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

Correct Answer: 5

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:
\[ R = 5 \, \Omega. \]