Question

$220\,V$, $50\,Hz$, AC source is connected to an inductance of $0.2\,H$ and a resistance of $20\,\Omega$ in series. What is the current in the circuit?
(A) $3.33\,A$
(B) $33.3\,A$
(C) $5\,A$
(D) $10\,A$

Answer 1

Hint
The current in the inductor is determined by using the relation of the current, voltage and the impedance. The impedance of the circuit is determined by using the inductive reactance and the resistance. The inductive reactance is determined by using the frequency and the inductance.
The inductive reactance is given by,
$\Rightarrow {X_L} = 2\pi fL$
Where, ${X_L}$ is the inductive reactance, $f$ is the frequency and $L$ is the inductance.
The total circuit impedance is given by,
$\Rightarrow Z = \sqrt {{{\left( {{X_L}} \right)}^2} + {R^2}}$
Where, $Z$ is the impedance, ${X_L}$ is the inductive reactance and $R$ is the resistance.
The current in the inductor is given by,
$\Rightarrow I = \dfrac{V}{Z}$
Where, $I$ is the current, $V$ is the voltage and $Z$ is the impedance.

Complete step by step answer
Given that, The voltage of the circuit is, $V = 220\,V$
The frequency of the circuit is, $f = 50\,Hz$
The inductance of the inductor is, $L = 0.2\,H$
The resistance of the circuit is, $R = 20\,\Omega$.
Now, The inductive reactance is given by,
$\Rightarrow {X_L} = 2\pi fL\,.................\left( 1 \right)$
By substituting the frequency and the inductance and the $\pi$ value in the above equation (1), then the above equation (1) is written as,
$\Rightarrow {X_L} = 2 \times 3.14 \times 50 \times 0.2$
On multiplying the above equation then the above equation is written as,
$\Rightarrow {X_L} = 62.8\,\Omega$
Now, The total circuit impedance is given by,
$\Rightarrow Z = \sqrt {{{\left( {{X_L}} \right)}^2} + {R^2}} \,....................\left( 2 \right)$
By substituting the inductive reactance and the resistance in the above equation (2), then the above equation (2) is written as,
$\Rightarrow Z = \sqrt {{{\left( {62.8} \right)}^2} + {{20}^2}}$
On squaring the both the terms inside the root, then the above equation is written as,
$\Rightarrow Z = \sqrt {3943.84 + 400}$
By adding the terms inside the root, then the above equation is written as,
$\Rightarrow Z = \sqrt {4343.84}$
On taking the square root, then the above equation is written as,
$\Rightarrow Z = 65.9\,\Omega$.
Now, The current in the inductor is given by,
$\Rightarrow I = \dfrac{V}{Z}\,.................\left( 3 \right)$
By substituting the current and the impedance in the above equation, then the above equation is written as,
$\Rightarrow I = \dfrac{{220}}{{65.9}}$
On dividing the above equation, then the above equation is written as,
$\Rightarrow I = 3.33\,A$
Hence, the option (A) is the correct answer.

Note
In the equation (2), the impedance is the square root of the sum of the individual square of the resistance and the total reactance. But here we use only the inductive reactance because in this circuit the inductor only used, so it is taken. If the circuit contains both the inductor and the capacitor, then the total reactance is taken.