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

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

Tardigrade Admin · Accepted Answer

Using P = I2R New power, P' = ( I - (0.5I/100) )2 R = I2 R ( 1 - (1/200) )2 = I2R ( (199/200) )2 = 0.99I2R = 0.99 P ∴ % decrease in power = (P' - P'/P) × 100 = (P - 0 .99P/P) × 100 = 1

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

0.5% increase

0.5% decrease

1% decrease

1.5% increase

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

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

0.5% increase

0.5% decrease

1% decrease

1.5% increase

Using P = I2R New power, P' = (I−1000.5I​)2R=I2R(1−2001​)2=I2R(200199​)2 = 0.99I2R= 0.99 P ∴ % decrease in power = PP′−P′​×100=PP−0.99P​×100 = 1

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