Basic Electrical Engineering MCQs Part 22

Contents

- 1 Two electric bulbs have tungsten filament of same length. If one of them gives 50 W and other one 100 W, then which of following is correct
- 2 A constant voltage is applied between two ends of uniform metallic wire. Some heat is developed. If both length and radius of wire are halved, heat developed in same duration is
- 3 Appliances based on heating effect of current work on
- 4 An electric kettle boils 1 kg of water in time t1 and another boils same amount of water in t2. When they are joined in series, the time t required to boil 1 kg of water
- 5 The current in 1000 W, 250 V heater operated at 250 V

## Two electric bulbs have tungsten filament of same length. If one of them gives 50 W and other one 100 W, then which of following is correct

Two electric bulbs have tungsten filament of same length. If one of them gives 50 W and other one 100 W, then which of following is correct:

- 50 W has thicker filament
- 100 W has thicker filament
- Both have filaments of same thickness
- Insufficient information

Correct answer: 2. 100 W has thicker filament

Solution:

From formula:

R = ρl/a

The filament of bulb having low resistance is thicked. Sice l is constant in both cases. The 100 W bulb has less resistance as compared to 50 W. Hence, 100 W has thicker filament.

## A constant voltage is applied between two ends of uniform metallic wire. Some heat is developed. If both length and radius of wire are halved, heat developed in same duration is

A constant voltage is applied between two ends of uniform metallic wire. Some heat is developed. If both length and radius of wire are halved, heat developed in same duration is:

- Twice
- Same
- Half
- One-fourth

Correct answer: 3. Half

## Appliances based on heating effect of current work on

Appliances based on heating effect of current work on:

- AC only
- DC only
- Both of these
- None of these

Correct answer: 3. Both of these

Explanation: The heating effect of current is independent of the direction of current. Therefore heating appliances can be used on AC as well as on DC.

## An electric kettle boils 1 kg of water in time t1 and another boils same amount of water in t2. When they are joined in series, the time t required to boil 1 kg of water

An electric kettle boils 1 kg of water in time t1 and another boils same amount of water in t2. When they are joined in series, the time t required to boil 1 kg of water:

- t1
- t2
- t1 + t2
- (t1*t2)/t1+t2

Correct answer: 3. t1 + t2

Explanation:

When heaters are connected in series, the total power P is mathematically written in the form:

P = P1P2/(P1 + p2)

If Q is heat required to boil 1 kg of water, then

Q = P1t1 = P2t2 = Pt

Q/t = {(Q/t1) * (Q/t2)} / {Q/t1 + Q/t2} = Q/(t1 + t2)

t = t1 + t2

## The current in 1000 W, 250 V heater operated at 250 V

The current in 1000 W, 250 V heater operated at 250 V:

- 1 A
- 2 A
- 3 A
- 4 A

Correct answer: 4. 4 A

Solution:

Resistance (R) = V²/P = 250²/1000 = 62.5 Ω

—————

Current (I) = Applied voltage/R = 250/62.5 = 4 A