Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

# Electrical Power, Energy, and Work MCQs.net

Multiple choice questions on Electrical, Power, Energy and Work

## The kilowatt-hour is the unit of

The kilowatt-hour is the unit of

1. Charge
2. Current
3. Energy
4. Power

Correct answer: 3. Energy

Explanation: kWh energy is the type of energy that is used by the electrical lights and other appliances in our homes and industries. For example, if you run a 100-W LED for 1 hour, the energy consumed is 100 watt-hour. (That is W = Pt = 100 W X 1 h = 100 Wh.

Whereas if you run the same 100-W LED for 12 hours, the energy consumed is 1200 Wh. Similarly, if you run an air conditioner having 2000-watt power consumption for 20 hours, the watt-hours will be 4000 Wh. Since watthour is a small unit of energy, kWh is the practical unit that is in practice.

kWh or kilowatt-hour is the unit that is used on utility bills.

## 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 statements 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:

1. 50 W has thicker filament
2. 100 W has thicker filament
3. Both have filaments of same thickness
4. Insufficient information

Correct answer: 2. 100 W has thicker filament

Solution:

From formula: R = ρl/a

Where

R = resistivity, measured in ohm-meters (Ω-m)

l = length, measured in meters (m)

a = cross-sectional area, in square meters (m2)

Since length l is constant in both cases and material of construction is also same, the filament of bulb having low resistance is thicker. The 100 W bulb has less resistance as compared to 50 W.  Hence, 100 W has thicker filament.

## Two bulbs 500 W and 300 W are manufactured to operate on 220 V line. The ratio of resistance of 500 W bulb to 300 W bulb is

Two bulbs 500 W and 300 W are manufactured to operate on 220 V line. The ratio of resistance of 500 W bulb to 300 W bulb is:

1. 4:3
2. 6:25
3. 3:5
4. 18:9

Correct answer: 3. 3:5

Solution:

From formula R = V²/P

Bulb1 R = 96.8 ohms

Bulb2 R = 161.33 ohms

———————

Bulb 1 R/Bulb 2 = 0.6 = 3:5

## 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:

1. Less than fan
2. More than fan
3. Same as fan
4. 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

## A 50 W tube light is connected in parallel to a room heater on main line. When happens to heater when light is switched off:

A 50 W tube light is connected in parallel to a room heater on main line. When happens to heater when light is switched off:

1. The heater output is smaller
2. The heater output is larger
3. The heater output remains the same
4. None of these

Correct answer: 3. The heater output remains the same

## If R1 and R2 are respectively the filament resistances of a 200 W and 100 W bulb designed to operate at same voltage then which of following mathematical equation is correct

If R1 and R2 are respectively the filament resistances of a 200 W and 100 W bulb designed to operate at same voltage then which of following mathematical equation is correct:

1. R1 = 2R2
2. R2 = 2R1
3. R1 =4R2
4. R2 = 4R1

Correct answer: 2. R2 = 2R1

Explanation:

Let’s consider voltage = 100 V.

For R1; R1 = V²/P = 100/200 = 0.5 ohms

For R2; R2 = V²/P = 100/100 = 1 ohm From above R2 = 2R1. Hence 2. Is correct answer

## The power rating of motor drawing a current of 50 A at 400 V is

The power rating of motor drawing a current of 50 A at 400 V is:

1. 5000
2. 10000
3. 12500
4. 20000

Correct answer: 4. 20000

Explanation:

From formula

P = VI = 400 * 50 = 20000 W

## 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:

1. 80 W
2. 200 W
3. Both will fuse
4. 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

## How long will it take for a 1 MΩ resistor to dissipate 1000 joules of heat if current in resistance is 20 µA

How long will it take for a 1 MΩ resistor to dissipate 1000 joules of heat if current in resistance is 20 µA:

1. 1200000 s
2. 2500000 s
3. 5000000 s
4. 10000000 s

Correct answer: 2. 2500000 s

Solution:

From formula: P = I²Rt

t = P/I²R = 1000/{(20*10^-6)² * (1 * 10^6) = 2500000 s

## Appliances based on heating effect of current work on

Appliances based on heating effect of current work on:

1. AC only
2. DC only
3. Both of these
4. 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.