Question

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

• A. $10\Omega$
• B. $62.5\Omega$
• C. $1/5\Omega$
• D. $5\Omega$

Answer 1

Answer: (D)

Resistance of heater,
$R_H=\frac{V^2}{P}=\frac{{\left(100\right)}^2}{1000}=10\Omega$
Total resistance of circuit,
$R^{\prime }=10+\left(\frac{10\times R}{10+R}\right)=\frac{100+20R}{10+R}$
current in heater,
$I_H=I\times \frac{R}{\left(10+R\right)}\times \frac{100}{\left(\frac{100+R}{10+R}\right)}\times \frac{R}{\left(10+R\right)}=\frac{5\times R}{5+R}$
So, Power,
$P=I_H^2{R}_H=\frac{25R}{{\left(5+R\right)}^2}\times 10=62.5$
$\therefore R=5\Omega$