PhysicsICSE Class 10 · Electrical Power and Energy

Power Rating of Appliances: R = V²/P, Current Rating and Bulb Brightness (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

A power rating such as '60 W, 240 V' means the appliance uses 60 W when connected to a 240 V supply. From it you can find the resistance, R = V²/P, and the normal current, I = P/V. On another voltage V′ the power becomes V′²/R, so at half the voltage it uses only a quarter of the power.

Resistance stays fixed, so halving the voltage cuts the power to one quarter; find R and rated current from the rating.

Illustration drawn for QuestionBuddy.

Understand Power Rating of Appliances

Reading a rating label. Every appliance carries a label giving its rated power and rated voltage, for example '1000 W, 230 V' on an iron. It means: when connected to a 230 V supply, the iron uses electrical energy at the rate of 1000 W (1000 J every second). It does not mean the iron always uses 1000 W, whatever the supply.

Two things to find from the rating. Treat the appliance as a resistor whose resistance stays the same:

Resistance: $P = \frac{V ^{2}}{R}$ , so $R = \frac{V ^{2}}{P}$ .
Normal (rated) current: $P = V I$ , so $I = \frac{P}{V}$ . This is the current rating of the appliance: the current it draws when working on its rated voltage.

Always use the rated $V$ and $P$ in these two formulas.

Higher wattage, lower resistance. For appliances rated at the same voltage, $R \propto \frac{1}{P}$ . A 100 W, 230 V bulb has a smaller resistance than a 25 W, 230 V bulb. If the filaments are of the same material and length, the 100 W filament is thicker ( $R \propto \frac{1}{A}$ ).

Working on a different voltage. The resistance does not change (ignoring heating effects), so on a supply $V^{'}$ :

$P^{'} = \frac{V ^{'2}}{R} = P (\frac{V ^{'}}{V})^{2} .$

At half the rated voltage, the power is one quarter: a 100 W, 230 V bulb on 115 V uses about 25 W and glows dimly.
On a higher voltage than rated, the power and current exceed their rated values and the appliance may burn out.
In reality, a filament on a lower voltage is cooler, so its resistance is a little lower and the actual power is a little more than this calculation gives.

Bulbs in parallel (normal household wiring). Each bulb gets the full supply voltage, so each works at its rated power: the higher-wattage bulb glows brighter. Switching one off does not affect the others, but the total current falls.

Bulbs in series. The same current flows through each, so $P = I^{2} R$ :

The bulb with the higher resistance (the lower-wattage one) uses more power and glows brighter.
Both glow less than normal, because the total resistance is larger and the current is smaller than either rated current.
Two identical bulbs in series on their rated voltage each get half the voltage, so each uses a quarter of its rated power, and together they use half the power of one bulb.

Series–parallel combinations. Work in this order: resistance of each bulb from $R = \frac{V ^{2}}{P}$ ; combine parallel groups; add series parts; total current $I = \frac{V}{R _{t o t a l}}$ ; then the current and power in each bulb. Adding another bulb in parallel to a group lowers that group's resistance; removing one raises it, which changes the current in the rest of the circuit.

Current rating and fuses. The fuse in an appliance's circuit must be rated slightly above the appliance's normal current $\frac{P}{V}$ . A fuse rated below the normal current melts every time the appliance is switched on; one rated far above gives no protection. When several appliances work in parallel, the currents add: the total must stay below the fuse rating.

Power Rating of Appliances: formulas and key facts

Resistance from the rating: $R = \frac{V ^{2}}{P}$; Rated voltage and rated power.
Normal (rated) current: $I = \frac{P}{V}$; Choose a fuse slightly above this.
Power on another voltage: $P^{'} = \frac{V ^{'2}}{R} = P (\frac{V ^{'}}{V})^{2}$; Resistance taken as constant.
Brightness rules: Parallel (same V): higher wattage brighter | Series (same I): higher resistance (lower wattage) brighter; Brightness follows the power actually used.

How to solve Power Rating of Appliances questions

Convert every rating into a resistance first, then treat the circuit as an ordinary resistor problem.

1For each appliance, write $R = \frac{V ^{2}}{P}$ and, if needed, $I = \frac{P}{V}$ using the rated values.
2If the appliance is on its rated voltage in parallel with others, it uses its rated power; the currents simply add.
3If the supply differs from the rated voltage, use $P^{'} = \frac{V ^{'2}}{R}$ or the ratio $(\frac{V ^{'}}{V})^{2}$ .
4For series or mixed circuits, find the total resistance and total current, then each bulb's current and $P = I^{2} R$ .
5Compare brightness by the power actually used, not by the wattage printed on the bulb.
6For a fuse, compare the normal current with the fuse rating and choose the nearest rating above it.

Power Rating of Appliances 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.

Appliance used on the wrong voltage

Suppose a relative brings home a hair dryer rated 1210 W, 110 V, made for a country with a 110 V supply, and plugs it into a 220 V socket.

Find the power it would use and explain what happens.

Try it first, then show the solutionSolution

$R = \frac{V ^{2}}{P} = \frac{11 0 ^{2}}{1210} = 10$ Ω.
On 220 V: $P^{'} = \frac{22 0 ^{2}}{10} = 4840$ W, four times its rating; the current would be $\frac{220}{10} = 22$ A.
Such a large power overheats the element and the motor, so the dryer burns out (or the fuse blows).

Answer: About 4840 W, four times the rating; it would be damaged.

Memory hook: Double the voltage, four times the power.

Low voltage in the evening

Imagine the supply in your area drops from 230 V to 207 V on a winter evening, when everyone switches on heaters. Your room heater is rated 1000 W, 230 V.

Estimate the heater's power at 207 V.

Try it first, then show the solutionSolution

The voltage is $\frac{207}{230} = 0.9$ of the rated value.
$P^{'} = 1000 \times 0. 9^{2} = 810$ W.

Answer: About 810 W, so the room warms more slowly.

Memory hook: 10% less voltage, about 19% less power.

Tricks to remember Power Rating of Appliances

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

Rating gives R and I
'R is V-squared over P; I is P over V': write both the moment you see a rating.
Use it when: Starting any rated-appliance problem.
Weak bulb wins in series
In series, the low-watt (high-resistance) bulb shines brighter; in parallel, the high-watt bulb shines brighter.
Use it when: Brightness comparisons.
Square of the voltage ratio
New power = old power × (new V ÷ old V)².
Use it when: Appliances on a lower or higher supply.
Fuse just above
Fuse rating: just above the normal current, never below it.
Use it when: Choosing or checking a fuse.

Power Rating of Appliances: solved examples

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

Example 1: Normal current from a rating · 1 mark · MCQBoard level

An appliance is rated 1000 W, 250 V. The current it draws when working normally is:
(a) 0.25 A
(b) 4 A
(c) 62.5 A
(d) 250 A

Try it first, then show the solutionSolution

$I = \frac{P}{V} = \frac{1000}{250} = 4$ A. (62.5 is $\frac{V ^{2}}{P}$ , the resistance in ohms, not a current.)

Answer: (b) 4 A

Example 2: Meaning of a bulb's rating · 2 marks · Short answerBoard level

A bulb is marked '25 W, 230 V'. (a) What does this mean? (b) Find the resistance of its filament when glowing normally.

Try it first, then show the solutionSolution

(a) When connected to a 230 V supply, the bulb uses electrical energy at the rate of 25 J every second. (1)
(b) $R = \frac{V ^{2}}{P} = \frac{23 0 ^{2}}{25} = 2116$ Ω. (1)

Answer: (a) 25 W when on 230 V (b) 2116 Ω

Example 3: Iron on a low supply · 3 marks · Short answerHarder board level

An electric iron is rated 880 W, 220 V. Find (a) its resistance, (b) its normal current, (c) the power it uses and the current it draws when the supply falls to 176 V (resistance unchanged).

Try it first, then show the solutionSolution

(a) $R = \frac{22 0 ^{2}}{880} = 55$ Ω. (1)
(b) $I = \frac{880}{220} = 4$ A. (1)
(c) $P^{'} = \frac{17 6 ^{2}}{55} = 563.2$ W (check: $880 \times 0. 8^{2} = 563.2$ W); $I^{'} = \frac{176}{55} = 3.2$ A. (1)

Answer: (a) 55 Ω (b) 4 A (c) 563.2 W, 3.2 A

Example 4: 60 W and 15 W bulbs in series · 4 marks · Long answerHarder board level

Bulbs rated 60 W, 240 V and 15 W, 240 V are connected in series to a 240 V supply. Find (a) the resistance of each, (b) the current, (c) the power used by each. (d) Which glows brighter? Which would glow brighter if they were connected in parallel?

Try it first, then show the solutionSolution

(a) $R_{60} = \frac{24 0 ^{2}}{60} = 960$ Ω; $R_{15} = \frac{24 0 ^{2}}{15} = 3840$ Ω. (1)
(b) $I = \frac{240}{960 + 3840} = \frac{240}{4800} = 0.05$ A. (1)
(c) $P_{60} = 0.0 5^{2} \times 960 = 2.4$ W; $P_{15} = 0.0 5^{2} \times 3840 = 9.6$ W. (1)
(d) In series the 15 W bulb glows brighter (higher resistance, more power). In parallel each gets 240 V and works at its rating, so the 60 W bulb glows brighter. (1)

Answer: (a) 960 Ω, 3840 Ω (b) 0.05 A (c) 2.4 W, 9.6 W (d) Series: 15 W bulb; parallel: 60 W bulb

Example 5: Choosing a fuse · 2 marks · Short answerHarder board level

A microwave oven is rated 1380 W, 230 V. Fuses of 3 A, 8 A and 15 A are available. Which should be used? Justify with a calculation.

Try it first, then show the solutionSolution

Normal current $I = \frac{P}{V} = \frac{1380}{230} = 6$ A. (1)
A 3 A fuse would melt at once; a 15 A fuse would let a dangerously large current pass. The 8 A fuse, just above 6 A, is correct. (1)

Answer: 8 A fuse (normal current 6 A)

Example 6: How many bulbs on one fuse? · 2 marks · Short answerHOTS / competency

A hall is wired with a 5 A fuse on a 220 V supply. What is the largest number of 40 W, 220 V bulbs that can be lit together in parallel without blowing the fuse?

Try it first, then show the solutionSolution

Current per bulb $= \frac{40}{220} = \frac{2}{11} \approx 0.18$ A. (½)
Number $= \frac{5}{2/11} = 27.5$ , so at most 27 bulbs (27 draw ≈ 4.91 A; 28 would draw ≈ 5.09 A). (1½)

Answer: 27 bulbs

Example 7: Case study: adding a bulb to a parallel group · 4 marks · Case-basedHOTS / competency

Bulb P is rated 120 W, 240 V and bulb Q is rated 60 W, 240 V.
(a) Find the resistance of each. Which has the thicker filament (same material and length)?
(b) P and Q are connected in series across 240 V. Find the power used by each and say which glows brighter.
(c) A second bulb identical to Q is now connected in parallel with Q, the pair still in series with P. Find the new current from the supply.
(d) Find the power now used by P and by each Q bulb, and state how the brightness of P changes.

Try it first, then show the solutionSolution

(a) $R_{P} = \frac{24 0 ^{2}}{120} = 480$ Ω; $R_{Q} = \frac{24 0 ^{2}}{60} = 960$ Ω. P has the thicker filament (lower resistance). (1)
(b) $I = \frac{240}{1440} = \frac{1}{6}$ A; $P_{P} = \frac{1}{36} \times 480 \approx 13.3$ W; $P_{Q} = \frac{1}{36} \times 960 \approx 26.7$ W. Q glows brighter. (1)
(c) Two 960 Ω bulbs in parallel give 480 Ω; total $= 480 + 480 = 960$ Ω; $I = \frac{240}{960} = 0.25$ A. Adding a parallel bulb lowered the resistance and raised the current. (1)
(d) $P_{P} = 0.2 5^{2} \times 480 = 30$ W, so P becomes brighter; each Q carries 0.125 A and uses $0.12 5^{2} \times 960 = 15$ W. (1)

Answer: (a) 480 Ω, 960 Ω; P (b) ≈ 13.3 W and 26.7 W; Q brighter (c) 0.25 A (d) P 30 W (brighter), each Q 15 W

Where students lose marks in Power Rating of Appliances

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

Mistake: Not finding R from the rating
The resistance is needed for any series or mixed circuit of rated bulbs.
Wrong working
Taking each 24 W bulb's resistance as 24 Ω, or using R = P/V.
Right working
R = V²/P with the rated values.
Do this: Write R = V²/P for every bulb before drawing the circuit.
Usually costs: Reported by CISCE examiners for 2025 Q8(iii).
Mistake: Rated current confused with circuit current
P/V is the current only on the rated voltage; in a series circuit the actual current is V/R_total.
Wrong working
Using P/V of one bulb as the current in a series–parallel circuit.
Right working
Find R_total, then I = V/R_total.
Do this: Rated values give R; circuit values come from the circuit.
Usually costs: Reported by CISCE examiners for 2025 Q8(iii).
Mistake: Missing the effect of removing a parallel bulb
Removing a bulb from a parallel group increases that group's resistance, so the total current falls.
Wrong working
'Removing one bulb reduces resistance, so the current increases.'
Right working
'Fewer parallel paths, larger resistance, smaller current.'
Do this: Recalculate the parallel group with one bulb fewer.
Usually costs: Reported by CISCE examiners for 2025 Q8(iii).
Mistake: Fuse rated below the normal current
The fuse must carry the normal current; it should melt only when the current exceeds it.
Wrong working
'Yes, a 5 A fuse can be used' with no calculation, or computing R instead of I.
Right working
I = P/V, compare with the fuse rating, then answer Yes or No.
Do this: Calculate the current first and give its unit (A).
Usually costs: Reported by CISCE examiners for 2024 Q3(ii).

Exam tips

Begin each answer with R = V²/P and I = P/V written out; these earn the first method marks.
Never call a bulb 'brighter' from its printed wattage alone in a series circuit; calculate the power actually used.
In a fuse question, answer 'Yes' or 'No' clearly and back it with the current in amperes.
When the supply differs from the rated voltage, say 'assuming resistance remains constant'.

Practice questions on Power Rating of Appliances

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
Four bulbs, all rated for 230 V, have powers of 25 W, 40 W, 60 W and 100 W. Which has the highest resistance?
(a) 25 W
(b) 40 W
(c) 60 W
(d) 100 W
Hint
$R = \frac{V ^{2}}{P}$ at the same V.
Answer
(a) 25 W (2116 Ω)
Q2 · 1 mark · Very short answerBoard level
Find the current rating of a mixer marked 460 W, 230 V.
Hint
$I = \frac{P}{V}$ .
Answer
$I = \frac{460}{230} = 2$ A.
Q3 · 1 mark · Very short answerBoard level
A geyser is marked '1.5 kW, 230 V'. State what this means.
Hint
Power at the stated voltage.
Answer
When connected to a 230 V supply, the geyser uses electrical energy at the rate of 1.5 kW (1500 J every second).
Q4 · 2 marks · Short answerHarder board level
A bulb rated 40 W, 240 V is connected to a 120 V supply. Find the power it uses (resistance constant).
Hint
Half the voltage.
Answer
$P^{'} = 40 \times (\frac{120}{240})^{2} = 10$ W.
Q5 · 2 marks · Short answerHarder board level
A room heater is rated 1.2 kW, 240 V. Find (a) its resistance, (b) its normal current.
Hint
Use 1200 W.
Answer
(a) $R = \frac{24 0 ^{2}}{1200} = 48$ Ω. (b) $I = \frac{1200}{240} = 5$ A.
Q6 · 2 marks · Short answerHarder board level
Can a 10 A fuse be used for a geyser rated 2.5 kW, 230 V? Justify with a calculation.
Hint
Find the normal current first.
Answer
$I = \frac{2500}{230} \approx 10.9$ A, which exceeds 10 A, so the fuse would melt when the geyser is switched on. No; a 15 A fuse should be used.
Q7 · 3 marks · Short answerHarder board level
A refrigerator (230 W), a television (115 W) and an iron (920 W), all rated for 230 V, work together on a 230 V supply. Find the current in each and the total current.
Hint
Parallel: the currents add.
Answer
Refrigerator 1 A, television 0.5 A, iron 4 A; total 5.5 A.
Q8 · 2 marks · Short answerHarder board level
Two bulbs are rated 100 W, 220 V and 25 W, 220 V. Find the ratio of their resistances. Which glows brighter when they are connected in parallel to 220 V?
Hint
$R \propto \frac{1}{P}$ at the same V.
Answer
$R_{100} : R_{25} = 25 : 100 = 1 : 4$ . In parallel each works at its rating, so the 100 W bulb glows brighter.
Q9 · 2 marks · Short answerHOTS / competency
A 100 W, 230 V filament bulb on a 115 V supply is calculated to use 25 W, but a meter shows slightly more. Explain why.
Hint
Temperature and resistance.
Answer
At 115 V the filament is much cooler than at 230 V. A metal's resistance falls as its temperature falls, so R is less than the rated-value resistance and $\frac{V ^{2}}{R}$ is a little more than 25 W.
Q10 · 2 marks · Short answerHOTS / competency
Two identical bulbs, each rated 60 W, 240 V, are connected in series across 240 V. Find the power used by each and the total power.
Hint
Each gets half the voltage.
Answer
Each gets 120 V, so each uses $60 \times (\frac{1}{2})^{2} = 15$ W; total 30 W, half the power of one bulb working normally.

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.Resistance from a rating?Show answer
R = V²/P.
2.Power at half the rated voltage?Show answer
One quarter of the rated power.
3.Brighter bulb in series: 40 W or 100 W (same voltage rating)?Show answer
The 40 W bulb.
4.Which has the thicker filament: 40 W or 100 W at 230 V?Show answer
The 100 W bulb.
5.Fuse rating relative to normal current?Show answer
Slightly above it.

Power Rating of Appliances: frequently asked questions

What does '100 W, 220 V' on a bulb mean?

It means the bulb uses electrical energy at the rate of 100 J per second when connected to a 220 V supply. Its resistance is R = V²/P = 484 Ω and its normal current is P/V ≈ 0.45 A.

Why does a low-wattage bulb glow brighter in series?

In series the same current flows through both bulbs, so the power used is I²R. The low-wattage bulb has the higher resistance (R = V²/P), so it uses more power and glows brighter.

What happens to an appliance used on a lower voltage?

Its resistance stays about the same, so its power falls as the square of the voltage: P′ = V′²/R. At half the rated voltage it uses about a quarter of its rated power.