A sinusoidal ac current flows through a resistor of resistance R . If the peak current is `I_(p)` , then the power dissipated is

Options

A. `I_(P)^(2)R cos theta`
B. `(1)/(2) I_(P)^(2)R`
C. `(4)/(pi)I_(P)^(2)R`
D. `(1)/(pi^(2))I_(P)^(2)R`

Answer

B

Explanation

Here,
`i = (e)/(sqrt(R^(2) + X_(L)^(2))) = (e)/(sqrt(R^(2) + omega^(2)L^(2))) = (e)/(sqrt(R^(2) + 4pi^(2)v^(2)L^(2)))`
`10 = (220)/(sqrt(64 + 4pi^(2)(50)^(2) L))`
`[because R = (V)/(I) = (80)/(10) = 8]`
On solving we get
L = 0.065 H