Question 311: AC fundamentals

Question

When a coil is connected to a d.c. source of e.m.f. 12 volt; then the current of 4 A flows in it . If the same coil is connected to a 12 volt, 50 Hz a.c. source, then the current flowing in it is 2.4 A . Then self-inductance of the coil will be

Accepted Answer

The self-inductance of the coil can be calculated using the relationship between the reactance in an AC (alternating current) circuit and the impedance in a DC (direct current) circuit. Here's how you can determine the self-inductance:
Resistance in DC Circuit
:
$    R = \frac{V_{	ext{dc}}}{I_{	ext{dc}}} = \frac{12}{4} = 3 \, \Omega    $
Impedance in AC Circuit
:
$    Z = \frac{V_{	ext{ac}}}{I_{	ext{ac}}} = \frac{12}{2.4} = 5 \, \Omega    $
Reactance in AC Circuit
:
Since the impedance in an AC circuit with resistance and inductance is a combination of both, expressed as:
$    Z = \sqrt{R^2 + (X_L)^2}    $
We know $ X_L $, the inductive reactance, and $ R $ the resistance, can be used to find:
$    X_L = \sqrt{Z^2 - R^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, \Omega    $
Self-Inductance Calculation
:
The formula for inductive reactance $ X_L $ is given by:
$    X_L = 2 \pi f L    $
Solving for $ L $:
$    L = \frac{X_L}{2 \pi f} = \frac{4}{2 \pi 	imes 50} = \frac{4}{100 \pi} = \frac{4}{\pi} 	imes 10^{-2} \, 	ext{H}    $
Thus, the self-inductance of the coil is given by option
C
:
$ \frac{4}{\pi} 	imes 10^{-2} \, 	ext{H} $

Answer

48 H

Answer

12 H

Answer

$\frac{4}{\pi} 	imes 10^{-2} \mathrm{H}$

Answer

$\frac{8}{\pi} 	imes 10^{-2} \mathrm{H}$

AC fundamentals

When a coil is connected to a d.c. source of e.m.f. 12 volt; then the current of 4 A flows in it . If the same coil is connected to a 12 volt, 50 Hz a.c. source, then the current flowing in it is 2.4 A . Then self-inductance of the coil will be

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