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A bulb of resistance $240$ $\Omega$ is connected across $120$ V. Current through it is:
A$2$ A, by $I = VR$ multiplying the two quantities
B$1$ A, the average shortcut for any small bulb
C$0.5$ A, by Ohm's law $I = V/R = 120/240$
D$0$ A, since the values do not match in scale
Answer & Solution
Correct answer: C. $0.5$ A, by Ohm's law $I = V/R = 120/240$
$I = V/R = 120/240 = 0.5$ A.
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