Question

An inductance of negligible resistance whose reactance is $100\,\Omega$ at $200\,Hz$ is connected to a $240\,V$, $50\,Hz$ power line. The current in the inductor is:
(A) $0.6\,A$
(B) $9.6\,A$
(C) $0.10\,A$
(D) $5.5\,A$

Answer 1

The inductor of the current line can be determined by using the reactance of the inductor formula, and by using the given information in the reactance of the inductor formula, then the inductor in the current line can be determined.

Formula used:
From the inductive reactance, the reactance of the inductor is given by,
${X_L} = 2\pi fL$
Where, ${X_L}$ is the inductive reactance in the inductor, $f$ is the frequency of the inductor and $L$ is the inductor.
The current in the inductor is given by,
$I = \dfrac{V}{{X_L^1}}$
Where, $I$ is the current, $V$ is the voltage and $X_L^1$ is the net reactance.
The net reactance is given by,
$X_L^1 = 2\pi FL$
Where, $X_L^1$ is the net reactance, $F$ is the frequency of the power line and $L$ is the inductor.

Complete step by step answer:
Given that,the inductive reactance of the inductor is, ${X_L} = 100\,\Omega$,
The frequency of the inductor is, $f = 200\,Hz$,
The voltage of the circuit is, $V = 240\,V$,
The frequency of the power line, $F = 50\,Hz$.
Now,
From the inductive reactance, the reactance of the inductor is given by,
${X_L} = 2\pi fL\,...............\left( 1 \right)$
By substituting the inductive reactance in the inductor and the frequency of the inductor in the above equation (1), then the equation (1) is written as,
$100 = 2 \times 3.14 \times 200 \times L$
By multiplying the terms in the above equation, then the above equation is written as,
$100 = 1256 \times L$
By keeping the terms inductor in one side and the other terms in the other side, then the above equation is written as,
$L = \dfrac{{100}}{{1256}}$
By dividing the terms in the above equation, then the above equation is written as,
$L = 0.079\,H$
Now,
The net reactance is given by,
$X_L^1 = 2\pi FL\,...............\left( 2 \right)$
By substituting the frequency and inductor in the above equation (2), then
$X_L^1 = 2 \times 3.14 \times 50 \times 0.079$
By multiplying the terms in the above equation, then the above equation is written as,
$X_L^1 = 24.8\,\Omega$
Now,
The current in the inductor is given by,
$I = \dfrac{V}{{X_L^1}}\,..............\left( 3 \right)$
By substituting the current and the net reactance in the above equation, then
$I = \dfrac{{240}}{{24.8}}$
By dividing the terms in the above equation, then the above equation is written as,
$\therefore I = 9.6\,A$

Hence, the option B is the correct answer.

Note: The current in the inductor is also determined by using the ohm’s law. The net reactance is assumed as the resistance, then by ohm’s law the current in the circuit is equal to the voltage in the circuit divided by the resistance or the reactance.