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 53o
A. $120\Omega$
B. $140\Omega$
C. $210\Omega$
D. $110\Omega$

Answer 1

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
XL is the reactance of inductor
XC 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 53O (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 37O and 53O. Also, this is a calculation intensive problem. Silly mistakes must be avoided at all costs.