Electric Power Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Electric power is a _______.

1. Scalar quantity
2. Vector quantity
3. Dimensionless quantity
4. None of the above

Ans. The correct answer is a. Scalar quantity

Explanation: Since electric power only has magnitude but no direction, it is a scalar quantity.

Ques. Power is given by the formula ______.

1. P = E + t
2. P = Et
3. P = E - t
4. P = E/t

Ans. The correct answer is d. P = E/t

Explanation: Electric power consumed by a device is given by the energy consumed per unit of time.

Ques. The energy consumed by an electric appliance of power 1 kW in 1 hr is

1. 36 × 10³ J
2. 360 J
3. 36 × 10⁶ J
4. 3.6 × 10⁶ J

Ans. The correct answer is d. 3.6 × 10⁶ J

Explanation: Energy, E = Pt

⇒ E = 1kW x 1 hour = 1000 W x 3600 seconds = 3.6 x 10⁶ J

Ques. Which of the following is not a unit of electric power?

1. Joule/sec
2. Volt-ampere
3. Joule-ampere
4. Watt

Ans. The correct answer is c. Joule-ampere

Explanation: Joule-ampere is not a unit of electric power. The SI unit of electric power is Watt (W) and other units are Joule/sec and Volt-ampere.

Ques. A Windmill has to generate……..power so the potential difference maintained by it can be 220 V and should able to produce a current of 2 A per unit of time.

1. 0.40 kW
2. 0.44 kW
3. 0.32 kW
4. 0.36 kW

Ans. The correct answer is a. 0.40 kW

Explanation: The formula of electric power is given by

P = VI

⇒ P = 220 x 2 = 440 W = 0.40 kW 

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Short Answers Questions [2 Marks Questions]

Ques. Define electric power.

Ans. The rate at which energy is transformed into an electrical circuit or used to produce work is known as electric power. It is a way to measure how much energy is consumed over a certain period of time.

The SI unit of electric power is watt (W).

Ques. List five applications of electric power.

Ans. The applications of electric power are

• Cooking
• Heating
• Healthcare
• Lighting
• Entertainment

Ques. Define one Watt.

Ans. Watt is the SI unit of electric power and it is given by

1 watt = 1 joule / 1 second

One watt is defined as the 1 joule of electrical energy consumed by the electrical appliance in one second.

Ques. Name other units of power.

Ans. The other units of electric power are

• Horsepower
• Metric horsepower
• Foot-pounds per minute
• Ergs per second.

Also Read:

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Long Answers Questions [3 Marks Questions]

Ques. The maximum power rating of a 30 Ω resistor is 2 kW. Would you connect the resistor directly across a 300 V DC source of negligible internal resistance?

Ans. The maximum power rating of the resistor tells that the given resistor is not supposed to handle power more than its rating.

Power to be handled by the resistance, P = V²/R = 300²/30 = 3 kW

Given that the maximum power rating of the resistor is 2 kW. So it should not be connected directly across a 300 V DC source of negligible internal resistance otherwise it will be damaged/burnt off.

Ques. Which of the two has higher resistance: a 1000 W heater or a 100 W tungsten bulb, both marked for 230 V?

Ans. We know power consumed by an electrical appliance is given by

P = V²/R

Where

• V is the operating voltage
• R is the resistance of the electrical appliance

From the above equation, we have

R = V²/P

Resistance of the heater, R_H = 230²/1000 = 52.9 Ω

Resistance of the bulb, R_B = 230²/100 = 529 Ω

Therefore the tungsten bulb has a higher resistance.

Ques. To produce 10³ Joule of heat in 10 seconds, how much voltage should be applied to 100-ohm resistance?

Ans. Given

• The heat energy to be produced by the resistance, H = 10³ J
• Time, t = 10 seconds
• Resistance, R = 100 Ω

Heat energy, H = V²t/R

⇒ V² = HR/t = (10³ x 100)/10 = 10⁴

⇒ V = 100 V

Hence the 100 V voltage should be applied to 100 ohm resistance to produce 10³ Joule of heat in 10 seconds.

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Very Long Answers Questions [5 Marks Questions]

Ques. Two room heaters are marked 220 V, 500 W and 220 V, 800 W. If the heaters are connected in series and the combination is connected to 220 V DC supply. Out of these two heaters which one will produce more heat?

Ans. Resistance of the first heater, R₁ = V²/P = 220²/500 = 96.8 Ω

Resistance of the second heater, R₂ = V²/P = 220²/800 = 60.5 Ω

When both the resistances are connected in series, the total resistance becomes

R = R₁ + R₂ = 96.8 + 60.5 = 157.3 Ω

Current flowing through the heaters, I = V/R = 220/157.3 = 1.4 A

Heat produced per second by the first heater

P₁ = I₂R₁ = 1.42 x 96.8 = 189.73 Joule s^-1

Heat produced per second by the second heater

P₂ = I₂R₂ = 1.42 x 60.5 = 118.58 Joule s^-1

So the first heater will produce more heat than the second one.

Ques. The particulars of electric items used in a house are listed below. What will be the electric bill of the house for two months (1 month = 30 days), if the rate of power is 1.35 Rs. per unit?

Ans. Energy consumed by the item can be calculated by

E = number of items x Power x time

Energy consumed by TV in 60 days, 

E₁ = 1 x 60 W x (2 x 60 hours) = 7200 Wh = 7.2 unit

Energy consumed by Fridge in 60 days, 

E₂ = 1 x 60 W x (24 x 60 hours) = 86400 Wh = 86.4 unit

Energy consumed by Tubes in 60 days, 

E₃ = 3 x 40 W x (4 x 60 hours) = 28800 Wh = 28.8 unit

Energy consumed by Bulbs in 60 days, 

E₄ = 2 x 60 W x (2 x 60 hours) = 14400 Wh = 14.4 unit

Energy consumed by Press in 60 days, 

E₅ = 1 x 600 W x (1/2 x 60 hours) = 18000 Wh = 18 unit

Total units of energy consumed in 2 months,

E = E₁ + E₂ + E₃ + E₄ + E₅

⇒ E = 7.2 + 86.4 + 28.8 + 14.4 + 18 = 154.8 unit

Given the cost of 1 unit = 1.35 Rs.

Therefore costs of 154.8 units = 154.8 x 1.35 = 208.98 Rs.

Ques. An electric motor of 1.0 kW operates at 220 V for 6 hours daily. What is the current drawn by the motor? What is the cost of running the motor for 30 days if the rate of electric energy is 50 paise per unit?

Ans. Given

• The power rating of the electric motor, P = 1.0 kW = 1000 W
• The operating voltage of the electric motor, V = 220 V

Electric power consumed by the electric motor is given by

P = VI

⇒ I = P/V = 1000/220 = 4.55 A

Hence the current drawn by the electric motor is 4.55 A

Electric energy consumed by the motor in a day = Power x Time = 1 x 6 = 6 kWh

Electric energy consumed by the motor in 30 days = 30 x 6 = 180 kWh = 180 unit

Given the cost of energy per unit = 50 paise

Therefore costs of 180 units = 180 x 50 = 9000 paise = 90 Rs.

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