Question

A resistor and a capacitor are connected in series to an AC source $v = v_m \sin \omega t$. Derive an expression for the impedance of the circuit.

Hint:

- The total impedance \( Z \) in a series R-C circuit is given by: \[ Z = \sqrt{R^2 + \frac{1}{\omega^2 C^2}} \] - The resistor's impedance is purely real (\( Z_R = R \)), while the capacitor's impedance is imaginary (\( Z_C = -j \frac{1}{\omega C} \)). - The magnitude of impedance is determined by combining the real and imaginary parts using: \[ |Z| = \sqrt{\text{Re}(Z)^2 + \text{Im}(Z)^2} \] - The total impedance determines the phase difference and voltage behavior in AC circuits involving resistors and capacitors.

Answer 1

Derivation of Impedance 
The total voltage in a series R-C circuit is given by: \[ v = v_R + v_C, \] where:  $v_R = i R$ is the voltage across the resistor,  $v_C = i \frac{1}{j\omega C}$ is the voltage across the capacitor.  The current in the circuit is the same through all components, and the impedance $Z$ of the circuit is given by: \[ Z = \frac{v}{i}. \] The impedance of the resistor is purely real, $Z_R = R$, and the impedance of the capacitor is purely imaginary, $Z_C = -j\frac{1}{\omega C}$. The total impedance is: \[ Z = Z_R + Z_C = R - j\frac{1}{\omega C}. \] The magnitude of the impedance is: \[ |Z| = \sqrt{\text{Re}(Z)^2 + \text{Im}(Z)^2}. \] Substituting the real and imaginary parts: \[ |Z| = \sqrt{R^2 + \left(-\frac{1}{\omega C}\right)^2} = \sqrt{R^2 + \frac{1}{\omega^2 C^2}}. \] Thus, the impedance of the circuit is: \[ Z = \sqrt{R^2 + \frac{1}{\omega^2 C^2}}. \]