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One 10V, 60W bulb is to be connected to ...

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 H

B

2.42 H

C

16.2 H

D

16.2 mH

Text Solution

Verified by Experts

The correct Answer is:
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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