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

Deriving an expression for the average power dissipated in a series LCR circuit: The voltage is given by: \[ v = v_m \sin(\omega t) \] The current is: \[ i = i_m \sin(\omega t + \varphi) \] The power \(P\) is given by 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 a cycle is given by averaging the two terms on the right-hand side of equation (1). The second term is time-dependent, while the first term is constant. Therefore, the average power \(P\) is: \[ P = \frac{v_m i_m}{2} \cos \varphi \] Obtaining the expression for the resonant frequency: At resonance, the reactance of the capacitor \(X_C\) equals the reactance of the inductor \(X_L\): \[ \frac{1}{\omega C} = \omega L \] This gives the angular frequency \(\omega\) as: \[ \omega = \frac{1}{\sqrt{LC}} \] Thus, the resonant frequency \(f\) is: \[ f = \frac{1}{2 \pi \sqrt{LC}} \]