Multiple choice questions on Electrical, Power, Energy and Work

## The kilowatt-hour is the unit of

The kilowatt-hour is the unit of

- Charge
- Current
- Energy
- 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:

- 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

Where

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

l = length, measured in meters (m)

a = cross-sectional area, in square meters (m^{2})

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:

- 4:3
- 6:25
- 3:5
- 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:

- Less than fan
- More than fan
- Same as fan
- 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:

- The heater output is smaller
- The heater output is larger
- The heater output remains the same
- 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:

- R1 = 2R2
- R2 = 2R1
- R1 =4R2
- 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:

- 5000
- 10000
- 12500
- 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:

- 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

## 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:

- 1200000 s
- 2500000 s
- 5000000 s
- 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:

- 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.

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