Question

A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 $\Omega$ and a resistance $R$, to a 100 V mains as shown in the figure. For the heater to operate at 62.5 W, the value of $R$ should be $\dots$ $\Omega$.

Answer 1

The resistance of the heater is:
$R_{\text{heater}} = \frac{V^2}{P} = \frac{(100)^2}{1000} = 10 \, \Omega.$
For the heater operating at $P = 62.5 \, \text{W}$, the voltage across the heater is:
$P = \frac{V^2}{R} \implies V = \sqrt{PR}.$
Substitute $P = 62.5 \, \text{W}$ and $R = 10 \, \Omega$:
$V = \sqrt{62.5 \times 10} = 25 \, \text{V}.$
In the circuit, the voltage drop across the $10 \, \Omega$ resistor is:
$V_R = 100 - 25 = 75 \, \text{V}.$
The current through the $10 \, \Omega$ resistor is:
$i_1 = \frac{V_R}{R} = \frac{75}{10} = 7.5 \, \text{A}.$
The current through the heater is:
$i_H = \frac{V}{R} = \frac{25}{10} = 2.5 \, \text{A}.$
The current through $R$ is:
$i_R = i_1 - i_H = 7.5 - 2.5 = 5 \, \text{A}.$
Using Ohm’s law for $R$:
$V = i_R R \implies R = \frac{V}{i_R}.$
Substitute $V = 25 \, \text{V}$ and $i_R = 5 \, \text{A}$:
$R = \frac{25}{5} = 5 \, \Omega.$
Thus, the value of $R$ is:
$R = 5 \, \Omega.$