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One `10V, 60W` bulb is to be connected to `100V` line. The required inductance coil has self-inductance of value `(f=50Hz)`A. 0.052 HB. 2.42 HC. 16.2 HD. 16.2 mH

Answer» Correct Answer - A
From P = VI = `V^(2)/R`
`implies I = P/V = 6A`
`implies R = V^(2)/P = 100/60 = 5/3Omega`
Now, `I = V/Z
6 = 100 / sqrt (5/3^(2)) + (2pifL)^(2)`
Solving this equation we get, L = 0.052 H


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