Question

How will you find the inductance of a choke coil needed to run an arc lamp with an AC source of $410V$ supply at $50Hz$ ? The arc runs at $10A$ current and has an effective resistance $40 \Omega$.

Answer 1

To find the value of inductance of a choke coil, first we should find the impedance. So, first we will write the formula of electric current in the terms of its voltage and impedance(as current and voltage is given). And then we will go through the formula of reactance of the choke coil to conclude the equation in terms of inductance. And then we will write the formula for the LR circuit.

Complete answer:
The current required by the arc lamp is $10A$ .
To find the inductance, we have to first find the impedance of the arc lamp and choke oil together.
Let $Z$ be the impedance of the arc lamp and choke coil together. Current $I$ is given by:
$\therefore I = \dfrac{V}{Z}$
where, $I$ is the electric current,
$V$ is the voltage and
$Z$ is the impedance.
$\Rightarrow 10 = \dfrac{{410}}{Z}$
$\Rightarrow Z = 41\Omega$ ……….(i)
And as we know that the reactance of the choke coil:
$\because \omega L = 2\pi f \times L$ ……….(ii)
where $f$ is the frequency of the voltage source.
Also, we know that in a LR circuit-
$\therefore {Z^2} = {R^2} + {(\omega L)^2}$ ……..(iii)
where, $Z$ is the impedance,
$R$ is the resistance,
$L$ is the inductance.
Now, put the value of impedance and resistance in eq(iii)-
$\Rightarrow {41^2} = {40^2} + {(\omega L)^2} \\\ \Rightarrow {(\omega L)^2} = 81 \\\ \Rightarrow \omega L = 9 \\\$
Now, we will put the above value in eq(ii):-
$\therefore 9 = 2\pi \times 50 \times L \\\ \Rightarrow L = 0.029H\,or\,29mH \\\$
Therefore, the required inductance of the choke coil is $29mH$.

Note:
A choke has a low resistance and a high inductance. Because an inductor is a non resistive device, it does not produce heat in the same way that a resistance does. To save electricity, we utilise a choke with a low resistance and a large inductance.