A 220 V Lamp on a 110 V Supply: Three Answers and One Filament
A secondary school laboratory is supplied with 110 V from a step-down transformer. A student is given a filament lamp marked “220 V, 60 W” and asked to find the power it consumes on the laboratory supply. He reasons as follows: the supply voltage is exactly half the rated voltage and power is proportional to voltage, so the lamp must consume 30 W. The teacher asks the class to check this with an ammeter in series and a voltmeter across the lamp. The meters read 110 V and 0.21 A. A second student multiplies and obtains 23.1 W, then declares that both the 30 W prediction and the usual P = V²/R calculation must be wrong, since neither gives 23 W. The teacher refuses to say who is right and asks for a physical explanation of the gap between the three numbers.
1. Using P = V²/R, show that even an ideal constant-resistance filament would not consume 30 W at 110 V, and name the error in the first student's proportional reasoning.
Read the worked answer
From the rating, the resistance at normal operating temperature is
R = V²/P = (220)²/60 = 806.7 Ω
If R stayed at that value, the power at 110 V would be
P = V²/R = (110)²/806.7 = 15.0 W
The first student's error is treating power as directly proportional to voltage. For a fixed resistance, P = V²/R, so power is proportional to the square of the voltage: halving the voltage quarters the power, giving 15 W, not 30 W. A 30 W answer would require the resistance to fall to 403 Ω, i.e. to roughly half its working value. Note also that 15 W is the lower bound here, because it is computed from the hot resistance, which is the largest the filament can have.
2. The measured value is 23.1 W, between 15 W and 30 W. Explain what assumption behind your 15 W figure fails, and in which direction it moves the answer.
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The assumption that fails is constant resistance. From the measurements,
R = V/I = 110/0.21 = 524 Ω
which is well below the 806.7 Ω the filament has at full brightness. The reason is that the filament is much cooler at 110 V than at 220 V, and the resistivity of tungsten falls with temperature. A cooler filament has less resistance, so it passes more current than the constant-R model predicts, and consumes more than 15 W.
The direction of the correction is therefore upward from 15 W. It cannot reach 30 W, because 30 W would require R = (110)²/30 = 403 Ω, a resistance lower than the measured 524 Ω; the filament would have to be cooler still than it actually is on this supply. The true value lies strictly between the two naive answers, and only measurement settles where.
3. Two identical lamps of the same rating are connected in series across the 220 V mains, so each receives about 110 V. Using the measured data, estimate the total power drawn from the mains and explain why it is not 60 W.
Read the worked answer
In series the same current flows through both lamps. Each lamp has about 110 V across it, and the measurement tells us that a lamp at 110 V draws 0.21 A. So the series current is 0.21 A and the total power drawn from the mains is
P = VI = 220 × 0.21 = 46.2 W, i.e. 23.1 W per lamp.
It is not 60 W because neither lamp is at its rated voltage, so neither filament reaches its rated temperature or resistance. It is not 30 W either, because that figure would come from assuming power falls linearly with voltage. Two cautions belong in the answer: the calculation assumes the lamps are truly identical, since only then do they share the 220 V equally; and the estimate is only as good as the single current measurement, so the class should verify the 0.21 A in the series arrangement rather than transfer it on trust.