Q.
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 $62.5W$
A) $10\Omega$
B) $62.5\Omega$
C) $\frac{1}{5}\Omega$
D) $5\Omega$
Explanation
(d) Resistance of heater
$R_{H} = \frac{V^{2}}{P} = \frac{(100)^{2}}{1000} = 10\ \Omega$
Total resistance of circuit
$R^{'} = 10 + \frac{10 \times R}{10 + R} = \frac{100 + 20R}{10 + R}$
Current in heater
$I_{H} = I.\frac{R}{10 + R} = \frac{100}{\frac{100 + 20R}{10 + R}} \times \frac{R}{(10 + R)} = \frac{5 \times R}{5 + R}$
$\therefore Power\ P = I_{H}^{2}R_{H} \Rightarrow \frac{25R}{(5 + R)^{2}} \times 10 = 62.5$
$\therefore R = 5\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.