AC Circuits — R, L, C and Impedance (Series LCR)

For an inductor $L$: $v=L{\displaystyle \frac{di}{dt}}$. With $v=V_0\sin (\omega t)$:

$$i=-\frac{V_0}{\omega L}\cos (\omega t)=\frac{V_0}{\omega L}\sin (\omega t-\frac{\pi }{2})$$

Inductive Reactance ($X_L$):

$$X_L=\omega L=2\pi fL\text{(unit: }\mathrm{\Omega }\text{)}$$

• Voltage leads current by ${90}^{\circ }$ (or current lags voltage by ${90}^{\circ }$).
• $X_L\propto f$ — reactance increases linearly with frequency. At $f=0$ (DC): $X_L=0$ (an ideal inductor acts as a short circuit for DC).
• At very high $f$: $X_L\to \mathrm{\infty }$ (an inductor acts as an open circuit for very high-frequency AC).
• Power: $P=0$ (no net energy dissipation — energy is merely stored and returned to the circuit).

Example: $L=0.1\text{ H}$, $f=50\text{ Hz}$: $X_L=2\pi \times 50\times 0.1=10\pi \approx \mathbf{31.4}\mathbf{\Omega }$

For a capacitor $C$: $i=C{\displaystyle \frac{dv}{dt}}$. With $v=V_0\sin (\omega t)$:

$$i=\omega C{V}_0\cos (\omega t)=\omega C{V}_0\sin (\omega t+\frac{\pi }{2})$$

Capacitive Reactance ($X_C$):

$$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}\text{(unit: }\mathrm{\Omega }\text{)}$$

• Current leads voltage by ${90}^{\circ }$ (or voltage lags current by ${90}^{\circ }$).
• $X_C\propto 1/f$ — reactance decreases inversely with frequency. At $f=0$ (DC): $X_C\to \mathrm{\infty }$ (a capacitor completely blocks DC).
• At very high $f$: $X_C\to 0$ (a capacitor passes high-frequency AC freely).
• Power: $P=0$ (no net energy dissipation — energy is stored in the electric field and returned).

Example: $C=100\mu \text{F}$, $f=50\text{ Hz}$: $X_C={\displaystyle \frac{1}{2\pi \times 50\times 100\times {10}^{-6}}}={\displaystyle \frac{1}{0.01\pi }}\approx \mathbf{31.8}\mathbf{\Omega }$

In a series LCR circuit with an applied voltage $v=V_0\sin (\omega t)$, the current is $i=I_0\sin (\omega t-\varphi )$ where:

Impedance ($Z$)

$$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$$

Phase Angle (voltage leads current by $\varphi$)

$$\tan \varphi =\frac{X_L-X_C}{R}=\frac{\omega L-1/(\omega C)}{R}$$

Phasor Diagram for Series LCR

Taking current $I$ as the reference phasor (along the x-axis):

• $V_R=I\times R$ — along current phasor (in phase).
• $V_L=I\times X_L$ — ${90}^{\circ }$ ahead of current (points up).
• $V_C=I\times X_C$ — ${90}^{\circ }$ behind current (points down).
• Net voltage: $V=\sqrt{V_R^2+(V_L-V_C{)}^2}$

Note: $V\ne V_R+V_L+V_C$ (algebraic sum). They must be added as phasors. This is exactly why individual voltmeter readings across an inductor or capacitor in a series LCR circuit can paradoxically exceed the total supply voltage.

Worked Example

A Series LCR circuit has $R=100\mathrm{\Omega }$, $X_L=200\mathrm{\Omega }$, $X_C=50\mathrm{\Omega }$, and $V_{\text{rms}}=$ 220 V.

$Z=\sqrt{{100}^2+(200-50{)}^2}=\sqrt{10000+22500}=\sqrt{32500}\approx \mathbf{180.3}\mathbf{\Omega }$

$I_{\text{rms}}={\displaystyle \frac{V_{\text{rms}}}{Z}}={\displaystyle \frac{220}{180.3}}\approx \mathbf{1.22}\mathbf{\text{ A}}$

$\tan \varphi ={\displaystyle \frac{200-50}{100}}={\displaystyle \frac{150}{100}}=1.5⟹\varphi \approx {\mathbf{56.3}}^{\circ }$ (inductive, voltage leads current).