# Basic Electrical Engineering MCQs Part 9

Basic Electrical Engineering MCQs Part 9

Contents

- 1 The time required for 2 kW heater for raising temperature of 10 litres of water through 10°C is
- 2 A heater coil rated at 1200 W, 220 V is connected to 100 V line. The power consumed is
- 3 A 8°C rise in temperature is observed in a conductor by passing current. When current is doubled, the rise in temperature will be
- 4 An electric heater, kettle, and oven are all rated at 1 kW, 230 V. Which of them has highest resistance
- 5 Certain tap supplies water at 20°C. A man takes 1 liter of water per minute at 39°C from geyser. The power of geyser is (Consider efficient to be 100%

## The time required for 2 kW heater for raising temperature of 10 litres of water through 10°C is

The time required for 2 kW heater for raising temperature of 10 litres of water through 10°C is:

- 185 s
- 210 s
- 420 s
- 520 s

Correct answer: 2. 210 s

Solution:

From equation

m * S * T = P * t

Where

m = mass in gram = 10 * 1000 = 10000 gram

—————-

T = Temperature difference = 10°C

S = Specific heat = Energy required in joules to raise temp of 1 gram of mass by 1 degree celsiuc.

In case of water S = 4.186 joules/gram/°C

P = Power of heater = 2 kW = 2000 watt = 2000 joules/sec

t = time is second

—————-

t = (m*S*t)/P = (10000 * 4.186 * 10)/1000 = 210 sec

## A heater coil rated at 1200 W, 220 V is connected to 100 V line. The power consumed is

A heater coil rated at 1200 W, 220 V is connected to 100 V line. The power consumed is:

- 139.22 W
- 247.9 W
- 315.4 W
- 448.9 W

Correct answer: 2. 247.9 W

Solution:

R = V²/P = 220²/1200 = 40.3 Ω

Power consumed at 100 V = (100)²/40.33 = 247.9 W

## A 8°C rise in temperature is observed in a conductor by passing current. When current is doubled, the rise in temperature will be

A 8°C rise in temperature is observed in a conductor by passing current. When current is doubled, the rise in temperature will be:

- 8°C
- 16°C
- 32°C
- 64°C

Correct answer: 3. 32°C

Solution:

From equation: H ∝ I²

———————-

H2/H1 = (2I/I)² = 4

———————-

Now θ ∝ H

θ2/θ1 = H2/H1 = 4

θ2 = 4θ1 = 4 * 8 = 32°C

## An electric heater, kettle, and oven are all rated at 1 kW, 230 V. Which of them has highest resistance

An electric heater, kettle, and oven are all rated at 1 kW, 230 V. Which of them has highest resistance:

- Heater
- Kettle
- Oven
- All have equal resistance

Correct answer: 4. All have equal resistance

Explanation: From formula R = V²/P, since all have same voltage and power ratings, all will have same resistance.

## Certain tap supplies water at 20°C. A man takes 1 liter of water per minute at 39°C from geyser. The power of geyser is (Consider efficient to be 100%

Certain tap supplies water at 20°C. A man takes 1 liter of water per minute at 39°C from geyser. The power of geyser is (Consider efficient to be 100%:

- 985 W
- 1330 W
- 1485 W
- 1965 W

Correct answer: 2. 1330 W

Solution:

Heat output of device in joules per second is the power of device.

Power of geyser = mc θ * 4.2 joules.sec = 1000/ * 1 * (39 – 20 ) = 1330 W