PhysicsICSE Class 10 · Electrical Power and Energy

Electrical Energy and Power: W = QV = VIt and P = VI = I²R = V²/R (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

Electrical energy is the work done in moving charge through a potential difference: W = QV = VIt, which with Ohm's law becomes I²Rt or (V²/R)t. Electrical power is the rate of using it: P = VI = I²R = V²/R. Its SI unit is the watt, 1 W = 1 J s⁻¹ = 1 V A.

Energy W = VIt; dividing by time gives power P = VI, and Ohm's law turns this into I²R or V²/R, measured in watts (J/s).

Illustration drawn for QuestionBuddy.

Understand Electrical Energy and Power: W = VIt and P = VI

Where the energy goes. A cell or the mains does work in pushing charge round a circuit. The appliance converts this electrical energy into another form:

Appliance	Main energy change
Heater, iron, kettle, geyser	Electrical → heat
Lamp (filament, LED, tube)	Electrical → light (and some heat)
Motor, fan, mixer	Electrical → mechanical (kinetic)
Loudspeaker	Electrical → sound

Electrical energy from the definition of pd. The potential difference between two points is the work done per unit charge: $V = \frac{W}{Q}$ . So the work done in moving charge $Q$ through pd $V$ is

$W = Q V .$

A steady current $I$ for time $t$ carries charge $Q = I t$ , so

$W = V I t .$

Using Ohm's law, $V = I R$ or $I = \frac{V}{R}$ :

$W = I^{2} R t = \frac{V ^{2}}{R} t .$

The energy is in joules when $V$ is in volts, $I$ in amperes, $R$ in ohms and $t$ in seconds. In a resistor (a heating coil, a filament) all of it appears as heat; this is the heating effect of current, $H = I^{2} R t$ .

Electrical power. Power is the rate at which electrical energy is consumed (or supplied):

$P = \frac{W}{t} = V I = I^{2} R = \frac{V ^{2}}{R} .$

SI unit: the watt (W). One watt is the power when one joule of energy is used each second: $1 W = 1 J s^{- 1}$ .
Since $P = V I$ : $1 W = 1 V \times 1 A$ . An appliance uses 1 W when a current of 1 A flows through it under a pd of 1 V.
Bigger units: 1 kilowatt (kW) $= 1000$ W; 1 megawatt (MW) $= 1 0^{6}$ W.

Which form to use. All three give the same answer for a resistor; pick the one that matches the data.

Known	Use	Typical situation
$V$ and $I$	$P = V I$	Meter readings; any device, including a motor
$I$ and $R$	$P = I^{2} R$	Resistors in series (same current)
$V$ and $R$	$P = \frac{V ^{2}}{R}$	Resistors in parallel (same pd)

Two consequences follow:

In series, $P \propto R$ : the larger resistor uses more power.
In parallel, $P \propto \frac{1}{R}$ : the smaller resistor uses more power.

Proportional changes. For a fixed resistor, $P \propto I^{2}$ and $P \propto V^{2}$ . Doubling the current (or the pd) makes the power, and the heat produced in a given time, four times. On a fixed supply voltage, $P \propto \frac{1}{R}$ : shortening a heating coil lowers $R$ and raises the power.

A motor is not a resistor. For a fan or a motor, $P = V I$ gives the electrical power taken in. Only the part $I^{2} R$ (with $R$ the coil's resistance) becomes heat; the rest becomes mechanical work. Using $\frac{V ^{2}}{R}$ with the coil's resistance gives a wrong, far larger value, because Ohm's law $V = I R$ does not hold for the whole motor while it is running.

Electrical Energy and Power: W = VIt and P = VI: formulas and key facts

Electrical energy: $W = Q V = V I t = I^{2} R t = \frac{V ^{2}}{R} t$; Joules, with t in seconds.
Electrical power: $P = \frac{W}{t} = V I = I^{2} R = \frac{V ^{2}}{R}$; Watts; I²R and V²/R hold for a resistor.
Heating effect: $H = I^{2} R t$; Heat produced in a resistor.
Units: $1 W = 1 J s^{- 1} ∣1 W = 1 V \times 1 A ∣1 k W = 1000 W ∣1 M W = 1 0^{6} W$; Power is a scalar.

How to solve Electrical Energy and Power: W = VIt and P = VI questions

List the known quantities, pick the matching form of the power or energy equation, and keep time in seconds for joules.

1Write down what is given: $V$ , $I$ , $R$ , $Q$ , $t$ . Convert minutes and hours to seconds, kW to W.
2If charge is given, use $W = Q V$ ; if current and time are given, use $W = V I t$ or $I^{2} R t$ .
3For power, choose $V I$ , $I^{2} R$ or $\frac{V ^{2}}{R}$ by which two quantities you know; for series resistors find the common current first, for parallel use the common pd.
4To find a missing quantity, rearrange: $I = \frac{P}{V}$ , $R = \frac{V ^{2}}{P}$ , $I = \frac{P}{R}$ .
5For 'what happens if' questions, use proportions: $P \propto I^{2}$ and $P \propto V^{2}$ for a fixed $R$ ; $P \propto \frac{1}{R}$ at fixed $V$ .
6Give units with every answer (J, W, A, Ω, V).

Electrical Energy and Power: W = VIt and P = VI in real life

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.

Why a kettle needs a thick cable

Suppose your electric kettle is marked 2.2 kW and your home supply is 230 V, while a phone charger lead next to it is very thin.

Find the current the kettle draws and explain why its cable is much thicker than the charger lead.

Try it first, then show the solutionSolution

$I = \frac{P}{V} = \frac{2200}{230} \approx 9.6$ A, a large current.
Heat produced in the cable is $I^{2} R t$ ; to keep it small, the cable must have a low resistance, so it is made thick (large area of cross-section).
A charger draws a tiny current, so a thin lead stays cool.

Answer: About 9.6 A; a thick cable has low resistance, so little heat is produced in it.

Memory hook: Big current, thick wire.

Shorter coil, hotter heater

Imagine the coil of an old room heater breaks, and someone joins the ends after cutting off a piece, making the coil shorter.

Will the heater now give more or less heat on the same supply? Is this safe?

Try it first, then show the solutionSolution

A shorter coil has a smaller resistance ( $R \propto l$ ).
At a fixed supply voltage, $P = \frac{V ^{2}}{R}$ , so the power rises.
The current $I = \frac{V}{R}$ also rises, which can overheat the coil and the wiring and blow the fuse.

Answer: More heat, but a larger current flows, so it is unsafe.

Memory hook: Same V, smaller R, more power.

Tricks to remember Electrical Energy and Power: W = VIt and P = VI

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

Three powers from one
Start from P = VI; swap V = IR to get I²R, or swap I = V/R to get V²/R.
Use it when: Recalling or deriving all three forms.
Series shares I, parallel shares V
Series: use I²R (bigger R, more power). Parallel: use V²/R (smaller R, more power).
Use it when: Comparing power in combined resistors.
Square law
Double I or V, four times the power; halve them, a quarter.
Use it when: Ratio questions.
Watt = joule per second
Power is how fast the joules go, like speed is how fast the metres go.
Use it when: Defining the watt.

Electrical Energy and Power: W = VIt and P = VI: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Unit of power · 1 mark · MCQBoard level

The SI unit of electric power is equivalent to:
(a) J s
(b) V A
(c) V A⁻¹
(d) C s⁻¹

Try it first, then show the solutionSolution

$P = V I$ , so 1 W = 1 V × 1 A = 1 V A. (V A⁻¹ is the ohm and C s⁻¹ is the ampere.)

Answer: (b) V A

Example 2: Deriving W = VIt and W = I²Rt · 3 marks · Short answerBoard level

Starting from the definition of potential difference, show that the electrical energy used when a current $I$ flows through a resistor $R$ for time $t$ is $W = V I t$ , and hence $W = I^{2} R t$ .

Try it first, then show the solutionSolution

The pd is the work done per unit charge: $V = \frac{W}{Q}$ , so $W = Q V$ . (1)
Current $I = \frac{Q}{t}$ , so $Q = I t$ and $W = V I t$ . (1)
By Ohm's law $V = I R$ , so $W = (I R) I t = I^{2} R t$ . (1)

Answer: W = QV = VIt = I²Rt

Example 3: Toaster: charge to energy and power · 3 marks · Short answerHarder board level

A charge of 1500 C flows through a toaster connected to a 230 V supply in 5 minutes. Find (a) the electrical energy used, (b) the power of the toaster, (c) the current.

Try it first, then show the solutionSolution

(a) $W = Q V = 1500 \times 230 = 345000$ J $= 3.45 \times 1 0^{5}$ J. (1)
(b) $t = 5 \times 60 = 300$ s; $P = \frac{W}{t} = \frac{345 000}{300} = 1150$ W. (1)
(c) $I = \frac{Q}{t} = \frac{1500}{300} = 5$ A (check: $V I = 230 \times 5 = 1150$ W). (1)

Answer: (a) 3.45 × 10⁵ J (b) 1150 W (c) 5 A

Example 4: Heat in a resistor · 3 marks · Short answerHarder board level

A 15 Ω resistor carries a steady current of 3 A for 2 minutes. Find (a) the pd across it, (b) the power, (c) the heat produced.

Try it first, then show the solutionSolution

(a) $V = I R = 3 \times 15 = 45$ V. (1)
(b) $P = I^{2} R = 3^{2} \times 15 = 135$ W. (1)
(c) $H = P t = 135 \times 120 = 16200$ J. (1)

Answer: (a) 45 V (b) 135 W (c) 16 200 J

Example 5: Same resistors, series then parallel · 4 marks · Long answerHarder board level

Resistors of 6 Ω and 12 Ω are connected to an 18 V supply of negligible internal resistance, (a) in series, (b) in parallel. Find the power used in each resistor in both cases, and state which resistor uses more power in each arrangement.

Try it first, then show the solutionSolution

(a) Total $R = 18$ Ω, so $I = \frac{18}{18} = 1$ A in both. $P_{6} = 1^{2} \times 6 = 6$ W; $P_{12} = 1^{2} \times 12 = 12$ W. The 12 Ω resistor uses more. (2)
(b) Each has the full 18 V: $P_{6} = \frac{1 8 ^{2}}{6} = 54$ W; $P_{12} = \frac{1 8 ^{2}}{12} = 27$ W. The 6 Ω resistor uses more. (1)
Same current in series, so $P \propto R$ ; same pd in parallel, so $P \propto \frac{1}{R}$ . Total power: 18 W in series, 81 W in parallel. (1)

Answer: (a) 6 W and 12 W (12 Ω more) (b) 54 W and 27 W (6 Ω more)

Example 6: Cutting a heating element · 3 marks · Short answerHOTS / competency

A heater is rated 800 W on 220 V. Its element is cut to half its length and used on the same supply. Find (a) the original resistance and current, (b) the new power and current. (c) Comment on the result.

Try it first, then show the solutionSolution

(a) $R = \frac{V ^{2}}{P} = \frac{22 0 ^{2}}{800} = 60.5$ Ω; $I = \frac{P}{V} = \frac{800}{220} \approx 3.64$ A. (1)
(b) Half the length gives $R = 30.25$ Ω; $P = \frac{22 0 ^{2}}{30.25} = 1600$ W; $I = \frac{220}{30.25} \approx 7.27$ A. (1)
(c) Power and current both double. The element runs much hotter and the larger current may overheat the wiring or blow the fuse. (1)

Answer: (a) 60.5 Ω, ≈ 3.64 A (b) 1600 W, ≈ 7.27 A (c) Power doubles; unsafe

Example 7: Case study: power of a motor · 4 marks · Case-basedHOTS / competency

A water-pump motor works on a 220 V supply and draws a current of 5 A. The resistance of its coil is 4 Ω.
(a) Find the electrical power taken by the motor.
(b) Find the power wasted as heat in the coil.
(c) Find the mechanical power output and the efficiency.
(d) A student calculates the power as $\frac{V ^{2}}{R} = \frac{22 0 ^{2}}{4}$ . Why is this wrong?

Try it first, then show the solutionSolution

(a) $P = V I = 220 \times 5 = 1100$ W. (1)
(b) Heat in the coil $= I^{2} R = 5^{2} \times 4 = 100$ W. (1)
(c) Mechanical power $= 1100 - 100 = 1000$ W; efficiency $= \frac{1000}{1100} \times 100% \approx 90.9%$ . (1)
(d) A running motor is not a pure resistor; most of the energy becomes mechanical work, so $V \neq = I R$ for it. Using $\frac{V ^{2}}{R}$ gives 12 100 W, far more than the 1100 W actually supplied. (1)

Answer: (a) 1100 W (b) 100 W (c) 1000 W, ≈ 90.9% (d) V²/R applies only to a resistor

Where students lose marks in Electrical Energy and Power: W = VIt and P = VI

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

Mistake: Not linking P, V, I and R
Most numericals need two steps: P = VI to get the current, and V = IR (or R = V²/P) to get the resistance.
Wrong working
Trying to find the current of a 12 V lamp from its power alone, or using P = V/I.
Right working
I = P/V, then R = V/I or R = V²/P.
Do this: Write P = VI and V = IR at the top of every power numerical.
Usually costs: Reported by CISCE examiners for 2025 Q8(iii).
Mistake: Time left in minutes or hours
Joules need seconds: W = VIt with t in s.
Wrong working
W = 12 × 0.5 × 1 = 6 J for one minute.
Right working
W = 12 × 0.5 × 60 = 360 J.
Do this: Convert time to seconds before substituting.
Usually costs: Usually costs the answer mark.
Mistake: Using I²R where the current differs
In parallel the pd is the same but the currents differ, so P = V²/R is quicker and safer.
Wrong working
Taking the total current through each of two parallel resistors.
Right working
Each parallel branch: P = V²/R with the common pd.
Do this: Series: find the common I. Parallel: use the common V.
Usually costs: Usually costs the method and answer marks.
Mistake: Applying V²/R to a motor
A running motor converts most energy to mechanical work, so Ohm's law does not apply to it as a whole.
Wrong working
Power of a 220 V motor with 4 Ω coil = 220²/4.
Right working
Input power = VI; heat in coil = I²R.
Do this: For motors and fans, use only P = VI for the input.
Usually costs: Gives an impossible answer.

Exam tips

Show the substitution step with units; a bare answer loses the method mark.
In a derivation, begin with V = W/Q and state Q = It and V = IR as you use them.
Check your answer with a second form of the power equation when time allows.
Write 1 W = 1 J s⁻¹ = 1 V A when asked to define or relate units.

Practice questions on Electrical Energy and Power: W = VIt and P = VI

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

Q1 · 1 mark · MCQBoard level
Which of these is NOT an expression for electrical power?
(a) $V I$
(b) $I^{2} R$
(c) $\frac{V ^{2}}{R}$
(d) $I R^{2}$
Hint
Substitute V = IR into VI.
Answer
(d) $I R^{2}$
Q2 · 1 mark · Very short answerBoard level
Define the watt.
Hint
Joules per second.
Answer
One watt is the power of an appliance that uses one joule of electrical energy per second (equivalently, that draws 1 A at a pd of 1 V).
Q3 · 1 mark · Very short answerBoard level
A 12 V battery sends a current of 0.5 A through a lamp for 1 minute. Find the electrical energy used.
Hint
$W = V I t$ with t in seconds.
Answer
$W = 12 \times 0.5 \times 60 = 360$ J.
Q4 · 2 marks · Short answerBoard level
Name the main energy conversion in (a) a loudspeaker, (b) an electric fan, (c) an electric iron, (d) an LED lamp.
Hint
What useful output does each give?
Answer
(a) Electrical → sound. (b) Electrical → mechanical (kinetic). (c) Electrical → heat. (d) Electrical → light.
Q5 · 2 marks · Short answerHarder board level
A resistor of 8 Ω uses power at the rate of 32 W. Find the current through it and the pd across it.
Hint
$I = \frac{P}{R}$ .
Answer
$I = \frac{32}{8} = 2$ A; $V = I R = 2 \times 8 = 16$ V.
Q6 · 3 marks · Short answerHarder board level
A kettle draws 2000 W from a 250 V supply. Find (a) the current, (b) the resistance of its element, (c) the energy used in 3 minutes.
Hint
Start from P = VI.
Answer
(a) $I = \frac{2000}{250} = 8$ A. (b) $R = \frac{V}{I} = \frac{250}{8} = 31.25$ Ω. (c) $W = P t = 2000 \times 180 = 3.6 \times 1 0^{5}$ J.
Q7 · 2 marks · Short answerHarder board level
A device working on 9 V uses 54 kJ of electrical energy in 10 minutes. Find (a) the charge that passed through it, (b) the current.
Hint
$W = Q V$ .
Answer
(a) $Q = \frac{W}{V} = \frac{54 000}{9} = 6000$ C. (b) $I = \frac{Q}{t} = \frac{6000}{600} = 10$ A.
Q8 · 2 marks · Short answerHarder board level
Resistors of 3 Ω and 6 Ω are connected in parallel across a 6 V battery of negligible internal resistance. Find the power used in each and the ratio of the powers.
Hint
Same pd: $P = \frac{V ^{2}}{R}$ .
Answer
$P_{3} = \frac{36}{3} = 12$ W; $P_{6} = \frac{36}{6} = 6$ W; ratio 2 : 1.
Q9 · 2 marks · Short answerHarder board level
A lamp working on 230 V has a resistance of 529 Ω when hot. Find its power and the current it draws.
Hint
$P = \frac{V ^{2}}{R}$ .
Answer
$P = \frac{23 0 ^{2}}{529} = 100$ W; $I = \frac{P}{V} = \frac{100}{230} \approx 0.43$ A.
Q10 · 2 marks · Short answerHOTS / competency
The current through a fixed resistor is doubled. By what factor does the heat produced in the same time change? What if, instead, the pd across it is halved?
Hint
$H = I^{2} R t$ and $P = \frac{V ^{2}}{R}$ .
Answer
Doubling the current makes the heat 4 times ( $H \propto I^{2}$ ). Halving the pd makes the power, and so the heat, $\frac{1}{4}$ ( $P \propto V^{2}$ ).
Q11 · 3 marks · Short answerHOTS / competency
Wires A and B are of the same material and length, but A has twice the diameter of B. Find the ratio of the powers used by A and B when they are connected (a) in parallel, (b) in series, to the same supply.
Hint
Area ∝ diameter², so $R_{A} = \frac{R _{B}}{4}$ .
Answer
$R_{A} : R_{B} = 1 : 4$ . (a) Parallel, same pd, $P \propto \frac{1}{R}$ : $P_{A} : P_{B} = 4 : 1$ . (b) Series, same current, $P \propto R$ : $P_{A} : P_{B} = 1 : 4$ .

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.Electrical energy in terms of Q and V?Show answer
W = QV.
2.Three expressions for power?Show answer
VI, I²R, V²/R.
3.1 W in terms of volt and ampere?Show answer
1 V A.
4.Which resistor uses more power in series?Show answer
The larger one.
5.Effect of doubling the current on heat produced?Show answer
Four times.

Electrical Energy and Power: W = VIt and P = VI: frequently asked questions

What is electrical power?

Electrical power is the rate at which electrical energy is used or supplied: P = W/t = VI. For a resistor it can also be written I²R or V²/R. Its SI unit is the watt (1 W = 1 J s⁻¹).

How is W = VIt derived?

From the definition of pd, V = W/Q, so W = QV. A current I for time t carries charge Q = It, so W = VIt. Using V = IR gives W = I²Rt = (V²/R)t.

When should I use I²R and when V²/R?

Use I²R when the current is known or shared (resistors in series) and V²/R when the pd is known or shared (resistors in parallel). For any device, VI gives the power taken in.