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Alternating Current question

2025 · 4 Apr · Shift 2 · Q72

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Alternating Current
2025 · 4 Apr · Shift 2 · Q72

Alternating Current question

2025 · 4 Apr · Shift 2 · Q72

JEE MainPhysicsAlternating CurrentNumerical+4 / −1

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

\underline{\hspace{2cm}}

A.

Numerical answer

View written solutionFree

Correct answer: 1

Given data

Inductance: $L = 1\,\text{H}$
Resistance: $R = 100\pi\,\Omega$
AC voltage: $V = 100\pi\,\text{V}$
Frequency: $f = 50\,\text{Hz}$

We interpret the given AC supply voltage as rms voltage, which is standard unless otherwise stated.

Find angular frequency

\omega = 2\pi f = 2\pi(50) = 100\pi\,\text{rad/s}

Find inductive reactance

X_L = \omega L = (100\pi)(1) = 100\pi\,\Omega

Find impedance of series $R$ - $L$ circuit

Z = \sqrt{R^2 + X_L^2}

Since $R = 100\pi$ and $X_L = 100\pi$ ,

Z = \sqrt{(100\pi)^2 + (100\pi)^2}

Z = 100\pi\sqrt{2}\,\Omega

Find rms current

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}}\,\text{A}

Find maximum current

For sinusoidal current,

I_0 = \sqrt{2}\, I_{\text{rms}}

So,

I_0 = \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 1\,\text{A}

Final answer

\boxed{1}

The maximum current flowing in the circuit is $1\,\text{A}$ .

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