Basic Electrical Engineering MCQs Part 21

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Basic Electrical Engineering MCQs Part 21

Three electric bulbs 30 W, 60 W, and 100 W are designed to work on 230 V mains. Which bulb will burn more brightly if they are connected in series across 220 V mains

Three electric bulbs 30 W, 60 W, and 100 W are designed to work on 230 V mains. Which bulb will burn more brightly if they are connected in series across 220 V mains:

  1. 30 W
  2. 60 W
  3. 100 W
  4. All will burn with equal brightness

Correct answer: 1. 30 W

Explanation:
Formula for resistance of bulb: R = V²/P

Since voltage is same, the 30 W bulb will have more resistance than others. Further to this, the bulbs are in series configuration, same current flows through them.

It is obvious that voltage drop across 40 V bulb will be greater as compared to other bulbs. Therefore 30 W will burn more brightly.

The power of heater is 600 W at 800°C. What will be its power at 200°C. The temperature coefficient of resistance is 0.0004/°C

The power of heater is 600 W at 800°C. What will be its power at 200°C. The temperature coefficient of resistance is 0.0004/°C.

  1. 322 W
  2. 359 W
  3. 744 W
  4. 691 W

Correct answer: 3. 744 W

Solution:

P200 = P800 [ 1 + α ( θ2 –  θ1)] = 600 [ 1 + 0.0004 * 600] = 744 W

A home is served by 230 V line. In a circuit protected by fuse of 10 A. The maximum number of 50 W lamps in parallel that can be turned on is:

A home is served by 230 V line. In a circuit protected by fuse of 10 A. The maximum number of 50 W lamps in parallel that can be turned on is:

  1. 19
  2. 46
  3. 52
  4. 69

Correct answer: 2. 46

Solution:

Consider n to be the required number of 50 W lamps connected in parallel.

Total power (P) = 50 n

——————-

From formula P = VI

50 * n = 230 * 10

n = (230 * 10)/50 = 46

A generator generates 200 kW of power at potential difference of 5 kV. The power transmission is done through cables of total resistance 5 Ω. The power loss in cables is

A generator generates 200 kW of power at potential difference of 5 kV. The power transmission is done through cables of total resistance 5 Ω. The power loss in cables is:

  1. 2000 W
  2. 4000 W
  3. 8000 W
  4. 16000 W

Correct answer: 3. 8000 W

Solution:

From formula I = P/V

200 kW/5 kV = 40 A

——————-

Power loss in cables = I²R = 40² * 5 = 8000 W

What is maximum safe current flow in 33 Ω,2 W resistor

What is maximum safe current flow in 33 Ω,2 W resistor.

  1. 0.125 A
  2. 0.25 A
  3. 0.5 A
  4. 0.75 A

Correct answer: 2. 0.25 A

Solution:

I = √P/R = √(2/33) = 0.25 A

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