An inductor of self inductance 1 H connected in series with a resistor of 100 π ohm and an ac supply of 100 π volt, 50 Hz. Maximum current flowing in the circuit is ________ A.

Correct answer: 0 Impedance of circuit aligned & Z= R^2+(X_ L )^2 = R^2+(ω_ L )^2 & = (100 π)^2+(2 π × 50 × 1)^2 & = (100 π)^2+(100 π)^2 & = 2 × 100 π & I_ rms =(V)/(2)= 100 π 2 × 100 π = 1 2 & I_ = 2 I_ rms = 2 × 1 2 =1 Ampere aligned Correct Answer : 1

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2025 · 4 April · Evening Shift · Physics · Alternating Current

An inductor of self inductance 1 H connected in series with a...

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

Step-by-step solution

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

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