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 $11\%$

Let the original length of filament be $l$.

The new length, $l^{\prime }=l-\frac{10}{100}l=0.9l$

Initial resistance was $R=\frac{\rho l}{A}$

The new resistance is $R^{\prime }=\frac{\rho l^{\prime }}{A}=\frac{0.9\rho l}{A}=0.9R$

Initial power, $P=\frac{V^2}{R}$

And final power will be

$P^{\prime }=\frac{V^2}{R^{\prime }}=\frac{V^2}{0.9R}=\frac{P}{0.9}=1.11P$

$\Rightarrow P^{\prime }=1.11P=P(1+\frac{11}{100})$

$\therefore P^{\prime }=P+\frac{11}{100}P$

This means power increased by $11\%$.

Hence, option $\left(b\right)$ is correct.

Alternative solution:

$P=\frac{V^2}{R}=\frac{V^2}{\rho \frac{l}{a}}\Rightarrow P\propto \frac{1}{l}$

For same voltage

$P^{\prime }\propto \frac{1}{l-\frac{10}{100}l}\propto \frac{1}{0.9l}$

$\therefore \frac{P^{\prime }}{P}=\frac{10}{9}$

So, percentage change in power will be

$\frac{P^{\prime }-P}{P}\times 100=(\frac{P^{\prime }}{P}-1)\times 100$

$\therefore \frac{\Delta P}{P}\times 100=(\frac{10}{9}-1)\times 100\approx 11$

Key concept: For same voltage $\text{Power}\propto \frac{1}{\text{Resistance}}\propto \frac{1}{\text{length}}$