JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2024
- Hard
- Numerical
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 \(R\), 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 _______ \(\Omega\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
\(R_{\text {heater }}=\frac{V^2}{P}=\frac{(100)^2}{1000}=10 \Omega\)
For heater \(P=\frac{V^2}{R} \Rightarrow V=\sqrt{P R}\)
\(\mathrm{V}=\sqrt{62.5 \times 10}\)
\(\mathrm{~V}=25 \mathrm{v}\)
(Image)
\(i_1=\frac{75}{10}=7.5 A, i_H=\frac{25}{10}=2.5 A\)
\(i_R=i_1-i_H=5\)
\(V=I R\)
\(R=\frac{25}{5}=5 \Omega\)
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