Question

A current of $4A$ flows in a coil when connected to a $12Vdc$ source . If the same coil is connected to a $12V,50rad/s$ a.c source, a current of $2.4A$ flows in the circuit. Determine the inductance of the coil.

Answer 1

Resistance of coil $R=\frac{V_{dc}}{I_{dc}}=\frac{12}{4}=3\Omega$
Rms value of AC source $V_{rms}=12V$
Rms value of alternating current $I_{rms}=2.4A$
Impedance of circuit $Z=\frac{V_{rms}}{I_{rms}}=\frac{12}{2.4}=5\Omega$
Using $Z^2=R^2+X_L^2$
We get $5^2=3^2+X_L^2$
$⟹X_L=\sqrt{4^2}=4\Omega$
Frequency of AC source $f=50Hz$
Thus inductance of circuit $L=\frac{X_L}{2\pi f}=\frac{4}{2\pi \times 50}=0.0127H=12.7mH$