JEE Main 2024 — 6 April, Shift 1 — Question 53

For DC voltage

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{5}=20 \Omega$

for AC voltage

$\mathrm{X}_{\mathrm{L}}=20 \sqrt{3} \Omega$

$\mathrm{R}=20 \Omega$ $\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}=\sqrt{3 \times 400+400}=40 \Omega$

Power $=\mathrm{i}_{\text {rms }}^{2} \mathrm{R}$

$$=\left(\frac{\mathrm{V}_{\text {rms }}}{\mathrm{Z}}\right)^{2} \times \mathrm{R}=\left(\frac{\frac{200}{\sqrt{2}}}{40}\right)^{2} \times 20=250 \mathrm{~W}$$