Work, Energy, and Power MCQs

Work, Energy, and Power MCQs.

The SI unit of electrical energy is

The SI unit of electrical energy is:

1. Watts (W)
2. Kilowatts (kW)
3. Kilowatt-hour (kWh)
4. Calories

Correct answer: 3. Kilowatt-hour (kWh)

Explanation: The unit of electrical energy is the kilowatt-hour (kWh). The unit finds its application particularly for utility bills. The utility bills coming to our homes use this unit.

One kilowatt-hour equals 3.6 megajoule.

1 kWh = 3.6 MJ

The SI unit of electrical power is

The SI unit of electrical power is:

1. Watts (W)
2. Kilowatt-hour (kWh)
3. Calories
4. Joules

Correct answer: 3. Watts (W)

Explanation: The SI unit of Electrical power is Watts. The name Watts is named to honor James Watt who was the inventor of the steam engine.

One Watt of electrical power equals one ampere under the pressure of one volt

The watt is a small amount of power. Small devices require only a few Watts to operate, however large devices required larger amounts of power. Since watts is too small for some devices, the power consumption of larger devices is measured in kilowatts (kW).

Where 1 kW = 1000 watts

Also Read: Electrical Power Generation

The SI unit of work is

The SI unit of work is:

1. Watts (W)
2. Kilowatts (kW)
3. Joule
4. All of these

Correct answer: 3. Joule

Explanation: The joule is the SI unit of energy. It is named after the English physicist James Prescott Joule. One joule can be defined as The work required to move an electric charge of one coulomb through an electrical potential difference of one volt, or one coulomb-volt (C⋅V).

Also see: Electricity consumption Sheet for Home Appliances

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

A 50 W bulb will give heat and light emery of

A 50 W bulb will give heat and light emery of:

1. 50 J/s
2. 100 J/s
3. 200 J/s
4. None of these

Correct answer: 1. 50 J/s

Also Read: Top 5 Properties associated with AC Waveform

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:

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 100 W, 250 V and 200 W, 250 V are connected in series to a 500 source. Which bulb is likely to fuse

Two bulbs 100 W, 250 V and 200 W, 250 V are connected in series to a 500 source. Which bulb is likely to fuse:

1. 100 W bulb
2. 200 W bulb
3. Both of these
4. None of these

Correct answer: 1. 100 W bulb

Solution:

From formula R = V²/P

Bulb1 R = 625 ohms

Bulb2 R = 312.5 ohms

———————

Voltage across Bulb 1 = 333 V

Voltage across Bulb 2 = 167 V

Since voltage across bulb 1 exceeds rated voltage, it will be fused.

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

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.

The power dissipated in a 1 kΩ resistor having voltage drop of 12 is

The power dissipated in a 1 kΩ resistor having voltage drop of 12 is:

1. 0.05 W
2. 0.144 W
3. 0.5 W
4. 1.2 W

Correct answer: 2. 0.144 W

Solution:

From formula P = V²/R = 12 V²/1 kΩ = 0.144 W

More MCQs on Electrical Engineering:

1. Basic Electrical Laws and Theorems
2. Basic Voltage and Current MCQs
3. High Voltage Engineering MCQs
4. Induction Motor MCQs
5. Power Electronics
6. Power Distribution MCQs
7. Power Generation MCQs
8. Power System Protection MCQs
9. Synchronous Generator MCQs
10. Synchronous motor MCQs
11. Transformer MCQs