Question

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

• A. $7\Omega$
• B. $5\Omega$
• C. $10\Omega$
• D. $14\Omega$

Answer 1

The correct option is B $5\Omega$

The given circuit can be visualized as,


Power rating of heater is $(1000\text{W},100\text{V})$

Power, $P=\frac{V^2}{R}$

$\Rightarrow 1000=\frac{(100{)}^2}{R_H}$

Where, $R_H$ is the resistance of a heater.

$\Rightarrow R_H=10\Omega$

Also $P=i^2{R}_H$

Thus current required through heater to generate $625\text{W}$ power will be,

$62.5=i^2\times 10$

Or, $i=\sqrt{6.25}=2.5\text{A}$

The equivalent resistance of the circuit is,

$R_{eq}=10+\frac{R_{H}R}{R_H+R}$

Or, $R_{eq}=10+\frac{10R}{10+R}=\frac{100+20R}{10+R}$

Current supplied by source, $I=\frac{100}{R_{eq}}=\frac{100}{\left(\frac{100+20R}{10+R}\right)}$

Or, $I=\frac{100(10+R)}{100+20R}$

Or, $I=\frac{10(10+R)}{10+2R}$

Now this current will distribute along the heater $R_H$ and resistance $R$ in the inverse ratio of their resistances.

Current through heater;

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

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

$\Rightarrow 2.5=\frac{10(10+R)}{(10+2R)}\times \left(\frac{R}{R+10}\right)$

$\Rightarrow 2.5=\frac{10R}{10+2R}$

Or, $25+5R=10R$

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

Hence, option (b) is the correct answer.

Why this question? Tip: In problems involving power consumption in a circuit containing some electrical elements, always remember that power consumed by an element is a fixed quantity as per its rating, thus the current flowing through it obey the relation $P=V\times i$.