Question

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

• A. 0.5% increase
• B. 0.5% decrease
• C. 1% decrease
• D. 1.5% increase

Answer 1

Answer: (C)

1% decrease

Answer 2

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