Q: An ideal choke coil takes a current of 8 ampere when connected to an AC supply of 100 volt and 50 Hz. A pure resistor under the same condition takes a current of 10 ampere. If the two are connected to an AC supply of 150 volts and 40 Hz. Then the current in a series combination of the above resistor and inductor is

Sol: For pure inductor,

$\large X_L = \frac{100}{8} = \frac{25}{2} ohm$

$\large \omega L = \frac{25}{2} ohm$

$\large L = \frac{25}{2 \omega } = \frac{25}{2 \times 2 \pi \times 50} $

= 1/8π H

$\large R = \frac{V}{I}= \frac{100}{10} $

= 10 ohm

For the combination, the supply is 150 v, 40 Hz

∴ X_{L} = ω L = 2π × 40 × (1/8π ) = 10 ohm

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

$\large Z = \sqrt{10^2 + 10^2 } $

= 10√2 ohm

$\large I_v = \frac{E_v}{Z} = \frac{150}{10\sqrt{2}} = \frac{15}{\sqrt{2}} A$