Question

A coil of inductance 0.50 H and resistance $100\Omega$ is connected to a 240 V, 50 Hz ac supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the current maximum?

Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Answer 1

Given:

The inductance of the coil is $0.50H$.

The resistance of the coil is $100\Omega$

The voltage supplied to the coil is $240V$.

a)

The peak voltage supplied to the coil is given by:

$V_o=\sqrt{2}V$

The maximum current through the coil is given by:

$I_o=\frac{V_o}{\sqrt{R^2+{\omega }^2{L}^2}}$

$I_o=\frac{\sqrt{2}\times 240}{\sqrt{{10}^4+(100\pi \times 0.5{)}^2}}$

$I_o=1.3\times {10}^{-2}A$

b)

The relation between the time lag and the inductance nad resistance of the coil is:

$tan\varphi =\frac{\omega L}{R}$

$tan\varphi =\frac{2\pi \times 50\times 0.5}{100}$

$tan\varphi =1.571$

$\Rightarrow \varphi ={57.5}^o$

$\omega t=57.5\pi /180$

$\Rightarrow t=3.19\times {10}^{-3}s$