A series RLC circuit has $R=40\Omega$, $L=0.25\text{H}$ and $C=80\mu \text{F}$. It is connected to a $230{\text{V}}_{\text{rms}}$ source operating at $f=50\text{Hz}$. Calculate (i) the magnitude of the total impedance $|Z|$, and (ii) the phase angle $\phi$ (in degrees) between the source voltage and the current.

Step-by-step solution

1. Compute the angular frequency: $\omega =2\pi f=2\pi \times 50=314.16\text{rad/s}$.
2. Find the inductive reactance: $X_L=\omega L=314.16\times 0.25=78.54\Omega$.
3. Find the capacitive reactance: $X_C=\frac{1}{\omega C}=\frac{1}{314.16\times 80\times 10^{-6}}=39.79\Omega$.
4. Determine the net reactance: $X=X_L-X_C=78.54-39.79=38.75\Omega$.
5. Calculate the magnitude of the impedance: $|Z|=\sqrt{R^2+X^2}=\sqrt{40^2+38.75^2}=\sqrt{3101.6}\approx 55.7\Omega$.
6. Find the phase angle using $\tan \phi =\frac{X}{R}$: $\phi ={\tan }^{-1}\left(\frac{38.75}{40}\right)\approx 44.4^{\circ }$.

Concept tested: Impedance and phase angle in series RLC

Published 20 September 2026 · How this solution was prepared and checked