Question

An inductor, a capacitor, and a resistor are connected in series with an AC source \( v = v_m \sin \omega t \). Derive an expression for the average power dissipated in the circuit. Also, obtain the expression for the resonant frequency of the circuit.

Hint:

The average power dissipated in an AC circuit depends only on the resistive part of the impedance, while the resonant frequency minimizes the overall impedance of the circuit.

Answer 1

Expression for Average Power Dissipated in a Series LCR Circuit: The voltage is defined as: \[ v = v_m \sin(\omega t) \] The current is given by: \[ i = i_m \sin(\omega t + \varphi) \] Power \(P\) is the product of voltage and current: \[ P = v \times i = (v_m \sin(\omega t)) \times (i_m \sin(\omega t + \varphi)) \] This simplifies to: \[ P = \frac{v_m i_m}{2} \left[ \cos \varphi - \cos(2\omega t + \varphi) \right] \tag{1} \] The average power over one cycle is obtained by averaging the two terms in equation (1). Since the second term is time-dependent and the first is constant, the average power \(P\) is: \[ P = \frac{v_m i_m}{2} \cos \varphi \] Expression for Resonant Frequency: At resonance, capacitive reactance \(X_C\) equals inductive reactance \(X_L\): \[ \frac{1}{\omega C} = \omega L \] This yields the angular frequency \(\omega\) as: \[ \omega = \frac{1}{\sqrt{LC}} \] Consequently, the resonant frequency \(f\) is: \[ f = \frac{1}{2 \pi \sqrt{LC}} \]