A 120V, 60W lamp is to be operated on 220V ac supply mains. calculate the value of non inductive resistance that would be required to ensure that the lamp is run on correct value
100Ω
200Ω
300Ω
500Ω
Correct answer is B
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 = \((440 - 240)\Omega\)
= \(200\Omega\)