Question

A heater is designed to operate with a power of $1000\text{W}$ in a $100\text{V}$ line. It is connected, in combination with a resistance $R$, to a $100\text{V}$ main supply as shown in Figure. What should be the value of $R$ such the heater may operate with a power of $62.5\text{W}$?

• A. $50\Omega$
• B. $25\Omega$
• C. $15\Omega$
• D. $5\Omega$

Answer 1

The correct option is D $5\Omega$

Given,

Power of the heater, $P=1000\text{W}$,

Voltage across it, $V=100\text{V}$,

So, the resistance of the heater is,

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


The power on which it operates is $P^{\prime }=62.5\text{W}$.

So, potential drop across heater,

$V^{\prime }=\sqrt{R\times P^{\prime }}=\sqrt{10\times 62.5}=\sqrt{625}=25\text{V}$

The potential drop across $AB$,

$V_{AB}=100-25=75\text{V}$

$\therefore$The current in $AB$,

$I=\frac{V_{AB}}{R}=\frac{75}{10}=7.5\text{A}$

This current divides into two parts.
Let $I_1$ be the current that passes through the heater. Therefore,

$25=I_1\times 10$

$\Rightarrow I_1=2.5\text{A}$

So, the current through $R$ is,

$(7.5-2.5=5\text{A})$.

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

$25=5\times R$

$\Rightarrow R=5\Omega$

Hence, option (d) is the correct answer.