An are lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V(rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

Options

A. `0.044 H`
B. `0.065 H`
C. `80 H`
D. `0.08H`

Answer

B

Explanation

Here
`i = (e)/(sqrt(R^(2) + X_(L)^(2))) = (e)/(sqrt(R^(2) + omega^(2)L^(2))) = (e)/(sqrt(R^(2) + 4 pi^(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