Basic Electrical Engineering MCQs Part 10

Contents

- 1 The heat produced (in k cals) in resistance R ohms when a current I amperes flow though it for t seconds is given be
- 2 An electric fan and electric kettle are marked as 200 W, 220 V and 1000 W, 220 V The resistance of kettle is
- 3 The total power supplied to an engine that drives an electric generator is 50 kW. The generator delivers 15 A to load of 100 Ω. The efficiency of system is
- 4 The utility company supplies power at 230 V. In a house having 11 bulbs of power rating 100 W the fuse rating should be:
- 5 A constant voltage is applied across ends of conductor. The heat produced is

## The heat produced (in k cals) in resistance R ohms when a current I amperes flow though it for t seconds is given be

The heat produced (in k cals) in resistance R ohms when a current I amperes flow though it for t seconds is given be:

- I²Rt
- I²Rt/5
- I²Rt * (4.2*10^3)
- I²Rt/(4.2*10^3)

Correct answer: 4. I²Rt/(4.2*10^3)

## An electric fan and electric kettle are marked as 200 W, 220 V and 1000 W, 220 V The resistance of kettle is

An electric fan and electric kettle are marked as 200 W, 220 V and 1000 W, 220 V The resistance of kettle is:

- Less than fan
- More than fan
- Same as fan
- Insufficient data

Correct answer: 1. Less than fan

Solution:

From formula R = V²/P

Resistance of fan = 220^2/200 = 242 ohms

Resistance of kettle = 220^2/1000 = 48.4 ohms

## The total power supplied to an engine that drives an electric generator is 50 kW. The generator delivers 15 A to load of 100 Ω. The efficiency of system is

The total power supplied to an engine that drives an electric generator is 50 kW. The generator delivers 15 A to load of 100 Ω. The efficiency of system is:

- 0.2
- 0.38
- 0.45
- 0.92

Correct answer: 3. 0.45

Solution:

Input to system = 50 kW = 50000 W

Output of system = I²R = 15² * 100 = 22500 W

——————–

System efficiency (η) = Output/Input = 22500/50000 = 0.45

## The utility company supplies power at 230 V. In a house having 11 bulbs of power rating 100 W the fuse rating should be:

The utility company supplies power at 230 V. In a house having 11 bulbs of power rating 100 W the fuse rating should be:

- 2.22 A
- 4.78 A
- 5.69 A
- 7.22 A

Correct answer: 2. 4.78 A

Solution:

P = VI

or I = P/V = 11 * 100/230 = 4.78 A

## A constant voltage is applied across ends of conductor. The heat produced is

A constant voltage is applied across ends of conductor. The heat produced is:

- Directly proportional to length
- Inversely proportional to length
- Directly proportional to square of length
- None of these

Correct answer: 2. inversely proportional to length