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Physics
A sinusoidal AC current flows through a resistor of resistance R . If the peak current is Ip, then the power dissipated is
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Q. A sinusoidal AC current flows through a resistor of resistance $R .$ If the peak current is $I_{p}$, then the power dissipated is
Alternating Current
A
$I_{p}^{2} R \cos \theta$
B
$\frac{1}{2} I_{p}^{2} R$
C
$\frac{4}{\pi} I_{p}^{2} R$
D
$\frac{1}{\pi} I_{p}^{2} R$
Solution:
Power $=I^{2} R=\left(\frac{I_{p}}{\sqrt{2}}\right)^{2} R=\frac{I_{p}^{2} R}{2}$