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
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Correct Answer: Option B
Explanation:
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 = \((440 - 240)\Omega\)
= \(200\Omega\)
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 = \((440 - 240)\Omega\)
= \(200\Omega\)