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Physics
A 100 W bulb is connected to an AC source of 220 V, 50 Hz. Then the current flowing through the bulb is
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Q. A 100 W bulb is connected to an AC source of 220 V, 50 Hz. Then the current flowing through the bulb is
KCET
KCET 2018
Alternating Current
A
$\frac{5}{11} A $
64%
B
$\frac{1}{2} A $
14%
C
$ 2 \, A $
14%
D
$\frac{3}{4} A $
7%
Solution:
$I = \frac{P}{V} = \frac{100}{220} $
$= \frac{5}{11} \,A $