NCERT · Class 12 · Physics · Alternating CurrentWhat is a series LCR alternating current circuit? Derive the expression for the total impedance ($Z$), phase angle ($\phi$) between voltage and current, and the resonant frequency ($fr$) of the circuit using a phasor diagram.

Series LCR Alternating Current Circuit

\nA series LCR circuit consists of an inductor ($L$), a capacitor ($C$), and a resistor ($R$) connected in series across an alternating voltage source given by: $$v = V_m \sin(\omega t)$$ \nSince all components are in series, the same instantaneous current $i = I_m \sin(\omega t - \phi)$ flows through each component.

Voltage Components & Phasor Diagram

\nLet the RMS current in the circuit be $I$. The voltages across the individual components are:

1. Voltage across Resistor ($V_R$): $V_R = I \cdot R$ (In phase with current $I$)
2. Voltage across Inductor ($V_L$): $V_L = I \cdot X_L$ (Leads current $I$ by $\pi/2$ radians or $90^\circ$)
3. Voltage across Capacitor ($V_C$): $V_C = I \cdot X_C$ (Lags behind current $I$ by $\pi/2$ radians or $90^\circ$) \nWhere:

• $X_L = \omega L$ (Inductive Reactance)
• $X_C = \frac{1}{\omega C}$ (Capacitive Reactance)

Derivation of Total Impedance ($Z$)

\nAssuming $V_L > V_C$, the net reactive voltage perpendicular to $V_R$ is $(V_L - V_C)$. \nUsing the Pythagorean theorem on the phasor triangle for result voltage $V$: $$V^2 = V_R^2 + (V_L - V_C)^2$$ \nSubstitute $V_R = I \cdot R$, $V_L = I \cdot X_L$, and $V_C = I \cdot X_C$: $$V^2 = (I \cdot R)^2 + (I \cdot X_L - I \cdot X_C)^2$$ $$V^2 = I^2 \left[ R^2 + (X_L - X_C)^2 \right]$$ $$V = I \sqrt{R^2 + (X_L - X_C)^2}$$ \nThe total effective resistance of an AC circuit is called Impedance ($Z$): $$Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2}$$ $$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$

Expression for Phase Angle ($\phi$)

\nFrom the voltage/impedance phasor triangle, the tangent of the phase angle $\phi$ is: $$\tan\phi = \frac{V_L - V_C}{V_R} = \frac{X_L - X_C}{R}$$ $$\phi = \tan^{-1}\left( \frac{\omega L - \frac{1}{\omega C}}{R} \right)$$

Resonant Frequency ($f_r$)

\nElectrical resonance occurs when the inductive reactance equals the capacitive reactance ($X_L = X_C$), resulting in minimum impedance ($Z = R$) and maximum current amplitude.

$$X_L = X_C$$ $$\omega_r L = \frac{1}{\omega_r C}$$ $$\omega_r^2 = \frac{1}{LC} \implies \omega_r = \frac{1}{\sqrt{LC}}$$ \nSince $\omega_r = 2\pi f_r$: $$2\pi f_r = \frac{1}{\sqrt{LC}}$$ $$f_r = \frac{1}{2\pi \sqrt{LC}}$$ \nWhere $f_r$ is the resonant frequency of the LCR series circuit.