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 Ω

Question

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

Answer 1

Correct answer is : 5 

\(\mathrm{R}_{\text {heater }}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(100)^{2}}{1000}=10 \Omega\)

For heater \(\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow \mathrm{V}=\sqrt{\mathrm{PR}}\)

\(\mathrm{V}=\sqrt{62.5 \times 10}\)

\(\mathrm{V}=25 \ \mathrm{v}\) 

\(\mathrm{i}_{1}=\frac{75}{10}=7.5 \mathrm{~A}, \quad \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{~A}\).

\(\mathrm{i}_{\mathrm{R}}=\mathrm{i}_{1}-\mathrm{i}_{\mathrm{H}}=5\)

\(\mathrm{V}=\mathrm{IR}\)

\(\mathrm{R}=\frac{25}{5}=5 \Omega\)