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Physics
The average power dissipated in a pure capacitance AC circuit is
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Q. The average power dissipated in a pure capacitance $AC$ circuit is
J & K CET
J & K CET 2009
Alternating Current
A
$ CV $
4%
B
zero
71%
C
$ \frac{1}{CV^{2}} $
13%
D
$ \frac{1}{4}CV^{2} $
11%
Solution:
Average power in AC circuit is given by
$P=V_{r m s} I_{r m s} \cos \phi$
For pure capacitive circuit $\phi=90^{\circ}$
so $P=0 .$