Question

If the length of the filament of a heater is reduced by 10%, the power of the heater will

• A. Increase by about 9%
• B. Increase by about 11%
• C. Increase by about 19%
• D. decrease by about 10%

Answer 1

The correct option is B Increase by about 11%

We know that:

$P=\frac{V^2}{R}$ and $R=\frac{\rho L}{A}$ where $\rho =$ resistivity , $L=$ length of filament and $A=$ cross sectional area of filament wire.
Thus, $P=\frac{V^{2}A}{\rho L}$
After reduced length $10$, the length becomes, $L^{\prime }=L-\frac{L}{10}=\frac{9}{10}L$
So, $P^{\prime }=\frac{V^{2}A}{\rho L^{\prime }}=\frac{10V^{2}A}{9\rho L}$
Thus, $\frac{P^{\prime }}{P}=\frac{10}{9}$
$\therefore$ % of power increase $=\frac{P^{\prime }-P}{P}\times 100=(\frac{P^{\prime }}{P}-1)\times 100=(\frac{10}{9}-1)\times 100=\frac{100}{9}\sim 11$%