Question

The current through a bulb is increased by $1 \%$. Assuming that the resistance of the filament remains unchanged the power of the bulb will

• A. increase by 1%
• B. decrease by 1%
• C. increase by 2%
• D. decrease by 2%

Answer 1

Answer: (C)

increase by 2%

Answer 2

Power $P=I^{2} R$
$\therefore P \propto I^{2}$
Then $\frac{R_{1}}{P_{2}}=\frac{I^{2}}{\left(\frac{101}{100}\right)^{2} I^{2}}$
Percentage increase in power
$\left(\frac{P_{2}}{R_{1}}-1\right) \%$
$=\left(\frac{101 \times 101}{100 \times 100}-1\right) \%$
$=2 \%$