Question

The reactance of the capacitor is $350\Omega $, $R = 180\Omega $. Find X$_L$ if in a series LCR circuit current leads the voltage by 53^o
A. $120\Omega $
B. $140\Omega $
C. $210\Omega $
D. $110\Omega $

Answer 1

Hint: LCR Circuit is an electronic circuit which consists of inductor, capacitor and a resistor connected in series or parallel. LCR circuits find a wide range of applications in Radio and Communication technology.

Complete step by step solution:
$\tan \phi $ in terms of impedance is given by the formula:
$\tan \phi = \dfrac{{{X_C} - {X_L}}}{R}$ Equation 1
Where,
$\phi $ is the phase difference between current and voltage
X_L is the reactance of inductor
X_C is the reactance of capacitor
R is the resistance of circuit

Insert all the values in the above formula and then compute the value of X$_L$.

Given in the question:
Reactance of the capacitor (X$_C$) as $350\Omega $
$R = 180\Omega $
Current leads voltage by 53^O (i.e. $\phi = {53^o}$)

Inserting all the values in Equation 1,
We get,
$ = > \tan {53^o} = \dfrac{{350 - {X_L}}}{{180}}$
The value of $\tan {53^o}$is $\dfrac{4}{3}$.
Inserting the value of $\tan {53^o}$,
We get,
$ = > \dfrac{4}{3} = \dfrac{{350 - {X_L}}}{{180}}$
$ = > \dfrac{4}{3} \times 180 = 350 - {X_L}$
$ = > 240 = 350 - {X_L}$
$ = > {X_L} = 110\Omega $

Hence Option (D) is correct.

Note: This is a tricky question often asked in competitive examinations. Solving such questions requires in-depth knowledge of LCR Circuits. One must also need to memorize the values of Sine and cosine 37^O and 53^O. Also, this is a calculation intensive problem. Silly mistakes must be avoided at all costs.