Question

When a coil is connected to a DC source of emf 12V, a current of 4 amp flows in it. If same coil is connected to 12V, 50Hz AC source, the current is 2.4 A. The self inductance of the coil is:

Answer 1

When connected to DC-

$R=\frac{V}{I}=\frac{12}{4}=3\Omega$

When connected to AC-

$\omega =2\pi f=100\pi$

$Z=\sqrt{R^2+X_L^2}$ and $X_L=\omega L$

$I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+{\omega }^2{L}^2}}$

$⟹R^2+{\omega }^2{L}^2=\frac{V^2}{I^2}=\frac{{12}^2}{{2.4}^2}=25$

$⟹9+(100\pi {)}^2{L}^2=25$

$⟹L=\sqrt{\frac{16}{(100\pi {)}^2}}=\frac{4}{100\pi }$

$⟹L=12.7mH$