The resistance of a 240V, 60 watt electric filament bulb is
-
A.
0.25(Omega) -
B.
480(Omega) -
C.
60(Omega) -
D.
240(Omega) -
E.
960(Omega)
Correct Answer: Option E
Explanation
power = IV
power = (frac{V^2}{R}) = 60
= (frac{240^2}{R})
= 960(Omega)
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