Basic Electrical Engineering MCQs Part 12
Contents
- 1 One kilowatt hour is equal to
- 2 Two bulbs 500 W and 300 W are manufactured to operate on 220 V line. The ratio of resistance of 500 W bulb to 300 W bulb is
- 3 Two coils connected in parallel across 200 V supply take 10 A from line. The power dissipated in first coil is 1000 W. The resistance of second coil is:
- 4 An electric motor operating at 120 V DC draws current of 12 A. The resistance of winding winding of motor if its efficiency is 30% is
- 5 The power dissipated in a 1 kΩ resistor having voltage drop of 12 is
One kilowatt hour is equal to
One kilowatt hour is equal to
- 36 * 10^2 J
- 36 * 10^3 J
- 36 * 10^4 J
- 36 * 10^5 J
Correct answer: 4. 36 * 10^5 J
Two bulbs 500 W and 300 W are manufactured to operate on 220 V line. The ratio of resistance of 500 W bulb to 300 W bulb is
Two bulbs 500 W and 300 W are manufactured to operate on 220 V line. The ratio of resistance of 500 W bulb to 300 W bulb is:
- 4:3
- 6:25
- 3:5
- 18:9
Correct answer: 3. 3:5
Solution:
From formula R = V²/P
Bulb1 R = 96.8 ohms
Bulb2 R = 161.33 ohms
———————
Bulb 1 R/Bulb 2 = 0.6 = 3:5
Two coils connected in parallel across 200 V supply take 10 A from line. The power dissipated in first coil is 1000 W. The resistance of second coil is:
Two coils connected in parallel across 200 V supply take 10 A from line. The power dissipated in first coil is 1000 W. The resistance of second coil is:
- 10 Ω
- 20 Ω
- 30 Ω
- 40 Ω
Correct answer: 4. 40 Ω
Solution:
From statement P = VI, current across 1st coil
I(coil 1) = P/V = 1000/200 = 5 A
——————
From KCL, current across coil 2
I(coil 2) = Incoming current – I(coil 1) = 10 A – 5 A = 5 A
——————
From Ohm’s law
R(coil 2) = V/I(coil 2) = 200/5 = 40 Ω
An electric motor operating at 120 V DC draws current of 12 A. The resistance of winding winding of motor if its efficiency is 30% is
An electric motor operating at 120 V DC draws current of 12 A. The resistance of winding winding of motor if its efficiency is 30% is:
- 3 Ω
- 5 Ω
- 7 Ω
- 12 Ω
Correct answer: 3. 7 Ω
Solution:
Input power = 120 * 12 = 1440 W
—————–
Power loss = 0.7 * 1440 W = 1008 W
—————–
P = I²R or R = P/I² = 1008/12² = 7 Ω
The power dissipated in a 1 kΩ resistor having voltage drop of 12 is
The power dissipated in a 1 kΩ resistor having voltage drop of 12 is:
- 0.05 W
- 0.144 W
- 0.5 W
- 1.2 W
Correct answer: 2. 0.144 W
Solution:
From formula P = V²/R = 12 V²/1 kΩ = 0.144 W