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
The correct answer is: 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 = \((440 - 240)\Omega\)
= \(200\Omega\)