Question

An arc lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:
A. 80H
B. 0.08H
C. 0.044H
D. 0.065H

Answer 1

For the AC circuit containing a resistor and an inductance, the Ohm’s law is given in terms of the impedance of the circuit. The resistance of the lamp can be found from the values of direct current and its voltage rating. By solving the final expression for inductance, we can find the required answer.

Complete answer:
We are given an arc lamp. We are given its current and voltage values. They are given as
$V = 80V \\\ I = 10A \\\$
Therefore, the resistance of the arc lamp can be calculated in the following way by using Ohm’s law.
$R = \dfrac{V}{I} = \dfrac{{80}}{{10}} = 8\Omega$
Now this lamp is connected to a 220V (rms), 50 Hz AC supply. Therefore, we have
$V' = 220V \\\ \nu = 50Hz \\\$
For a circuit containing a resistance and an inductance, the impedance of the circuit is given as
$Z = \sqrt {{R^2} + {{\left( {\omega L} \right)}^2}} = \sqrt {{R^2} + {{\left( {2\pi \nu L} \right)}^2}}$
Here L is the inductance of the inductor of the given inductor which we need to find out.
So, the Ohm’s law can be written in the following way.
$V' = IZ = I\sqrt {{R^2} + {{\left( {2\pi \nu L} \right)}^2}}$
Now we can insert all the known values in this expression. Doing so, we get
$\Rightarrow 220 = 10\sqrt {{8^2} + {{\left( {2\pi \times 50L} \right)}^2}} \\\ \Rightarrow \sqrt {64 + {{\left( {314L} \right)}^2}} = 22 \\\$
Squaring both sides, we get
$\Rightarrow 64 + {\left( {314L} \right)^2} = 484 \\\ \Rightarrow 98596{L^2} = 484 - 64 = 420 \\\ \Rightarrow {L^2} = \dfrac{{420}}{{98596}} = 0.00425 \\\ \Rightarrow L = \sqrt {0.00425} \simeq 0.065H \\\$

So, the correct answer is “Option D”.

Additional Information:
In case, if the given circuit also includes a capacitor along with a resistance and an inductance then the impedance of the circuit can be written as
$Z = \sqrt {{R^2} + {{\left( {\omega L - \dfrac{1}{{\omega C}}} \right)}^2}}$
Here C is the capacitance in the LCR circuit.

Note:
1. The amount of current will remain the same in the circuit whether we use an AC source or a DC source.
2. The impedance of the given AC circuit is resultant of the resistance and the impedance of the inductor connected to the circuit. With increase in the frequency of the AC signal, the impedance due to the inductor increases.