Power of the lamp = 60W; Â
\(I = {\frac{P}{V}}\)
\(I = {\frac{60}{120}}\) = \(0.5A\)
\(R = {\frac{V}{I}}\)
\(R_mains\) = \(\frac{220}{0.5}\) = \(440\Omega\)
\(R_lamp\) = \(\frac{120}{0.5}\) = \(240\Omega\)
\(\therefore\) The non-inductive resistance to keep the lamp = \((440 - 240)\Omega\)
= \(200\Omega\)