LCR Circuit
By A series LCR circuit consists of an inductor (L), a capacitor (C), and a resistor (R) connected in series to an AC source. The circuit exhibits resonance at the resonant frequency \begin{align} \omega_0=\frac{1}{\sqrt{LC}} \end{align} At resonance, the impedance of the circuit is minimum and the current through it is the maximum. The energy oscillates between the inductor and the capacitor.
When an AC voltage $V=V_0\sin\omega t$ is applied, a time varying current $i$ flows through the circuit. Apply Kirchhoff's law to get \begin{align} L\frac{\mathrm{d}i}{\mathrm{d}t}+iR+\frac{q}{C}=V \end{align} where $q$ is the charge on the capacitor.
The capacitive reactance is $X_C=\frac{1}{\omega C}$ and the inductive reactance is $X_L=\omega L$. The impedance of the circuit is \begin{align} Z&=\sqrt{R^2+\left(X_C-X_L\right)^2}\\ &=\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2} \end{align} At resonance, $X_C=X_L$ and the impedance is minimum.
The phase difference between the voltage and current is given by \begin{align} \tan\phi=\frac{X_L-X_C}{R} \end{align}
If $X_L > X_C$ then the voltage leads the current and the circuit is inductive. If $X_C > X_L$ then the voltage lags the current and the circuit is said to be capacitive. If $X_C = X_L$ then the voltage and the current are in phase and the circuit is resistive.
The current flowing in the circuit is \begin{align} i & =i_0\cos(\omega t+\phi) \\ &=\frac{V_0}{Z}\sin(\omega t+\phi) \end{align}
Resonance in series LCR circuit
An interesting characteristic of the series LCR circuit is the phenomenon of resonance. The natural frequency of series LCR circuit is \begin{align} \omega_0=\frac{1}{\sqrt{LC}}. \end{align} The circuit resonates when applied voltage's frequency is equal to the circuit's natural frequency. The current becomes maximum at the resonance. The maximum current is given by \begin{align} i_0&=\frac{V_0}{Z} \\ &=\frac{V_0}{\sqrt{R^2+(X_L-X_C)^2}}\\ &=\frac{V_0}{R}. \end{align}
Bandwidth of the series LCR circuit
The current in LCR circuit varies with the frequency of applied voltage. It attains a maximum value, $i_\text{max}=V_\text{max}/R$, at the resonance frequency $\omega_0$. The current becomes $0.707i_\text{max}$ at the two frequencies $\omega_1$ and $\omega_2$. These frequencies lies symmetrically on two sides of $\omega_0$. The bandwidth of the circuit is given by \begin{align} \Delta\omega=\omega_2-\omega_1=\frac{R}{L}. \end{align}
The average power in the circuit varies with the frequency. It is the maximum at the resonance frequency and reduces to half of the maximum value at the frequencies $\omega_1$ and $\omega_2$.
Quality factor of the series LCR circuit
The quality factor (Q) is defined as the ratio of the resonant frequency $\omega_0$ and the bandwidth $\Delta\omega$, \begin{align} Q=\frac{\omega_0}{\omega_2-\omega_1}=\frac{\omega_0 L}{R}. \end{align} It is the ratio of the stored energy in the circuit to the energy lost per cycle. Higher Q values a sharper resonance peak. The quality factor is a dimensionless parameter.
Problems from IIT JEE
Problem (IIT JEE 2017) In the circuit shown, $L=1\,\mu H$, $C=1\,\mu F$ and $R=1\, k \Omega$. They are connected in series with an AC source $V=V_0\sin\omega t$ as shown. Which of the following option(s) is(are) correct?
- At $\omega\approx0$ the current flowing through the circuit becomes nearly zero.
- The frequency at which the current will be in phase with the voltage is independent of $R$.
- The current will be in phase with the voltage if $\omega=10^4$ rad/s.
- At $\omega\gg 10^6$ rad/s, the circuit behaves like a capacitor.
Solution: In a series LCR circuit, the current is in phase with the voltage at resonant frequency \begin{align} \omega_r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(10^{-6})(10^{-6})}}=10^6\,\mathrm{rad/s}.\nonumber \end{align} The resonant frequency is independent of $R$.
If the frequency is very low ($\omega\approx 0$) then inductive reactance $X_L=\omega L\to 0$ and capacitive reactance $X_C={1}/{(\omega C)}\to\infty$. Thus, the impedance of the circuit becomes very large and hence the current flowing through the circuit becomes nearly zero. Thus, at low frequencies, the circuit behaves like a capacitor as its behaviour is largely controlled by $X_C$.
If the frequency is very large ($\omega\gg \omega_r$) then inductive reactance $X_L=\omega L\to \infty$ and capacitive reactance $X_C={1}/{(\omega C)}\to 0$. Again, the impedance of the circuit becomes very large and the current flowing through the circuit becomes nearly zero. Thus, at high frequencies, the circuit behaves like an inductor as its behaviour is largely controlled by $X_L$.
We encourage you to deduce these results from the expressions of phase $\phi$ and current $i$, \begin{align} \phi&=\tan^{-1}\left(\frac{\frac{1}{\omega C}-\omega L}{R} \right), \nonumber \\ i&=\frac{V_0}{\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L \right)^2}}\sin(\omega t+\phi).\nonumber \end{align}
The variation of phase difference (between the current and the voltage), $\phi$, with the frequency $\omega$ is shown in the first figure. Note that $\phi\to\pi/2$ when $\omega\to 0$ (capacitive) and $\phi\to-\pi/2$ when $\omega\to \infty$ (inductive). At the resonant frequency $\omega_r$, the current is in phase with the voltage i.e., $\phi=0$ at $\omega=\omega_r$. The second figure shows the variation of the magnitude of the current ($i_0$) with frequency. The magnitude of the current attains its maximum value, $i_r=V_0/R$, at the resonant frequency $\omega_r$.
