P = V²/R so R = V²/P = 1210Ω. Then I = P/V = 40/220 ≈ 0.18 A. Real-life calculation.

A 40W Bulb on 220V Supply — Find Current and Resistance

Question

A bulb is rated 40 W, 220 V. Find:

The resistance of the bulb
The current drawn when connected to a 220 V supply

Solution — Step by Step

Power rating $P = 40$ W, Voltage rating $V = 220$ V. The bulb is connected at exactly its rated voltage, so we can use the rating directly. We need resistance $R$ and current $I$ .

The key formula here is:

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

Why this one and not $R = V/I$ ? Because we don’t know $I$ yet — but we do know both $V$ and $P$ . Substituting:

R = \frac{(220)^2}{40} = \frac{48400}{40} = 1210 \, \Omega

Now that we have $R$ , we can get $I$ two ways. The cleanest is:

I = \frac{P}{V} = \frac{40}{220} = \frac{2}{11} \approx 0.18 \, \text{A}

Quick sanity check: $I = V/R = 220/1210 = 2/11 \approx 0.18$ A. Matches. We’re good.

Answers: $R = 1210 \, \Omega$ , $I \approx 0.18$ A

Why This Works

The rating “40 W, 220 V” means the bulb is designed to dissipate 40 W when exactly 220 V is applied across it. So the rated voltage and rated power are not two separate facts — they define the bulb’s resistance at operating temperature.

The formula $R = V^2/P$ comes from combining $P = VI$ and $V = IR$ . Eliminate $I$ and you get $P = V^2/R$ , so $R = V^2/P$ . This is the go-to formula whenever you know power and voltage but not current.

Notice the resistance of the bulb filament (1210 Ω) is the hot resistance — when the tungsten filament is glowing. At room temperature, the resistance is much lower. This is why a bulb draws a surge of current the instant you switch it on.

Alternative Method

We can find $I$ first, then get $R$ from it.

I = \frac{P}{V} = \frac{40}{220} \approx 0.18 \text{ A}

Then using Ohm’s Law:

R = \frac{V}{I} = \frac{220}{0.18} \approx 1210 \, \Omega

Same answer, different order. In the exam, either route is fully valid — pick whichever feels more natural.

Memorise the three forms of the power formula as a triangle: $P = VI = I^2R = V^2/R$ . Depending on which two quantities are given, pick the form that uses exactly those two. No rearranging needed mid-step.

Common Mistake

A very common error: using $R = V/P$ instead of $R = V^2/P$ . Students see “220 V” and “40 W” and write $R = 220/40 = 5.5 \, \Omega$ — which is completely wrong. The correct relation is $P = V^2/R$ , so you must square the voltage. The units give it away: $V^2/P = \text{V}^2/\text{W} = \text{V}^2/(\text{V·A}) = \text{V/A} = \Omega$ . If you skip the square, the units don’t even work out to ohms.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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