Question

An LCR circuit contains resistance of $100\,{\text{ohm}}$ and a supply of $200\,{\text{volt}}$ at $300\,{\text{radian}} \cdot {{\text{s}}^{ - 1}}$ angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by $60^\circ$. If on the other hand only the inductor is taken out, the current leads by $60^\circ$ with applied voltage. The current flowing in the circuit is
A. $1\,{\text{A}}$
B. $1.5\,{\text{A}}$
C. $2\,{\text{A}}$
D. $2.5\,{\text{A}}$

Answer 1

Use the formula for phase of the current in the circuit. This formula gives the relation between the capacitive reactance, inductive reactance and resistance in the circuit. Obtain this formula when the capacitor is taken out from the circuit and when the inductor is taken out from the circuit. Show that the resonance condition is obtained in the circuit. Lastly, use the formula for the root mean square current in the circuit to determine the final answer.

Formulae used:
The phase $\phi$ of the current in the circuit is given by
$\tan \phi = \dfrac{{{X_C} - {X_L}}}{R}$ …… (1)
Here, ${X_C}$ is capacitive reactance, ${X_L}$ is the inductive reactance and $R$ is resistance in the circuit.
The expression for root mean square current ${I_{rms}}$ is
${I_{rms}} = \dfrac{{{V_{rms}}}}{z}$ …… (2)
Here, ${V_{rms}}$ is the root mean square voltage and $z$ is the impedance.

Complete step by step answer:
We have given that the resistance, root mean square voltage and angular frequency in the circuit is $100\,\Omega$, $200\,{\text{V}}$ and $300\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$.
$R = 100\,\Omega$
$\Rightarrow {V_{rms}} = 200\,{\text{V}}$
$\Rightarrow \omega = 300\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$
If only a capacitor is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by $60^\circ$. Hence, the value of capacitive reactance also becomes zero.
$\phi = - 60^\circ$
$\Rightarrow {X_C} = 0$

Substitute $- 60^\circ$ for $\phi$ and $0$ for ${X_C}$ in equation (1).
$\tan \left( { - 60^\circ } \right) = \dfrac{{0 - {X_L}}}{R}$
$\Rightarrow - \tan 60^\circ = \dfrac{{ - {X_L}}}{R}$
$\Rightarrow \tan 60^\circ = \dfrac{{{X_L}}}{R}$ …… (3)
If only the inductor is taken out from the circuit and the rest of the circuit is joined, current leads the voltage by $60^\circ$. Hence, the value of inductive reactance also becomes zero.
$\phi = 60^\circ$
${X_L} = 0$
Substitute $60^\circ$ for $\phi$ and $0$ for ${X_L}$ in equation (1).
$\tan 60^\circ = \dfrac{{{X_C} - 0}}{R}$
$\Rightarrow \tan 60^\circ = \dfrac{{{X_C}}}{R}$ …… (4)
From equations (3) and (4), we can conclude that
$\dfrac{{{X_L}}}{R} = \dfrac{{{X_C}}}{R}$
$\Rightarrow {X_L} = {X_C}$

The capacitive reactance is equal to the inductive reactance only when there is resonance.For the resonance condition, the impedance in the circuit is equal to resistance in the circuit.
$z = R$
Hence, the equation (2) for the resonance condition becomes
${I_{rms}} = \dfrac{{{V_{rms}}}}{R}$
Substitute $200\,{\text{V}}$ for ${V_{rms}}$ and $100\,\Omega$ for $R$ in the above equation.
${I_{rms}} = \dfrac{{200\,{\text{V}}}}{{100\,\Omega }}$
$\therefore {I_{rms}} = 2\,{\text{A}}$
Therefore, the current flowing in the circuit is $2\,{\text{A}}$.

Hence, the correct option is C.

Note: One can also solve the same question by another method. One can show that the phase difference of the current flowing the circuit is equal to zero. Hence, there is a resonance condition in the circuit. Then use the formula for root mean square current and determine the value of the electric current flowing in the circuit at the resonance condition.