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

Medium Level (Application & Explanation)

Q1. Define electric power and explain why it is called the “rate of doing electrical work.” Use two real-life examples to show how power helps compare appliances.

Answer: Electric power is the rate at which electrical energy is used or produced. In simple words, it tells how fast a device uses energy to do work like lighting, heating, or rotating a fan. The keyword is “rate,” which means energy per unit time. A device with higher power converts more energy in the same time. For example, a 100 W bulb converts more electrical energy to light and heat per second than a 60 W bulb. Similarly, a 2000 W microwave heats food faster than a 60 W fan because its power is higher. Power is measured in watt (W), where 1 W = 1 J/s. This helps us compare appliances: higher power generally means faster work and more energy consumption per second.

Q2. List the three main formulas of electric power and explain when to use each. Support your explanation with a short calculation in each case.

Answer: Electric power can be calculated using different forms depending on what is known:

  • P = V × I: Use when you know the potential difference and current. Example: A device at 220 V drawing 0.5 A has P = 220 × 0.5 = 110 W.
  • P = I²R: Use when current and resistance are known, useful for resistive heating devices. Example: I = 2 A, R = 20 Ω gives P = 2² × 20 = 80 W.
  • P = V²/R: Use when voltage and resistance are known, often in fixed-voltage circuits. Example: V = 100 V, R = 50 Ω gives P = 10000/50 = 200 W. All three are equivalent (from Ohm’s law V = IR) but choosing the right one reduces steps and mistakes. Always keep units consistent: V in volts, I in amperes, R in ohms, and P in watts.

Q3. A room heater is rated 1500 W and is used for 2.5 hours. Calculate the energy consumed in kWh and joules. If electricity costs ₹8 per unit, what is the cost? Explain each step and unit clearly.

Answer:

  • Power rating = 1500 W = 1.5 kW. Time used = 2.5 hours.
  • Electrical energy (in kWh) = Power (kW) × Time (h) = 1.5 × 2.5 = 3.75 kWh.
  • Since 1 kWh = 3.6 × 10^6 J, energy in joules = 3.75 × 3.6 × 10^6 = 13.5 × 10^6 J = 13.5 MJ.
  • Cost calculation: 1 kWh = 1 “unit” on the bill. Therefore, 3.75 kWh = 3.75 units.
  • Total cost = 3.75 × ₹8 = ₹30. This shows how power (watts) relates to energy over time (kWh) and to billing. The higher the power or the longer the usage, the more energy is consumed and the higher the cost.

Q4. Distinguish between electric power and electrical energy with suitable examples. Why is electricity billing done in kWh and not in watts?

Answer:

  • Electric power (W) is the rate of energy consumption; it tells “how fast” energy is used at a moment. Electrical energy (Wh or kWh) is the total amount used over time.
  • Example: A 100 W bulb has a power of 100 W, meaning it uses 100 J each second. If it runs for 10 hours, energy = 100 W × 10 h = 1000 Wh = 1 kWh.
  • Another example: A 2000 W microwave used for 0.5 hours consumes 2 kW × 0.5 h = 1 kWh. Although their power is different, they can consume the same energy if used for different durations.
  • Billing is in kWh because it measures the total energy consumed over time, which directly reflects the electricity supplied by the provider. Watts alone do not include time, so they cannot show how much energy you used or how much you should pay.

Q5. Two resistors, 50 Ω and 200 Ω, are connected separately to the same 200 V supply. Calculate the current and power for each. Which one gets hotter and why?

Answer:

  • Using I = V/R:
    • For 50 Ω: I = 200/50 = 4 A. Power P = V × I = 200 × 4 = 800 W; alternatively, P = V²/R = 40000/50 = 800 W.
    • For 200 Ω: I = 200/200 = 1 A. Power P = 200 × 1 = 200 W; or P = 40000/200 = 200 W.
  • The 50 Ω resistor consumes 800 W, while the 200 Ω resistor consumes 200 W at the same voltage. More power means more energy converted to heat per second, so the 50 Ω resistor gets much hotter.
  • Key idea: At fixed voltage, P = V²/R. Lower resistance draws higher current and thus higher power, resulting in greater heating. This is why low-resistance devices like heaters can produce more heat.

High Complexity (Analytical & Scenario-Based)

Q6. A family uses the following daily: six LED bulbs of 9 W for 5 hours, three fans of 75 W for 8 hours, and a 150 W TV for 4 hours. Calculate the monthly energy consumption (30 days) in kWh and the bill at ₹8 per unit. Suggest one saving tip using the concept of power.

Answer:

  • Daily energy:
    • LEDs: 6 × 9 W × 5 h = 270 Wh = 0.27 kWh
    • Fans: 3 × 75 W × 8 h = 1800 Wh = 1.8 kWh
    • TV: 150 W × 4 h = 600 Wh = 0.6 kWh
    • Total/day = 0.27 + 1.8 + 0.6 = 2.67 kWh
  • Monthly (30 days) = 2.67 × 30 = 80.1 kWh (≈ 80.1 units)
  • Bill = 80.1 × ₹8 = ₹640.8 (≈ ₹641)
  • Saving tip: Reduce fan usage by 2 hours/day or shift to more efficient fans. For example, saving 3 fans × 75 W × 2 h = 450 Wh/day = 0.45 kWh/day means 13.5 kWh/month saved, reducing the bill by about ₹108. Lowering usage time directly reduces energy because Energy = Power × Time.

Q7. Two irons are available: 500 W and 1000 W. The 500 W iron needs 90 minutes to complete your clothes; the 1000 W iron finishes in 45 minutes. Compare energy used and cost at ₹6 per unit. Which should you choose and why?

Answer:

  • Energy used:
    • 500 W for 1.5 h: Energy = 0.5 kW × 1.5 h = 0.75 kWh
    • 1000 W for 0.75 h: Energy = 1.0 kW × 0.75 h = 0.75 kWh
  • Cost for each: 0.75 kWh × ₹6 = ₹4.50. Both cost the same because Energy = Power × Time is equal in both cases.
  • Choice depends on time and
    convenience
    , not cost. The 1000 W iron saves time (finishes in half the time) with the same energy consumption for this task. However, if you tend to leave the iron on unnecessarily, the higher power can waste energy faster. Conclusion: For disciplined use, choose 1000 W for speed; otherwise, 500 W may be safer to avoid accidental overuse.

Q8. A heater of resistance 115 Ω is connected to a 230 V supply. If the voltage temporarily rises to 250 V, calculate the current and power in both cases and discuss the effect of voltage fluctuation on power and safety.

Answer:

  • At 230 V:
    • I = V/R = 230/115 = 2 A
    • P = V × I = 230 × 2 = 460 W (also P = V²/R = 52900/115 = 460 W)
  • At 250 V:
    • I = 250/115 ≈ 2.17 A
    • P = V²/R = 62500/115 ≈ 543.5 W
  • Observation: Power increases from 460 W to about 544 W when voltage rises. Because P ∝ V² for a fixed R, a small increase in voltage causes a larger increase in power and heating. This can overheat devices and reduce lifespan. Conclusion: Voltage fluctuations can significantly increase power and heat in resistive devices, so stable supply and protective devices (like proper ratings) are important to ensure safety.

Q9. Two bulbs of 60 W and 100 W are connected to the same 230 V supply. Calculate their resistances and currents. Which bulb has higher resistance and why? Relate your answer to power formulas.

Answer:

  • Using R = V²/P and I = P/V:
    • 60 W bulb:
      • R = 230²/60 = 52900/60 ≈ 881.7 Ω
      • I = 60/230 ≈ 0.261 A
    • 100 W bulb:
      • R = 230²/100 = 52900/100 = 529 Ω
      • I = 100/230 ≈ 0.435 A
  • The 60 W bulb has higher resistance (≈ 882 Ω) than the 100 W bulb (≈ 529 Ω). At the same voltage, a lower power rating means less current and therefore higher resistance (since P = V²/R). Conversely, the higher-power bulb draws more current and has lower resistance. This explains why the 100 W bulb is brighter and consumes more energy per second, while the 60 W bulb limits current due to its larger resistance.

Q10. A student claims, “If resistance doubles, power doubles.” Analyse this statement for two cases: (a) constant voltage and (b) constant current. Use formulas to justify the correct conclusion.

Answer:

  • Case (a) Constant voltage (typical home supply): P = V²/R. If resistance doubles (R → 2R), then P becomes V²/(2R) = P/2. Power halves, not doubles. So under a fixed voltage, increasing resistance reduces power and heating.
  • Case (b) Constant current (less common in homes): P = I²R. If resistance doubles, P becomes I²(2R) = 2P. Here, power doubles.
  • Conclusion: The statement is only true for constant current conditions. For normal household circuits where voltage is fixed, increasing resistance decreases current and thus decreases power. Understanding which quantity is held constant (V or I) is essential to predict how power changes. This highlights why P = V × I, P = I²R, and P = V²/R must be applied thoughtfully based on the given situation.

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