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Alternating Current question

2024 · 9 Apr · Shift 1 · Q89

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Alternating Current
2024 · 9 Apr · Shift 1 · Q89

Alternating Current question

2024 · 9 Apr · Shift 1 · Q89

JEE MainPhysicsAlternating CurrentNumerical+4 / −1

When a coil is connected across a

20 \mathrm{~V}

dc supply, it draws a current of

5 \mathrm{~A}

. When it is connected across

20 \mathrm{~V}, 50 \mathrm{~Hz}

ac supply, it draws a current of

4 \mathrm{~A}

. The self inductance of the coil is

\underline{\hspace{2cm}}\mathrm{mH}

. (Take

\pi=3

)

Numerical answer

View written solutionFree

Correct answer: 10

Find the resistance of the coil from DC supply

When connected to a DC source, the inductive reactance is zero in steady state, so only resistance acts.

Given:

DC voltage $V = 20\,\text{V}$
DC current $I = 5\,\text{A}$

So,

R = \frac{V}{I} = \frac{20}{5} = 4\,\Omega

Find the impedance of the coil from AC supply

Given for AC:

AC voltage $V = 20\,\text{V}$
AC current $I = 4\,\text{A}$

Thus impedance,

Z = \frac{V}{I} = \frac{20}{4} = 5\,\Omega

Use the relation between impedance, resistance, and inductive reactance

For a coil,

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

Substitute $Z=5$ and $R=4$ :

5^2 = 4^2 + X_L^2

25 = 16 + X_L^2

X_L^2 = 9

X_L = 3\,\Omega

Relate inductive reactance to self inductance

We know,

X_L = \omega L = 2\pi f L

Given:

$f = 50\,\text{Hz}$
$\pi = 3$

So,

3 = 2 \times 3 \times 50 \times L

3 = 300L

L = \frac{3}{300} = 0.01\,\text{H}

Convert into mH:

0.01\,\text{H} = 10\,\text{mH}

Final Answer

\boxed{10}

The derived answer matches the stored correct answer.

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