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 .................. $\Omega$.
JEE MAIN 2024, Diffcult
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$R_{\text {heater }}=\frac{V^2}{P}=\frac{(100)^2}{1000}=10 \Omega$
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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