A heater is designed to operate with a power of $1000 W$ in a $100V$ line. It is connected in combination with a resistance of $10\Omega$ and a resistance $R$, to a $100 V$ mains as shown in the figure. What will be the value of $R$ so that the heater operates with a power of $62.5W$?

From $P=\frac{V^2}{R}$

Resistance of heater,

From Current required across heater for power of $62.5W$,

$R=\frac{V^2}{P}=\frac{(100{)}^2}{1000}=10\Omega$

$P=i^{2}R$

$i=\sqrt{\frac{P}{R}}=\sqrt{\frac{62.5}{10}}=2.5A$

Main current in the circuit, $I=\frac{100}{10+\frac{10R}{10+R}}=\frac{100(10+R)}{100+20R}=\frac{10(10+R)}{10+2R}$

This current will distribute in inverse ratio of resistance between heater and $R$

$i=\left(\frac{R}{10+R}\right)I$

or

$2.5=\left(\frac{R}{10+R}\right)\left[\frac{10(10+R)}{10+2R}\right]=\frac{10R}{10+2R}$

Solving this equation, we get

$R=5\Omega$