A coil of negligible resistance is connected in series with 90 \Omega resistor across 120 ~V, 60 ~Hz supply. A voltmeter reads 36 ~V across resistance. Inductance of the coil is :

Alternating Current PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q79 with Solution

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Correct answer: B

Given data

Resistance in series: $R = 90\,\Omega$
Supply voltage: $V = 120\,\text{V}$
Frequency: $f = 60\,\text{Hz}$
Voltmeter reading across resistor: $V_R = 36\,\text{V}$
Coil has negligible resistance, so it is a pure inductor of inductance $L$

Find the current in the circuit

Since the resistor and inductor are in series, the same current flows through both.

Across the resistor,

V_R = IR

So,

I = \frac{V_R}{R} = \frac{36}{90} = 0.4\,\text{A}

Find the total impedance

For the whole series circuit,

V = IZ

Thus,

Z = \frac{V}{I} = \frac{120}{0.4} = 300\,\Omega

Use impedance relation for series $R-L$ circuit

For a resistor and pure inductor in series,

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

where $X_L = \omega L$ is inductive reactance.

So,

300 = \sqrt{90^2 + X_L^2}

Squaring both sides,

300^2 = 90^2 + X_L^2

90000 = 8100 + X_L^2

X_L^2 = 81900

X_L = \sqrt{81900} \approx 286.2\,\Omega

Calculate inductance

We know,

X_L = \omega L = 2\pi f L

Hence,

L = \frac{X_L}{2\pi f}

Substitute values:

L = \frac{286.2}{2\pi \times 60}

L \approx \frac{286.2}{376.99} \approx 0.76\,\text{H}

Check options

A: $0.91\,\text{H}$
B: $0.76\,\text{H}$
C: $2.86\,\text{H}$
D: $0.286\,\text{H}$

Thus, the correct option is:

\boxed{0.76\,\text{H}}

which is Option B.

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View written solutionFree

Correct answer: B

Given data

Resistance in series: $R = 90\,\Omega$
Supply voltage: $V = 120\,\text{V}$
Frequency: $f = 60\,\text{Hz}$
Voltmeter reading across resistor: $V_R = 36\,\text{V}$
Coil has negligible resistance, so it is a pure inductor of inductance $L$

Find the current in the circuit

Since the resistor and inductor are in series, the same current flows through both.

Across the resistor,

V_R = IR

So,

I = \frac{V_R}{R} = \frac{36}{90} = 0.4\,\text{A}

Find the total impedance

For the whole series circuit,

V = IZ

Thus,

Z = \frac{V}{I} = \frac{120}{0.4} = 300\,\Omega

Use impedance relation for series $R-L$ circuit

For a resistor and pure inductor in series,

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

where $X_L = \omega L$ is inductive reactance.

So,

300 = \sqrt{90^2 + X_L^2}

Squaring both sides,

300^2 = 90^2 + X_L^2

90000 = 8100 + X_L^2

X_L^2 = 81900

X_L = \sqrt{81900} \approx 286.2\,\Omega

Calculate inductance

We know,

X_L = \omega L = 2\pi f L

Hence,

L = \frac{X_L}{2\pi f}

Substitute values:

L = \frac{286.2}{2\pi \times 60}

L \approx \frac{286.2}{376.99} \approx 0.76\,\text{H}

Check options

A: $0.91\,\text{H}$
B: $0.76\,\text{H}$
C: $2.86\,\text{H}$
D: $0.286\,\text{H}$

Thus, the correct option is:

\boxed{0.76\,\text{H}}

which is Option B.

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