⇒ I2 = 1 + I1 = 2 A

We know that, I2 R2 = 6

⇒ R2 = 3 Ω

QUESTION: 7

The Thevenin equivalent resistance (RTh) for the circuit shown in fig. is: Solution:  By KCL at point ‘O’

I1 = I2 + 2 = 0.5 + 2 = 2.5 QUESTION: 8

Applying Norton’s Theorem, the Norton’s equivalent circuit to the left of the terminals “a” and “b” in the below circuit is having equivalent current source (IN) and equivalent resistance (RN) Solution:

To find Norton’s resistance, we need to short circuit the voltage source and open circuit the current source RN = (4 + 4) || 8 = 4 Ω

To find the Norton’s current, we need to find short circuit current across terminals a-b   QUESTION: 9

Find the value of Req in the figure given below Solution: By converting Delta (ABC) in to star   Req = 18 || (14 + 4) = 9 Ω

QUESTION: 10

Consider the following circuit and find the current through the 6Ω resistor. Solution:

Let the current passing through 6 Ω resistor is I.

By source transformation,  By source transformation,  By current division, QUESTION: 11

The Norton current at terminals a and b of the circuit shown at Figure: Solution:

To find Norton’s current, we need to find the short circuit current across the terminals a and b.  By source transformation, By KVL,

-180 + 20 ISC + 120 +40 ISC = 0

⇒ ISC = 1 A

QUESTION: 12

Given condition justifies which network theorems- the load impedance should be complex conjugate of the internal impedance of the active network.

Solution:

According to maximum power transfer theorem, in AC circuits the load impedance should be complex conjugate of the internal impedance of the active network QUESTION: 13

Three capacitors of 10μF, 20μF and 40μF are connected in parallel across 100 V. The total charge stored in capacitors is:

Solution: Ceq = 10 + 20 + 40 = 70 μF

Charge, Q = CV = (70 μ F) (100 V) = 7 × 10-3 C

QUESTION: 14

An ac voltage is described by v(t) = 10 cos (400 πt), find out frequency and RMS value of voltage.

Solution:

V(t) = 10 cos (400 π t)

ω = 400 π

⇒ 2 π f = 400 π

⇒ f = 200 Hz

Vm = 10 V QUESTION: 15

An inductive coil of 0.2 H is connected to 200 V, 50 Hz source. What are the inductive reactance and RMS current in the circuit?

Solution:

Given that, inductance = 0.2 H

Voltage = 200 V

Inductive reactance = 2 π fL = 2 π × 50 × 0.2 = 62.8 Ω QUESTION: 16

A series resonant circuit has R = 2Ω, L = 1mH and C = 0.1 μF, the value of quality factor Q is:

Solution:

Given that, R = 2 Ω, L = 1 mH, C = 0.1 μ F

For series resonant circuit, QUESTION: 17

A ramp voltage V(t) = 100V is applied to an RC differentiating circuit with R = 5kΩ and C = 4μF. The maximum output voltage is

Solution:     Maximum value of V(t) = 2 V

QUESTION: 18

In an a.c. circuit, v = 50sin(ωt+60∘),i = 10cos(ωt)then the power factor of a.c. circuit will be:

Solution:

Given that, V = 50 sin (ω t + 60°)

i = 10 cos (ω t)

i = 10 cos (ω t)

i = 10 sin (ω t + 90°)

Phase difference (ϕ) = 30°  QUESTION: 19

A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.

Solution:

Z = 8 +j 6  = 4229.11 W

QUESTION: 20

What will be the bandwidth of a series resonant circuit provided it has an inductive reactance of 1000 Ohm, a capacitive reactance of 1000 Ohm a resistance of 0.1 Ohm? It also know that the resonant frequency is 10MHz.

Solution:

Given that, XL = 1000 Ω

XC = 1000 Ω

R = 0.1 Ω

F0 = 10 MHz  