Q51 / 75 · JEE Main 2025 (4 Apr Shift 2) · Physics
An inductor of self inductance 1 H connected in series with a resistor of $100 \pi \mathrm{ohm}$ and an ac supply of $100 \pi$ volt, 50 Hz. Maximum current flowing in the circuit is ________ A.
Correct answer: 1
Solution
Impedance of circuit
$\begin{aligned}<br/>& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}\right)^2} \\ & =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \\ & =\sqrt{(100 \pi)^2+(100 \pi)^2} \\ & =\sqrt{2} \times 100 \pi \\ & \mathrm{I}_{\mathrm{rms}}=\frac{V}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \\ & \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }<br/>\end{aligned}$
Correct Answer : 1
$\begin{aligned}<br/>& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}\right)^2} \\ & =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \\ & =\sqrt{(100 \pi)^2+(100 \pi)^2} \\ & =\sqrt{2} \times 100 \pi \\ & \mathrm{I}_{\mathrm{rms}}=\frac{V}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \\ & \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }<br/>\end{aligned}$
Correct Answer : 1
Hence, the correct answer is 1.