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  4. If the current in an electric bulb decreases by 0.5% then power in the bulb changes by

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Q. If the current in an electric bulb decreases by 0.5% then power in the bulb changes by

Current Electricity

Solution:

Using P = $I^2R $ New power, P' = $\left( I - \frac{0.5I}{100} \right)^2 R = I^2 R \left( 1 - \frac{1}{200} \right)^2 = I^2R \left( \frac{199}{200} \right)^2$ = $0.99I^2R = $ 0.99 P $\therefore $ % decrease in power = $\frac{P' - P'}{P} \times 100 = \frac{P - 0 .99P}{P} \times 100$ = 1