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

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q48 – JEE-MAIN 2025

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Home Physics Alternating CurrentQuestion #15

An inductor of self inductance

1

H is 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.

Enter Calculated Numerical Value:

Clue 1 of 3: Circuit Analysis

Identify the components and their arrangement in the AC circuit.

🧠 Full Solution Path

Step 1: Calculate Inductive Reactance✦ Active

First, calculate the angular frequency $ω$ and then the inductive reactance $X_{L}$ for the given inductor.

ω = 2 π f = 2 π (50 Hz) = 100 π rad/s

Inductive Reactance, X_{L} = ω L = (100 π rad/s) (1 H) = 100 π ohm

Step 2: Calculate Total Impedance○ Expand

For a series R-L circuit, the total impedance $Z$ is the vector sum of the resistance $R$ and the inductive reactance $X_{L}$ .

Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{(100 π)^{2} + (100 π)^{2}} = \sqrt{2 \cdot (100 π)^{2}} = 100 π \sqrt{2} ohm

Step 3: Calculate Maximum Current○ Expand

The given AC supply voltage is typically the RMS value. To find the maximum current $I_{m a x}$ , we need to convert the RMS voltage $V_{r m s}$ to peak voltage $V_{0}$ and then use Ohm's law for AC circuits.

V_{0} = V_{r m s} \sqrt{2} = (100 π V) \sqrt{2} = 100 π \sqrt{2} V

Maximum Current, I_{m a x} = \frac{V_{0}}{Z} = \frac{100 π \sqrt{2} V}{100 π \sqrt{2} ohm} = 1 A

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Physics Question 48 – JEE-MAIN 2025

Clue 1 of 3: Circuit Analysis

🧠 Full Solution Path