A bulb is rated $60$ W, $240$ V. Find the current it draws when operating normally.
A$0.4$ A
B$4$ A
C$14400$ A
D$0.25$ ACORRECT
JUPEBPhysicsCurrent Electricity
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Why the answer is D, and why the others tempt you.
$P = IV$, so $I = \dfrac{P}{V} = \dfrac{60}{240} = 0.25$ A. Its resistance when hot is $\dfrac{V^2}{P} = 960\ \Omega$. Inverting the ratio gives 4 A.
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