A series LCR circuit has $R=15,\Omega$, $L=0.3,\text{H}$, $C=120,\mu \text{F}$ and is driven at $f=60,\text{Hz}$ by a $250,\text{V rms}$ source. Calculate the magnitude $|Z|$ of the impedance and the phase angle $\phi$ (in degrees) between the source voltage and current.

Step-by-step solution

1. Compute angular frequency $\omega =2\pi f=2\pi \times 60\approx 377,\text{rad/s}$.
2. Inductive reactance $X_L=\omega L=377\times 0.3=113.1,\Omega$.
3. Capacitive reactance $X_C=1/(\omega C)=1/(377\times 120\times 10^{-6})=22.09,\Omega$.
4. Net reactance $X=X_L-X_C=113.1-22.09=91.01,Omega$.
5. Impedance magnitude $|Z|=\sqrt{R^2+X^2}=\sqrt{15^2+91.01^2}approx92.2,Omega$.
6. Phase angle $\phi ={\tan }^{-1}!\left(\frac{X}{R}\right)={\tan }^{-1}!\left(\frac{91.01}{15}\right)\approx 80.5^{\circ }$ (lagging).

Concept tested: Impedance magnitude and phase angle

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked