Question

Establish the expression for impedance of the circuit when elements X, Y, and Z are connected in series to an AC source. Show the variation of current in the circuit with the frequency of the applied AC source.

Hint:

The current in a series R-L-C circuit varies with frequency due to the combination of resistive, inductive, and capacitive reactances. At resonance, the impedance is minimum, and the current is maximum.

Answer 1

The three elements X, Y, and Z are connected in series across an AC source. From the previous question, we know: - Element X is a resistor, with impedance \( Z_X = R \). - Element Y is an inductor, with impedance \( Z_Y = j\omega L \), where \( \omega = 2\pi f \) is the angular frequency and \( L \) is the inductance. - Element Z is a capacitor, with impedance \( Z_Z = \frac{1}{j\omega C} \), where \( C \) is the capacitance. The total impedance \( Z_{\text{total}} \) of the series circuit is the sum of the individual impedances: \[ Z_{\text{total}} = Z_X + Z_Y + Z_Z \] \[ Z_{\text{total}} = R + j\omega L + \frac{1}{j\omega C} \] \[ Z_{\text{total}} = R + j\left( \omega L - \frac{1}{\omega C} \right) \] Thus, the impedance of the series circuit is: \[ Z_{\text{total}} = R + j \left( \omega L - \frac{1}{\omega C} \right) \] ### Step 1: Variation of current with frequency The current \( I \) in the series circuit is given by Ohm's law: \[ I = \frac{V}{Z_{\text{total}}} \] Since \( Z_{\text{total}} \) depends on the frequency \( f \), the current will vary with \( f \). - When \( f \) is low, the capacitive reactance \( \left( \frac{1}{\omega C} \right) \) is high, so the current is lower. - When \( f \) increases, the inductive reactance \( (\omega L) \) increases, and the capacitive reactance decreases, so the current starts increasing until it reaches a peak. - Beyond a certain frequency, the inductive reactance dominates, and the current decreases. Thus, the current shows a resonant behavior with frequency, with the impedance being lowest at a certain resonant frequency \( f_0 \). % Correct Answer Correct Answer:} - Impedance of the series circuit: \[ Z_{\text{total}} = R + j \left( \omega L - \frac{1}{\omega C} \right) \]