By BYJU'S Exam Prep

Updated on: September 25th, 2023

Electrical Questions and Answers and detailed solutions are available here. MCQ-based Electrical Questions and Answers will provide help in preparing for the upcoming GATE EE exam. We have provided 5 MCQ-based questions here.

Check out the detailed solution for the Electrical Questions and Answers, as it will help you clear blockage in solving similar types of questions. Objective-based MCQ and answers will help in the complete study of the electrical topics.

Table of content

Electrical Question 1

The number of kilowatts in 135 milliwatts is

1. 1.35 × 10–4 kW
2. 135 × 10–3 kW
3. 0.0135 kW
4. 0.00135 kW

Answer: A. 1.35 × 10–4 kW

Solution

The number of kilowatts in 135 milliwatts is

1 kW = 1000 watt

1 Watt = 103

>10-3

Kilowatt

1 milliwatt = 103

>10-3

watt

135 milliwatt = 135×103

>×10-3

watt

= 135×106

>×10-6

Kilowatt

= 1.35×104

>×10-4

Kilowatt

Electrical Question 2

An electric heater draws 3.5 A from a 110 V source. The resistance of the heating element is approximately-

1. 385 Ω
2. 38.5 Ω
3. 3.1 Ω
4. 31 Ω

Solution

An electric heater is an example of a resistive load. We can apply Ohms law to any passive element.

According to Ohms law,

V=I.R

>V = I.R

Current drawn from the supply

I=3.5 A

>I = 3.5 A

and V=110 V

>V = 110 V

R=VI

>R = V/I

R=1103.5=31.428 Ω31Ω

>R = 110/3.5 = 31.428 Ω ≡ 31Ω

Electrical Question 3

A certain series circuit consists of a 1/8 W resistor, a 1/4 W resistor, and a 1/2 W resistor. The total resistance is 1200 Ω

. If each resistor is operating in the circuit at its maximum power dissipation, the total current flow is

1. 27 mA
2. 2.7 mA
3. 19 mA
4. 190 mA

Solution

Total power consumption in the circuit is the sum of all the elements of power consumption.

Total power consumption =18+14+12=0.875 Watt

>= 1/8+1/4+1/2=0.875 Watt

The total resistance given in the circuit is 1200 Ω

Power consumption

P=V2R

>P = V2/R

0.875×1200=V2

>×1200 = V2

V2=1050

>V= 1050

V=1050=32.4 Volt

>V = √1050=32.4 Volt

Total current= Total voltage in the circuitTotal resistance in the circuit

>Total current= Total voltage in the circuit/Total resistance in the circuit

I=32.41200=0.027 A=27 mA

>I = 32.4/1200 = 0.027 A = 27 mA

Electrical Question 4

A series circuit consists of a 4.7 kΩ

, a 12 kΩ

, and a 2.2 kΩ

>Ω

resistor. The resistor that has the most voltage drop is

1. 12 kΩ
2. 2.2 kΩ
3. 4.7 kΩ
4. Impossible to determine from the given information

Solution

A series circuit consists of a 4.7 kΩ

, a 12 kΩ

, and a 2.2 kΩ resistor. The resistor that has the most voltage drop is dependent on two factors.

From Ohms law

V=IR

The voltage drop is dependent on two factors in a series circuit

• Current flowing in the element
• Resistance value

But in a series circuit, all the elements have the same current, so the maximum voltage drop occurs at, the resistance which is having more value among all.

∴ 12 KΩ resistance has highest voltage drop.

Electrical Question 5

The total resistance of a circuit is 680Ω. The percentage of the total voltage appearing across a 47 Ω resistor that makes up part of the total series resistance is

1. 68%
2. 47%
3. 69%
4. 6.91%

Solution

Total voltage = V

The total resistance in series combination is R+47 = 680

R = 680-47 = 633 Ω

From Ohms law

V=IR

The two-element series circuit is having the same current i.e., total current.

The total current I=V680

>I=V/680

The voltage drops across 47 Ω is

V47=V680×47

>V47 = V/680×47

The percentage of the total voltage appearing across a 47 Ω

V47V×100=47680×100=6.91%

>V47/V×100=47/680×100=6.91%

☛ Related Questions:

POPULAR EXAMS
SSC and Bank
Other Exams
GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com