Basic Electrical Engineering MCQs Part 8

Contents

- 1 Heat produced in conductor depends
- 2 The energy consumed in kWh when 10, 20 W lamps are operated for 10 hours per day in November is
- 3 A resistor coil of 120 Ω is immersed in 42 kg of water. If a steady current of 7 A is made to flow through the coil, the rise temperature of water per minute is
- 4 Certain motor when operated for 45 minutes consumes 0.8 kWh of energy. During that time, the total energy loss is 300000 J. The total joules of work performed in 45 minutes are
- 5 Two electric lamps, each of 50 W are connected in parallel. The power consumed by combination is

## Heat produced in conductor depends

Heat produced in conductor depends:

- Directly on time
- Inversely on time
- Both of these
- None of these

Correct answer: 1. Directly on time

## The energy consumed in kWh when 10, 20 W lamps are operated for 10 hours per day in November is

The energy consumed in kWh when 10, 20 W lamps are operated for 10 hours per day in November is:

- 15 kWh
- 30 kWh
- 60 KWh
- 120 kWh

Correct answer: 3. 60 KWh

Solution:

Watts consumed by 10 lamps = 10 * 20 = 200 W

When operated for 10 hours = 2000 wh

For 30 days of November = 2000 * 30 = 60000 Wh = 60 kWh

## A resistor coil of 120 Ω is immersed in 42 kg of water. If a steady current of 7 A is made to flow through the coil, the rise temperature of water per minute is

A resistor coil of 120 Ω is immersed in 42 kg of water. If a steady current of 7 A is made to flow through the coil, the rise temperature of water per minute is:

- 1°C
- 2°C
- 3°C
- 4°C

Correct answer: 2. 2°C

Solution:

Heat produced per minute (H) = I² Rt/J = {(7)² * 120 * 60 = 84000 cal

From formula

H = ms θ or θ = H/ms = 84000/(42 * 1000 * 1) = 2°C

## Certain motor when operated for 45 minutes consumes 0.8 kWh of energy. During that time, the total energy loss is 300000 J. The total joules of work performed in 45 minutes are

Certain motor when operated for 45 minutes consumes 0.8 kWh of energy. During that time, the total energy loss is 300000 J. The total joules of work performed in 45 minutes are:

- 1130000 J
- 1920000 J
- 2580000 J
- 3220000 J

Correct answer: 3. 2580000 J

Solution:

Motor input = 0.8 kWh = 0.8 * 36 * 10^5 J = 2880000 J

———–

Motor output = Input – Loss = 2880000 – 300000 = 2580000 J

## Two electric lamps, each of 50 W are connected in parallel. The power consumed by combination is

Two electric lamps, each of 50 W are connected in parallel. The power consumed by combination is:

- 25 W
- 50 W
- 100 W
- None of these

Correct answer: 3. 100 W

Explanation:

PT = P1 + P2 = 50 W + 50 W = 100 W