JEE Main 2025 — Alternating Current Question with Solution
JEE Main 2025 (4 Apr Shift 2)
Question
An inductor of self inductance 1 H connected in series with a resistor of and an ac supply of volt, 50 Hz. Maximum current flowing in the circuit is ________ A.
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Show full solutionCorrect answer: 1
Correct answer
1
Step-by-step explanation
Impedance of circuit
$\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}\right)^2} \\ & =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \\ & =\sqrt{(100 \pi)^2+(100 \pi)^2} \\ & =\sqrt{2} \times 100 \pi \\ & \mathrm{I}_{\mathrm{rms}}=\frac{V}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \\ & \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }
\end{aligned}$
Correct Answer : 1
$\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}\right)^2} \\ & =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \\ & =\sqrt{(100 \pi)^2+(100 \pi)^2} \\ & =\sqrt{2} \times 100 \pi \\ & \mathrm{I}_{\mathrm{rms}}=\frac{V}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \\ & \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }
\end{aligned}$
Correct Answer : 1
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This is a previous-year question from JEE Main 2025, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.