Electrical Power and Energy ICSE Class 10: Formulas, Power Ratings and the kWh

What Electrical Power and Energy is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Electrical Power and Energy is in the 4. Electricity and Magnetism (ii) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Electrical energy W = VIt

By definition of pd, the work done in moving charge $Q$ through pd $V$ is $W=QV$. Since $Q=It$:
$W=VIt$.
Using Ohm's law ($V=IR$):
$W=I^{2}Rt=\frac{V^2}{R}t$.

This energy appears as heat (heater, iron), light (lamp), mechanical energy (motor, fan) or sound (loudspeaker).

Worked example

A heater draws 5 A from a 220 V supply for 10 minutes. Find the energy used in joules and in kW h.

Try it first, then show the solutionSolution

1. $W=VIt=220\times 5\times 600=660000$ J.
2. In kW h: $\frac{660000}{3.6\times 10^6}\approx 0.183$ kW h.

Answer: 6.6 × 10⁵ J ≈ 0.183 kW h

Electrical power

Power is the rate at which electrical energy is used: $P=\frac{W}{t}$.
$P=VI=I^{2}R=\frac{V^2}{R}$.
Unit: watt (W); 1 W = 1 J s⁻¹ = 1 V A. Also kW (1000 W).

Choose the form that fits the data: with the same current (series), $P=I^{2}R$; with the same voltage (parallel), $P=\frac{V^2}{R}$.

Worked example

A 12 V, 24 W lamp works normally. Find the current it draws and its resistance.

Try it first, then show the solutionSolution

1. $I=\frac{P}{V}=\frac{24}{12}=2$ A.
2. $R=\frac{V^2}{P}=\frac{144}{24}=6$ Ω.

Answer: 2 A; 6 Ω

Power rating of appliances

A label such as '100 W, 220 V' means the appliance uses 100 W when connected to 220 V. From it:

• resistance $R=\frac{V^2}{P}$ (assumed constant);
• normal current $I=\frac{P}{V}$ (the current rating).

On a different supply $V^{\prime }$, the power becomes $P^{\prime }=\frac{V^{\prime 2}}{R}$: at half the voltage, a quarter of the power.

Typical ratings: LED lamp 10 W, fan 60–75 W, TV 100–150 W, refrigerator 150–250 W, iron 750–1000 W, geyser or AC 1–2 kW.

Worked example

A bulb is rated 100 W, 220 V. Find its resistance and normal current, and its power on a 110 V supply.

Try it first, then show the solutionSolution

1. $R=\frac{220^2}{100}=484$ Ω; $I=\frac{100}{220}\approx 0.45$ A.
2. On 110 V: $P=\frac{110^2}{484}=25$ W.

Answer: 484 Ω; about 0.45 A; 25 W

The kilowatt hour (commercial unit)

Electricity boards charge for energy in kilowatt hours (kW h), called 'units'. One kW h is the energy used by a 1 kW appliance in 1 hour:
$1\text{ kW h}=1000\text{ W}\times 3600\text{ s}=3.6\times 10^6$ J.

Energy in kW h $=$ power in kW $\times$ time in hours. The electricity meter (kW h meter) that records a home's consumption reads kW h (energy), not kW (power).

Worked example

A 2 kW heater runs 3 hours a day for 30 days. Find the energy used and the cost at ₹7 per unit.

Try it first, then show the solutionSolution

1. Energy $=2\times 3\times 30=180$ kW h.
2. Cost $=180\times 7=₹1260$.

Answer: 180 kW h; ₹1260

Household consumption

To find a household's energy use:
1. For each appliance: (number) × (power in kW) × (hours per day).
2. Add for the daily total in kW h; multiply by the number of days.
3. Multiply by the rate per unit for the cost.

Keep power in kW (divide watts by 1000) and time in hours.

Worked example

A home uses four 60 W lamps for 5 h a day, one 75 W fan for 8 h a day and one 120 W TV for 4 h a day. Find the energy used in 30 days and the cost at ₹7 per unit.

Try it first, then show the solutionSolution

1. Lamps: $4\times 0.06\times 5=1.2$ kW h; fan: $0.075\times 8=0.6$ kW h; TV: $0.12\times 4=0.48$ kW h.
2. Daily total $=2.28$ kW h; 30 days: $68.4$ kW h.
3. Cost $=68.4\times 7=₹478.80$.

Answer: 68.4 kW h; ₹478.80

Bulbs in series and in parallel

For bulbs rated at the same voltage, the lower-wattage bulb has the higher resistance ($R=\frac{V^2}{P}$).

• In parallel (same voltage): $P=\frac{V^2}{R}$, so the higher-wattage bulb glows brighter (normal operation).
• In series (same current): $P=I^{2}R$, so the higher-resistance (lower-wattage) bulb glows brighter, and both are dimmer than normal.

Removing a bulb from a parallel group increases the total resistance of that group; removing one from a series string decreases the total resistance.

Worked example

A 100 W and a 40 W bulb, both rated 220 V, are joined in series across 220 V. Which glows brighter?

Try it first, then show the solutionSolution

1. $R_{100}=484$ Ω; $R_{40}=1210$ Ω.
2. $I=\frac{220}{1694}\approx 0.130$ A.
3. $P_{40}=I^2\times 1210\approx 20.4$ W; $P_{100}=I^2\times 484\approx 8.2$ W.

Answer: The 40 W bulb glows brighter in series.

Current rating and choosing a fuse

The normal current of an appliance is $I=\frac{P}{V}$. A fuse should be rated slightly above this current: the appliance can take its normal current, but a dangerous larger current melts the fuse.

A fuse rated below the normal current would blow every time the appliance is switched on; one rated far above it would not protect the circuit.

Total current in a circuit is the sum of the currents of all appliances in parallel.

Worked example

An electric iron is rated 1100 W, 220 V. Find its normal current. Is a 3 A fuse suitable? A 6 A fuse?

Try it first, then show the solutionSolution

1. $I=\frac{1100}{220}=5$ A.
2. 3 A is below 5 A: it would melt at once, so it is not suitable.
3. 6 A is just above 5 A: suitable.

Answer: 5 A; 3 A no, 6 A yes

Physics in real life: Electrical Power and Energy

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Electrical Energy and Power
• Power Rating of Appliances
• Kilowatt Hour and Electricity Bills

Electrical Power and Energy formula sheet

Electrical energy

$W=QV=VIt=I^{2}Rt=\frac{V^2}{R}t$

In joules with t in seconds.

Electrical power

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

In watts.

From a power rating

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

Use the rated values.

Commercial unit

$1\text{kW h}=3.6\times 10^6\text{J}$

Energy (kW h) = power (kW) × time (h).

Power on another voltage

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

R taken as constant.

How the board asks Electrical Power and Energy

The question patterns to expect, and the method that works for each.

Tricks to remember Electrical Power and Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

• PIV

  Power = I × V, like 'P-I-V'.

  Use it when: The basic power formula.

• Series loves I²R, parallel loves V²/R

  Same current → I²R; same voltage → V²/R.

  Use it when: Comparing brightness.

• Low watt, high ohm

  R = V²/P: the smaller the wattage, the bigger the resistance.

  Use it when: Rated bulb problems.

• kilo × hours = units

  Units on the bill = kilowatts × hours.

  Use it when: Household bills.

Electrical Power and Energy: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Electrical Power and Energy

The slips examiners see most often, and what to write instead.

• ✗Mistake: Not finding R from the rating

  CISCE examiners' comments, 2025: for identical rated bulbs, candidates did not use R = V²/P and made calculation errors.

  ✓Do this: Start every rated-appliance problem with R = V²/P and I = P/V.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Series and parallel confused

  CISCE examiners' comments, 2025: candidates wrote wrong expressions for the total resistance of a series–parallel combination of bulbs.

  ✓Do this: Redraw the circuit; combine parallel groups first, then add series parts.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Current rating vs actual current

  CISCE examiners' comments, 2025: candidates confused a bulb's rated current with the current actually drawn in the circuit, and did not see that removing a bulb from a parallel group increases the net resistance and reduces the current.

  ✓Do this: The actual current depends on the whole circuit: I = V/R_total.

  Usually costs: 1 mark.

• ✗Mistake: Fuse rated below the normal current

  CISCE examiners' comments, 2024: candidates believed a fuse should be rated lower than the appliance's current, made calculation errors or computed R instead of I, and answered without supporting calculation.

  ✓Do this: Calculate I = P/V first; the fuse must be slightly above it.

  Usually costs: 2 marks.

• ✗Mistake: kW and kW h confused

  CISCE examiners' comments, 2025: candidates gave the wrong unit for the reading of the electricity meter; it shows energy in kW h, not power in kW.

  ✓Do this: kW = power; kW h = energy (what the electricity meter counts and what you pay for).

  Usually costs: 1 mark.

Practice questions on Electrical Power and Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The commercial unit of electrical energy is:
   (a) watt
   (b) joule
   (c) kilowatt hour
   (d) kilowatt

   Hint

   On the bill.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   The power of an appliance connected to a pd $V$ and carrying current $I$ is:
   (a) $\frac{V}{I}$
   (b) $VI$
   (c) $V^{2}I$
   (d) $\frac{I}{V}$

   Hint

   PIV.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   A bulb rated 100 W, 220 V is connected to a 110 V supply. Its power is:
   (a) 100 W
   (b) 50 W
   (c) 25 W
   (d) 200 W

   Hint

   P ∝ V².

   Answer

   (c)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): A household electricity meter measures energy in kW h.
   Reason (R): The bill depends on the total energy used, which is power multiplied by time.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Energy vs power.

   Answer

   (c)

5. Q5 · 2 marks · Short answerHarder board level

   A coil of resistance 20 Ω carries a current of 2 A for 30 s. Find the heat produced.

   Hint

   I²Rt.

   Answer

   $H=I^{2}Rt=4\times 20\times 30=2400$ J

6. Q6 · 2 marks · Short answerHarder board level

   A bulb is rated 40 W, 200 V. Find its resistance and normal current.

   Hint

   R = V²/P; I = P/V.

   Answer

   $R=\frac{200^2}{40}=1000$ Ω; $I=\frac{40}{200}=0.2$ A

7. Q7 · 3 marks · Long answerBoard level

   (a) Write three expressions for electrical power. (b) Define one kilowatt hour and express it in joules.

   Hint

   VI, I²R, V²/R.

   Answer

   (a) $P=VI=I^{2}R=\frac{V^2}{R}$. (b) The energy used by a 1 kW appliance in 1 hour; $1\text{ kW h}=3.6\times 10^6$ J.

8. Q8 · 3 marks · Long answerHarder board level

   A 2 kW geyser is used for 1.5 hours daily and a 0.5 kW iron for 1 hour daily. Find the energy used in 30 days and the cost at ₹8 per unit.

   Hint

   Daily units first.

   Answer

   Daily: $3+0.5=3.5$ kW h; 30 days: 105 kW h; cost $=₹840$.

9. Q9 · 4 marks · Long answerHOTS / competency

   Two bulbs rated 100 W, 220 V and 40 W, 220 V are connected in series to 220 V. Which glows brighter and why? Which glows brighter in parallel?

   Hint

   R = V²/P; same current in series.

   Answer

   Series: the 40 W bulb (1210 Ω) gets more power because P = I²R with the same current. Parallel: the 100 W bulb, because both get 220 V and P = V²/R is larger for the smaller resistance.

10. Q10 · 3 marks · Long answerHarder board level

    A 5 A fuse is available. Can it be used with (a) a 1760 W, 220 V room heater, (b) a 900 W, 220 V mixer? Justify each with a calculation.

    Hint

    I = P/V for each.

    Answer

    (a) $I=\frac{1760}{220}=8$ A, more than 5 A: no, the fuse would blow. (b) $I=\frac{900}{220}\approx 4.09$ A, just below 5 A: yes.

11. Q11 · 1 mark · MCQHOTS / competency

    Two heating coils of equal resistance are connected first in series, then in parallel, to the same supply. The ratio of heat produced per second (series : parallel) is:
    (a) 1 : 2
    (b) 2 : 1
    (c) 1 : 4
    (d) 4 : 1

    Hint

    P = V²/R_total: 2R vs R/2.

    Answer

    (c)

12. Q12 · 4 marks · Long answerHOTS / competency

    Three identical bulbs, each rated 12 W, 6 V, are connected to a 6 V battery: bulb A in series with a parallel pair B and C. (a) Find the resistance of each bulb. (b) Find the current from the battery. (c) Which bulb is brightest? (d) If C is removed, does A become brighter or dimmer?

    Hint

    R = 36/12 = 3 Ω.

    Answer

    (a) 3 Ω each. (b) Total $=3+1.5=4.5$ Ω; $I=\frac{6}{4.5}\approx 1.33$ A. (c) A (it carries the whole current; B and C share it). (d) Total becomes 6 Ω, current falls to 1 A, so A becomes dimmer.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Formula for electrical energy?Show answer

   W = VIt (= I²Rt = V²t/R).

2. 2.Formula for electrical power?Show answer

   P = VI (= I²R = V²/R).

3. 3.1 kW h in joules?Show answer

   3.6 × 10⁶ J

4. 4.Resistance from a rating?Show answer

   R = V²/P

5. 5.Normal current from a rating?Show answer

   I = P/V

6. 6.Which bulb glows brighter in series, 40 W or 100 W (same voltage rating)?Show answer

   40 W.

7. 7.How should a fuse be rated?Show answer

   Slightly above the normal current of the circuit.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write all three forms of P and W?
• ☐Can you get R and I from '100 W, 220 V'?
• ☐Can you convert kW h to J?
• ☐Can you calculate a monthly bill?
• ☐Do you know which bulb is brighter in series?
• ☐Can you justify a fuse choice with I = P/V?

Electrical Power and Energy: frequently asked questions