JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 27)
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 _______ $$\Omega$$.
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Answer
5
Explanation
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$$\begin{gathered} R_H=\frac{100 \times 100}{1000}=10 \Omega \\ i_H=\sqrt{\frac{62.5}{10}}=2.5 \mathrm{~A} \\ V_H=25 \mathrm{~V} \\ \Rightarrow \quad i_b=\frac{75}{10}=7.5 \mathrm{~A} \\ \Rightarrow \quad i_R=5 \mathrm{~A} \\ R=\frac{25}{5}=5 \Omega \end{gathered}$$
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