JEE Main 2024 — 8 April, Shift 2 — Question 51

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

For heater $P=\frac{V^{2}}{R} \Rightarrow V=\sqrt{P R}$ $\mathrm{V}=\sqrt{62.5 \times 10}$

$\mathrm{V}=25 \mathrm{v}$

$\mathrm{i}_{1}=\frac{75}{10}=7.5 \mathrm{~A}, \quad \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{~A}$. $\mathrm{i}_{\mathrm{R}}=\mathrm{i}_{1}-\mathrm{i}_{\mathrm{H}}=5$

$\mathrm{V}=\mathrm{IR}$

$\mathrm{R}=\frac{25}{5}=5 \Omega$