Question

When a coil is connected across a 20 V dc supply, it draws a current of 5 A. When it is connected across 20 V, 50 Hz ac supply, it draws a current of 4 A. The self inductance of the coil is ______ mH.
(Take $\pi$ = 3)

Answer 1

Case 1: DC Circuit When the coil is connected to a DC supply, the inductive reactance is 0, so:

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

Case 2: AC Circuit When the coil is connected to an AC supply:

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

where $Z$ is the impedance of the coil, given by:

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

Substitute $R = 4 \, \Omega$:

$5 = \sqrt{4^2 + X_L^2} \implies X_L^2 = 25 - 16 = 9 \implies X_L = 3 \, \Omega.$

The inductive reactance $X_L$ is related to the self-inductance $L$ by:

$X_L = 2 \pi f L \implies L = \frac{X_L}{2 \pi f}.$

Substitute $X_L = 3 \, \Omega$, $f = 50 \, \text{Hz}$, and $\pi = 3$:

$L = \frac{3}{2 \cdot 3 \cdot 50} = \frac{3}{300} = 0.01 \, \text{H} = 10 \, \text{mH}.$

Final Answer: 10 mH.