Question

A heater is designed to operate 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 $100V$ mains as shown in the figure. What will be the value of $R$ so that the heater operates with a power of $62.5W$?

• A. $15\Omega$
• B. $10\Omega$
• C. $5\Omega$
• D. $25\Omega$

Answer 1

Answer: (C)

$5\Omega$

Given that,

Power rating of a heater, $P=1000W$

Voltage rating of heater, $V=100V$

$\therefore$ Resistance of heater, $R=\frac{V^2}{P}=\frac{100\times 100}{1000}=10\Omega$

Given that, Power at which heater operates ,$P^{\prime }=62.5W$

Hence, Voltage drop across the heater, $V^{\prime }=\sqrt{R{P}^{\prime }}=\sqrt{10\times 62.5}=25V$

Since the voltage drop across the heater is $25V$ hence voltage drop across $10\Omega$ resistor is $,V_r=(100-25)=75V$

Current in the circuit, $I=\frac{V_r}{R}=\frac{75}{10}=7.5A$

Further, the current divides into two parts.

Let $I_1$ be the current that passes through the heater.

$\therefore 25=I_1\times 10⟹I_1=2.5A$

Hence, current through resistance $R$ is, $i=7.5-2.5=5A$

Applying Ohm's law across $R$, we get

$\therefore iR=25⟹R=\frac{25}{5}=5\Omega$

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