An electric bulb, rated as 200W at 100V, is used in a circuit having a 200V supply. The resistance R, that must be put in series with the bulb, so that the
bulb delivers the same power isΩ)
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Solution
Power, P=V2R RB=V21P=100×100200=50Ω
For the 200V supply, power is still equal to 200W
To deliver same power, the potential difference across the bulb should be equal to 100V. This will happen by connecting an equal resistance in series with the bulb, so that the potential difference across both, the bulb and the resistance, is 100V, each. ∴R=50Ω