Electric Power – Long Answer Questions (CBSE Class 10 Science: Electricity)

Medium Level (Application & Explanation)

Q1. Explain with examples the three formulas you can use to calculate electric power in a circuit. When would you use each formula?

Answer:
Electric power can be found using three main formulas: $P = VI$ , $P = I^2R$ , and $P = \frac{V^2}{R}$ .

Use $P = VI$ when you know both voltage (V) and current (I) are given or can be easily measured.
If current (I) and resistance (R) are known, use $P = I^2R$ because it avoids finding voltage separately.
When you have voltage (V) and resistance (R), use $P = \frac{V^2}{R}$ ; this formula skips the need to know the current.
For example, a lamp connected directly to 220 V and drawing 0.5 A: $P = VI = 220 \times 0.5 = 110\,W$ .
A heater of 15 Ω working at 5 A: $P = I^2R = 25 \times 15 = 375\,W$ .
A 60 Ω resistor across 120 V: $P = V^2/R = 14400/60 = 240\,W$ .
Choosing the right formula makes calculations simpler and quicker.

Q2. An electric iron of power rating 1000 W is used for 3 hours daily. Find: (a) Total energy consumed in one day in kWh and joules (b) Cost for one month (30 days) if the rate is ₹7 per unit.

Answer:
(a) Energy per day = Power × Time = 1000 W × 3 h = 3 kWh (since 1000 W = 1 kW).
Convert to joules: 1 kWh = 3.6 × 10⁶ J, so 3 kWh = 3 × 3.6 × 10⁶ = 10.8 × 10⁶ J.
(b) For 30 days: Energy = 3 kWh/day × 30 = 90 kWh.
Cost = 90 × ₹7 = ₹630.
Thus, the iron uses 3 units (kWh) per day and costs ₹630 in a month.

Q3. Define kilowatt-hour. Why is it used as a commercial unit of energy instead of joule? Show how to convert kWh to joules.

Answer:
A kilowatt-hour (kWh) is the energy consumed by using 1 kW of power for 1 hour.
It is used in homes and industries to measure large amounts of energy easily.
Joule is very small for practical billing, as many appliances use millions of joules daily.
1 kWh = 1000 W × 1 hour = 1000 W × 3600 s = 3,600,000 J or 3.6 × 10⁶ J.
So, to convert kWh to joules, multiply the value in kWh by 3.6 × 10⁶.
Example: 5 kWh = 5 × 3.6 × 10⁶ = 18 × 10⁶ J.

Q4. A 200 W television is switched on for 5 hours every day. (a) How much electricity does it consume in a week (in kWh)? (b) How much energy in joules would this be?

Answer:
(a) Power = 200 W = 0.2 kW.
Daily consumption = 0.2 kW × 5 h = 1 kWh.
In one week (7 days): 1 × 7 = 7 kWh.
(b) Total energy in joules: 7 × 3.6 × 10⁶ = 25.2 × 10⁶ J.
Thus, the TV uses 7 units (kWh) per week, equal to 25.2 million joules.

Q5. An electric fan has a resistance of 80 Ω and runs on a potential difference of 200 V. Calculate the power consumed by the fan. Also, if it operates for 10 hours, find the total energy used in units.

Answer:
Power, $P = V^2/R = (200)^2/80 = 40000/80 = **500\,W**$ .
Convert to kW: 500 W = 0.5 kW.
Energy for 10 hours = 0.5 × 10 = 5 kWh (units).
Therefore, the fan uses 500 W of power and 5 units of energy in 10 hours.

High Complexity (Analysis & Scenario-Based)

Q6. If two bulbs, A (60 W, 220 V) and B (100 W, 220 V), are connected in parallel in a household circuit, calculate the total current drawn from the mains. Also, explain why household appliances are connected in parallel.

Answer:
For parallel connection:
Current by A = Power/Voltage = 60/220 ≈ 0.27 A
Current by B = 100/220 ≈ 0.45 A
Total current = 0.27 + 0.45 = 0.72 A

Reason for parallel connection:

Each appliance gets the same voltage (220 V).
If one stops working, others still operate.
Power and brightness remain same for each.
It allows independent switching and prevents overloading of single circuit.

Q7. A geyser contains a heating element of resistance 44 Ω, and it works on 220 V mains. Calculate: (a) Current drawn (b) Power consumed (c) Energy used in 2 hours (in kWh).

Answer:
(a) Current, $I = V/R = 220/44 = **5\,A**$
(b) Power, $P = V \times I = 220 \times 5 = **1100\,W**$
(c) Power in kW: 1100 W = 1.1 kW
Energy in 2 hours = 1.1 × 2 = 2.2 kWh
Thus, the geyser draws 5 A, uses 1100 W power, and consumes 2.2 units in 2 hours.

Q8. A school wants to reduce its electricity bill. They decide to replace 20 old 60 W bulbs, each used for 6 hours a day, with energy-efficient 9 W LED bulbs for the same hours. Calculate the total energy saved in a month (30 days) in kWh and cost saved if cost per unit is ₹5.

Answer:
Old bulbs’ daily energy: $20 \times 60 \times 6 = 7200\,Wh = 7.2\,kWh$ .
In 30 days: $7.2 \times 30 = 216\,kWh$ .
LED bulbs’ daily energy: $20 \times 9 \times 6 = 1080\,Wh = 1.08\,kWh$ .
In 30 days: $1.08 \times 30 = 32.4\,kWh$ .
Energy saved: 216 – 32.4 = 183.6 kWh.
Money saved: 183.6 × ₹5 = ₹918.
So, by switching, the school saves 183.6 units and ₹918 in a month.

Q9. Two heaters, $H_1$ (rating 1200 W, 200 V) and $H_2$ (1000 W, 200 V) are connected in series to a 200 V supply. Calculate the total power consumed by them together. Is this more or less than when they are connected individually to 200 V, and why?

Answer:
Series resistance:
$R_1 = V^2/P = 200^2/1200 ≈ 33.33\,\Omega$
$R_2 = 200^2/1000 = 40\,\Omega$
Total $R = 73.33\,\Omega$ , total power:
$P = V^2/R = 200^2/73.33 ≈ 545\,W$
Individually, their powers are 1200 W and 1000 W, much more than 545 W.
Reason: In series, current is same and shared between resistances, so each gets less power than their ratings and total power is reduced compared to parallel or solo connections.

Q10. A factory uses a machine of 5 kW for 8 hours daily. In the next month, they need to operate two such machines for 10 hours each day. Calculate the difference in the monthly (30 days) electricity bill, if 1 unit = ₹6.

Answer:
Current month:
Daily energy = 5 kW × 8 h = 40 kWh
Monthly = 40 × 30 = 1200 kWh
Cost = 1200 × ₹6 = ₹7200

Next month:
Daily energy = 2 × 5 kW × 10 h = 10 × 10 = 100 kWh
Monthly = 100 × 30 = 3000 kWh
Cost = 3000 × ₹6 = ₹18,000

Difference: 18,000 – 7,200 = ₹10,800
So, bill increases by ₹10,800 next month due to more machines and hours.