Correct Answer: Option B

Explanation

Power of the lamp = 60W;  

\(I = {\frac{P}{V}}\)

\(I = {\frac{60}{120}}\) = \(0.5A\)

\(R = {\frac{V}{I}}\)

\(R_mains\) = \(\frac{220}{0.5}\) = \(440\Omega\)

\(R_lamp\) = \(\frac{120}{0.5}\) = \(240\Omega\)

\(\therefore\) The non-inductive resistance to keep the lamp = \((440 - 240)\Omega\)

= \(200\Omega\)

There is an explanation video available below.

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