Basic Electrical Engineering MCQs Part 11
Contents
- 1 Two identical heater coils are connected in parallel across mains. If one of coil breaks the other
- 2 What maximum voltage can be provided across a 1 W, 225 Ω resistor without exceeding rated power dissipation
- 3 Two electric lamps of 40 W, 220 V each are connected in series across 220 V source. The power consumed by the combination of two lamps is:
- 4 Lamps used for lighting homes are connected in
- 5 Two lamps of 80 W and 200 W rated for 220 V are connected in series and a 440 volts are applied across them. Which lamp will fuse
Two identical heater coils are connected in parallel across mains. If one of coil breaks the other
Two identical heater coils are connected in parallel across mains. If one of coil breaks the other:
- Develops lower temperature
- Develops higher temperature
- Develops same temperature as before
- None of these
Correct answer: 3. Develops same temperature as before
What maximum voltage can be provided across a 1 W, 225 Ω resistor without exceeding rated power dissipation
What maximum voltage can be provided across a 1 W, 225 Ω resistor without exceeding rated power dissipation:
- 5 W
- 10 W
- 15 W
- 20 W
Correct answer: 3. 15 W
Solution:
V = √P * R = √(1 * 225) = 15 W
Two electric lamps of 40 W, 220 V each are connected in series across 220 V source. The power consumed by the combination of two lamps is:
Two electric lamps of 40 W, 220 V each are connected in series across 220 V source. The power consumed by the combination of two lamps is:
- 10 W
- 20 W
- 40 W
- 80 W
Correct answer: 2. 20 W
Explanation:
PT = (P1 * P2)/(P1 + P2) = 20 W
Lamps used for lighting homes are connected in
Lamps used for lighting homes are connected in:
- Series configuration
- Parallel configuration
- Series-parallel configuration
- None of above
Correct answer: 2. Parallel configuration
Two lamps of 80 W and 200 W rated for 220 V are connected in series and a 440 volts are applied across them. Which lamp will fuse
Two lamps of 80 W and 200 W rated for 220 V are connected in series and a 440 volts are applied across them. Which lamp will fuse:
- 80 W
- 200 W
- Both will fuse
- Both will work fine
Correct answer: 1. 80 W
Resistance of 80 Watt lamp for rated voltage = V²/P1 = 220²/80 = 605 Ω
Resistance of 200 Watt lamp for rated voltage = V²/P1 = 220²/200 = 242 Ω
———————
Total circuit resistance (Rt) = R1 + R2 = 605 Ω + 242 Ω= 847 Ω
———————
Circuit current (I) = 440 V/847 Ω = 0.519A
———————
Voltage across 80 W lamp = IR1 = 0.519 * 605 Ω = 314 V
Voltage across 200 W lamp = IR1 = * 242 Ω = 126 V
———————
Since voltage across 80 W lamp is more than rated voltage, the 80 W lamp will fuse.