Electric Power – Detailed Explanation for Class 10 Science

When an electric current flows through a conductor, electrical energy is used up and we say that the current is doing work.

We know that the rate of doing work is called power, so electric power is the electrical work done per unit time. That is, $$\text{Power} = \frac{\text{Work Done}}{\text{Time Taken}} \quad \text{or} \quad P = \frac{W}{t}$$

⚙️ SI Unit of Electric Power

We calculated power as: $$P = \frac{W}{t}$$

The SI unit of work is joule and time is in seconds, so the unit of power is: $$\text{Power} = \frac{\text{Joule}}{\text{Second}} = \text{Watt (W)}$$

Thus, $$\ \text{Watt} = \frac{1 \ \text{Joule}}{1 \ \text{Second}}$$

As watt is a small unit, we often use the kilowatt: $$ \ \text{kW} = 1000 \ \text{W}$$

Electric power is also the rate at which electrical energy is consumed:

When an appliance consumes 1 joule of electrical energy per second, its power is said to be 1 watt.

🔣 Formula for Calculating Electric Power

From earlier, $$P = \frac{W}{t}$$

And we know: $$W = V \times I \times t$$

Substitute into the power formula: $$P = \frac{V \times I \times t}{t} = V \times I$$

🔄 So,

Power, $$\text{Power}, \ P = V \times I \ \text{watts}$$

where:

• V = Potential difference (Volts)
• I = Current (Amperes)

Also written as:

Electric power = Voltage × Current

Now defining 1 watt again: $$1 \ \text{W} = 1 \ \text{V} \times 1 \ \text{A} = 1 \ \text{VA}$$

🧮 Alternate Formulae for Electric Power

We’ve seen: $$P = V \times I \quad \text{…(1)}$$

Using Ohm’s Law: $$V = I \times R$$

(i) When I and R are known:

Substitute into (1): $$P = I \times (I \times R) = I^2 \times R$$

(ii) When V and R are known:

From Ohm’s Law: $$I = \frac{V}{R}$$

Substitute into (1): $$P = V \times \frac{V}{R} = \frac{V^2}{R}$$

It is clear from equation that power is inversely proportional to the resistance. Thus, the resistance of high power devices is smaller than the low power ones. For example, the resistance of 100 watt (220 volt) bulb is smaller than that of a 60 watt (220 volt) bulb (see Figure.

So, we now have:

📌 From this, it’s clear that power is inversely proportional to resistance.

🧾 Power-Voltage Rating of Electrical Appliances

Every appliance has a label like:
“100 W – 220 V”

• 100 W → power consumed
• 220 V → operating voltage

To find the current drawn: $$P = V \times I \Rightarrow I = \frac{P}{V}$$

Common Ratings:

🧠 Sample Problems

Q1. Find the Current drawn by a 40 W bulb at 220 V?

Solution: $$P = V \times I $$

$$ I = \frac{P}{V} $$

$$ I = \frac{40}{220} = \frac{2}{11} \approx 0.18 \ \text{A}$$

Q2. A 220 V, 100 W bulb is operated at 110 V. What is Power consumed by bulb?

Step 1: Calculate resistance from rating of bulb : $$P = \frac{V^2}{R} \Rightarrow R = \frac{220^2}{100} = 484 \ \Omega$$

Step 2: Power consumed by bulb operated at 110 V :

$$P = \frac{V^2}{R}$$

$$P = \frac{110^2}{484} = \frac{12100}{484} = 25 \ \text{W}$$

Answer: (d) 25 W

Q3. Which of the following is NOT a power formula?

• (a) $I^2 R$ ✔️
• (b) $I R^2$ ❌
• (c) $VI$ ✔️
• (d) $\frac{V^2}{R}$ ✔️
  Answer: (b) $I R^2$

⚡ Calculating Electrical Energy

We will now derive a formula for calculating electrical energy in terms of power and time. We have already studied that :

$$\text{Electric Power} = \frac{\text{Work done by electric current}}{\text{Time taken}} $$

Now, according to the law of conservation of energy, Work done by electric current = Electric energy consumed So, we can now write down the above relation as :

$$\text{Power} = \frac{\text{Electrical Energy}}{\text{Time}} $$

$$ \text{Electrical Energy} = \text{Power} \times \text{Time}$$

or $$E = P \times t$$

Hence the electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used.

From this we conclude that the electrical energy consumed by an electrical appliance depends on two factors :

(i) power rating of the appliance, and

(ii) time for which the appliance is used

🔁 SI Units:

If:

• P in watts
• t in hours
• Then, E is in watt-hour (Wh)

One Watt-hour (Wh) is the energy consumed by 1 W device in 1 hour.

🔷 Conceptual Questions with Answers – Electric Power

Q1. Why is electric power written as $P = I^2 R$ or $P = \frac{V^2}{R}$?

Answer:
From Ohm’s Law : $$V = IR$$
We substitute this in the basic formula : $$P = VI$$

$$P = IV = I \cdot IR = I^2 R$$

Alternatively, $$I = \frac{V}{R}$$

$$ P = V \cdot \frac{V}{R} = \frac{V^2}{R}$$
These alternative forms are useful based on the known variables in a circuit.

Q2. If the voltage is constant, what happens to electric power when the current is doubled?

Answer:
Electric power $$P = VI$$
If V is constant and I doubles then P also doubles.
So, power increases proportionally with current when voltage is constant.

Q3. Two electric bulbs are rated 60 W and 100 W respectively. Which one will glow brighter when connected to the same voltage? Why?

Answer:
The 100 W bulb will glow brighter because it consumes more power, meaning it converts more electrical energy into light and heat per second.

Q4. Why is electrical energy measured in kilowatt-hour (kWh) and not in joules for practical purposes?

Answer:
Electrical appliances consume large amounts of energy, and using joules (which is a small unit) would result in very large numbers.
1 kWh = 3.6 × 10⁶ J
So, kWh is a more convenient and practical unit for everyday usage.

Q5. Why does a fuse blow out when too many appliances are connected to a single socket?

Answer:
Connecting multiple appliances increases the current.
If current exceeds the fuse’s rated limit, the fuse wire melts due to overheating (since P=I²R), breaking the circuit to prevent damage.

Q6. What is the significance of the power rating on an electrical appliance?

Answer:
It indicates how much power the appliance consumes when operated at the rated voltage.
For example, a 60 W bulb at 220 V means it uses 60 joules of energy per second when plugged into a 220 V supply.

Q7. How does electric power depend on resistance when current is kept constant?

Answer:
From P = I² R, when current I is constant:

• Power increases with an increase in resistance.
• So, higher resistance → more power dissipated (as heat).

Q8. Why do metallic wires become hot when electric current passes through them?

Answer:
Due to the Joule heating effect:

• When current flows through a resistance, electrical energy is converted into heat.
• The power loss as heat is given by P = I² R, causing the wire to heat up.

Q9. Can you use a 100 W, 110 V bulb in a 220 V circuit safely? Explain.

Answer:
No.

• At 220 V, the power consumed will be much more than 100 W, possibly around 400 W using $P = \frac{V^2}{R}$.
• This will overheat and damage the bulb. Always match voltage rating.

Q10. How does reducing resistance affect power consumption if voltage is constant?

Answer:
From $$P = \frac{V^2}{R}$$

• Lower resistance → higher power consumption.
• So, more energy is consumed per second at lower resistance.

Here are Multiple Choice Questions (MCQs) with explanations for the topic Electric Power – suitable for Class 10 Science students:

🔷 MCQs with Explanation – Electric Power

Q1. An electric bulb is rated 100 W, 220 V. What is the current flowing through the bulb?

A. 0.45 A
B. 0.27 A
C. 0.55 A
D. 2.2 A

Answer: ✅ A. 0.45 A
Explanation:
Use the formula P = VI ⇒ I = P/V = 100/220 ≈ 0.45 A

Q2. Which of the following is not a correct unit of power?

A. Watt
B. Kilowatt
C. Horsepower
D. Kilowatt-hour

Answer: ✅ D. Kilowatt-hour
Explanation:
Kilowatt-hour (kWh) is a unit of energy, not power.
Power units include watt (W), kilowatt (kW), and horsepower (HP).

Q3. If the resistance of a device is doubled, and the voltage remains constant, what happens to the power consumed?

A. Becomes double
B. Becomes half
C. Becomes one-fourth
D. Remains the same

Answer: ✅ B. Becomes half
Explanation:
From P = V²/R, if R doubles, power becomes half.

Q4. Which device is used to measure electric power?

A. Ammeter
B. Voltmeter
C. Wattmeter
D. Ohmmeter

Answer: ✅ C. Wattmeter
Explanation:
A wattmeter is specifically designed to measure electric power in a circuit.

Q5. An electric iron of 1 kW is used for 2 hours daily for 30 days. What is the energy consumed?

A. 30 kWh
B. 60 kWh
C. 1 kWh
D. 100 kWh

Answer: ✅ B. 60 kWh
Explanation:
Energy = Power × Time
Energy = 1 kW × 2 h/day × 30 days = 60 kWh

Q6. If an electric heater draws 10 A current at 220 V, what is its power?

A. 2.2 kW
B. 22 W
C. 2200 kW
D. 220 W

Answer: ✅ A. 2.2 kW
Explanation:
P = V I = 220 × 10 = 2200 W = 2.2 kW

Q7. Which of the following correctly represents power in terms of resistance and current?

A. P = I R
B. P = R²/I
C. P = I² R
D. P=I²/R

Answer: ✅ C. P = I² R
Explanation:
This is a derived formula of power using Ohm’s Law:
P = VI, and V = IR ⇒ P = I²R

Q8. A bulb is rated 60 W, 240 V. What is its resistance?

A. 480 Ω
B. 960 Ω
C. 640 Ω
D. 320 Ω

Answer: ✅ B. 960 Ω
Explanation:
Using $$P = \frac{V^2}{R} $$

$$R = \frac{V^2}{P} = \frac{240^2}{60} = 960\,\Omega$$

Q9. Which quantity determines the brightness of an electric bulb?

A. Voltage
B. Resistance
C. Power
D. Current

Answer: ✅ C. Power
Explanation:
Brightness of a bulb depends on the power it consumes. More power means more energy is converted into light per second.

Q10. A 500 W heater is used for 10 hours. How much energy is consumed in kWh?

A. 0.5 kWh
B. 5 kWh
C. 50 kWh
D. 500 kWh

Answer: ✅ B. 5 kWh
Explanation:
Energy = Power × Time = 500 x 10 = 5 x 1000 = 5 kWh

Q11. What is the commercial unit of electrical energy used by electricity boards?

A. Joule
B. Watt
C. Kilowatt-hour
D. Calorie

Answer: ✅ C. Kilowatt-hour
Explanation:
Electricity boards use kWh (unit) to measure energy consumption for billing.

Q12. What is the power of an appliance if it draws 0.5 A current from a 220 V source?

A. 110 W
B. 220 W
C. 440 W
D. 55 W

Answer: ✅ B. 110 W
Explanation:
P = V I = 220 × 0.5 = 110 W

Q13. The resistance of a 100 W, 220 V bulb is:

A. 220 Ω
B. 440 Ω
C. 484 Ω
D. 100 Ω

Answer: ✅ C. 484 Ω
Explanation:
Use R = V²/P = 220²/100 = 484 Ω

Q14. If the current in a circuit is 3 A and the resistance is 10 Ω, what is the power consumed?

A. 30 W
B. 90 W
C. 100 W
D. 9 W

Answer: ✅ B. 90 W
Explanation:
Use P = I² R = 3² × 10 = 9 × 10 = 90 W

Q15. In which of the following does the appliance consume the maximum power?

A. 220 V, 2 A
B. 220 V, 1 A
C. 110 V, 2 A
D. 110 V, 1 A

Answer: ✅ A. 220 V, 2 A
Explanation:
P = VI, so:

• (A) = 220 × 2 = 440 W
• (B) = 220 × 1 = 220 W
• (C) = 110 × 2 = 220 W
• (D) = 110 × 1 = 110 W

Q16. What is the relation between energy, power, and time?

A. Energy = Power ÷ Time
B. Energy = Time ÷ Power
C. Energy = Power × Time
D. Energy = Power + Time

Answer: ✅ C. Energy = Power × Time
Explanation:
This is the fundamental formula for electrical energy.

🔷 Assertion and Reason Based MCQs – Electric Power

Q1.

Assertion (A): Electric power is given by the product of voltage and current.
Reason (R): Power = V × I is the fundamental formula of electric power.

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation: The formula P = VI defines electric power. R correctly explains A.

Q2.

Assertion (A): A 60 W bulb consumes more energy per second than a 100 W bulb.
Reason (R): Power is the rate of energy consumption.

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is false but R is true. ✅
• (D) Both A and R are false.

Answer: ✅ (C)
Explanation: Power indicates energy consumed per second. A 100 W bulb consumes more than 60 W. So A is false, but R is true.

Q3.

Assertion (A): Power dissipated in a conductor increases with increase in resistance if current is constant.
Reason (R): Power P = I² R

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation: With constant current, P = I² R, so power increases with resistance.

Q4.

Assertion (A): A fuse wire has high resistance and low melting point.
Reason (R): Fuse wire melts easily to break the circuit when excess current flows.

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation: High resistance increases heating effect, and low melting point ensures quick melting.

Q5.

Assertion (A): Electric energy is measured in watt.
Reason (R): Watt is the SI unit of power.

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is false but R is true. ✅
• (D) Both A and R are false.

Answer: ✅ (C)
Explanation: Electric energy is measured in joules or kilowatt-hours, not watts. R is true; A is false.

Q6.

Assertion (A): The commercial unit of electrical energy is kilowatt-hour.
Reason (R): 1 kWh = 3600 J

• (A) Both A and R are true, and R is the correct explanation of A.
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false. ✅
• (D) A is false but R is true.

Answer: ✅ (C)
Explanation: 1 kWh = 3.6 × 10⁶ J (not 3600 J). So A is correct, R is incorrect.

Q7.

Assertion (A): A 220 V, 100 W bulb has less resistance than a 220 V, 60 W bulb.
Reason (R): Resistance R=V²/P

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation:

• $R = \frac{220^2}{100} = 484\,\Omega$
• $R = \frac{220^2}{60} ≈ 807\,\Omega$
  So 100 W bulb has less resistance.

Q8.

Assertion (A): Electric power consumed is inversely proportional to resistance, if voltage is constant.
Reason (R): P = V²/R

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation: From P = V²/R, if voltage is constant, higher resistance leads to lower power.

Q9.

Assertion (A): 1 kWh is a bigger unit than 1 joule.
Reason (R): 1 kWh = 3.6 × 10⁶ joules.

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation: 1 kWh is 3.6 million joules. Clearly, it is a much larger unit than 1 J.

Q10.

Assertion (A): The resistance of a conductor remains constant when the current through it changes.
Reason (R): Ohm’s Law is valid only when the temperature remains constant.

• (A) Both A and R are true, and R is the correct explanation of A. ✅
• (B) Both A and R are true, but R is not the correct explanation of A.
• (C) A is true but R is false.
• (D) A is false but R is true.

Answer: ✅ (A)
Explanation: Resistance depends on material and temperature. If temperature is constant, resistance stays constant with changing current.

📝 Worksheet – Electric Power (Class 10 Science)

Instructions: Attempt all questions. Show proper steps for numericals.

Section A: Fill in the Blanks

1. Electric power is the rate at which __________ is consumed.
2. SI unit of electric power is __________.
3. 1 kWh = __________ joules.
4. Power consumed by a device is given by the formula P=___×___P = \_\_\_ \times \_\_\_.
5. If voltage is constant, power is __________ proportional to resistance.

Section B: True or False

1. A 100 W bulb consumes more power than a 60 W bulb.
2. Power = V²R.
3. 1 watt = 1 joule/second.
4. More resistance always means more power consumed.
5. The energy consumed in 2 hours by a 1000 W device is 2 kWh.

Section C: Match the Columns

Section D: Numerical Questions

1. Calculate the power of a device that draws 5 A current from a 220 V supply.
2. Find the resistance of a bulb rated 100 W, 200 V.
3. How much energy is consumed by a 2000 W appliance in 2.5 hours?

📄 Test Paper – Electric Power

Subject: Science (Physics)
Class: 10
Chapter: Electricity – Electric Power
Max Marks: 25
Time: 45 minutes

Section A: Very Short Answer (1 mark each × 5 = 5 marks)

1. Define electric power.
2. What is the SI unit of energy?
3. Write the commercial unit of electric energy.
4. State the formula for electric power in terms of resistance.
5. What is the relation between power, energy, and time?

Section B: Short Answer (2 marks each × 4 = 8 marks)

6. A 60 W bulb is used for 5 hours. Calculate the energy consumed in kWh.
7. What will be the power consumed by a device if it draws 3 A current and 120 V voltage?
8. Why does a fuse wire melt when excess current flows?
9. Write any two formulas for calculating power.

Section C: Long Answer (4 marks each × 3 = 12 marks)

10. Derive the expression P=I² R and P = V²/R.
11. A heater of 1500 W is used for 3 hours daily for 30 days. Calculate:
    a) Total energy consumed in kWh
    b) Electricity bill at ₹6 per unit
12. Two bulbs are rated 60 W and 100 W respectively. Both are connected to 220 V.
    a) Which bulb has more resistance?
    b) Which bulb will glow brighter?