Put a resistor, an inductor and a capacitor in series on an AC source. The same current flows through all three, but their voltages peak at different times, a quarter cycle apart. To add them you need a way to add oscillations that are out of step: phasors.
Key idea
$$Z = \sqrt{R^2 + (X_L - X_C)^2}, \qquad \tan\phi = \frac{X_L - X_C}{R}, \qquad I_{\text{rms}} = \frac{V_{\text{rms}}}{Z},$$
where $\phi$ is the angle by which the source voltage leads the current. The impedance $Z$ plays the part of resistance for the whole circuit.
Why it works
Picture each oscillating voltage as an arrow (a phasor) spinning at $\omega$, whose shadow on the vertical axis is the actual voltage. Take the current's arrow as the reference. $V_R = IR$ points along it; $V_L = IX_L$ points $90^\circ$ ahead; $V_C = IX_C$ points $90^\circ$ behind, so it is opposite to $V_L$. The source voltage is the vector sum: $V_L$ and $V_C$ partly cancel, leaving $I(X_L - X_C)$ at right angles to $IR$. Pythagoras gives $V = I\sqrt{R^2 + (X_L - X_C)^2}$, and the angle of the sum is $\phi$.
Worked examples
Worked example: From reactances
$R = 30\ \Omega$, $X_L = 80\ \Omega$ and $X_C = 40\ \Omega$ in series. Find $Z$ and $\phi$.
- $Z = \sqrt{30^2 + (80 - 40)^2} = \sqrt{900 + 1600} = 50\ \Omega$.
- $\tan\phi = 40/30$, so $\phi = 53.1^\circ$: the voltage leads, since the inductor dominates.
Answer: $Z = 50\ \Omega$, $\phi \approx 53.1^\circ$
Worked example: A full circuit
$R = 40\ \Omega$, $L = 0.10\ \text{H}$ and $C = 50\ \mu\text{F}$ in series on $120\ \text{V}$ rms at $60\ \text{Hz}$. Find $Z$, the current and $\phi$.
- $X_L = 2\pi(60)(0.10) = 37.7\ \text{Ω}$; $X_C = \dfrac{1}{2\pi(60)(50 \times 10^{-6})} = 53.1\ \text{Ω}$.
- $Z = \sqrt{40^2 + (37.7 - 53.1)^2} = 42.8\ \text{Ω}$, so $I_{\text{rms}} = 120/42.8 = 2.80\ \text{A}$.
- $\tan\phi = \dfrac{-15.4}{40}$, so $\phi = -21.0^\circ$: negative, the circuit is more capacitive and the voltage lags the current.
Answer: $Z \approx 42.8\ \text{Ω}$, $I \approx 2.80\ \text{A}$
Worked example: More voltage than the source
In the same circuit, find the rms voltages across the inductor and the capacitor.
- $V_L = IX_L = (2.80)(37.7) = 106\ \text{V}$ and $V_C = IX_C = 149\ \text{V}$.
- $V_C$ is larger than the source's $120\ \text{V}$! It is no contradiction: $V_L$ and $V_C$ are opposite in phase and mostly cancel, so the three voltages do not add as plain numbers.
Answer: $V_L \approx 106\ \text{V}$, $V_C \approx 149\ \text{V}$
Common mistake
Adding $R$, $X_L$ and $X_C$ as plain numbers. That treats three voltages that peak at different times as if they peaked together. In the first example it would give $150\ \Omega$ instead of $50\ \Omega$. Check: $Z$ can never be larger than $R + |X_L - X_C|$, and it equals $R$ only when $X_L = X_C$.