In a series RLC circuit, the impedance depends on frequency, because $X_L$ rises and $X_C$ falls. At one special frequency they cancel. There the circuit responds most strongly, the same way a swing goes highest when you push it at its natural rhythm. That is the natural frequency of the RLC circuit, now driven from outside.
Key idea
Resonance at $\omega_0 = \dfrac{1}{\sqrt{LC}}$: $X_L = X_C$, $Z = R$, the current is largest ($V/R$) and in step with the voltage.
Average power: $\bar{P} = I_{\text{rms}}V_{\text{rms}}\cos\phi = I_{\text{rms}}^2R$, with power factor $\cos\phi = R/Z$. The sharpness of the resonance is measured by $Q = \omega_0L/R$.
Average power: $\bar{P} = I_{\text{rms}}V_{\text{rms}}\cos\phi = I_{\text{rms}}^2R$, with power factor $\cos\phi = R/Z$. The sharpness of the resonance is measured by $Q = \omega_0L/R$.
Why it works
Only the resistor turns electrical energy into heat. An inductor or capacitor stores energy for a quarter cycle and hands it back the next quarter: its current and voltage are $90^\circ$ apart, so their product averages to zero. For the whole circuit, only the part of the current in step with the voltage, $I\cos\phi$, delivers energy on average. And since $\cos\phi = R/Z$ and $V = IZ$, $IV\cos\phi = I^2R$. At resonance $\phi = 0$ and the circuit acts like the resistor alone.
Worked examples
Worked example: At resonance
$R = 20\ \Omega$, $L = 0.10\ \text{H}$ and $C = 100\ \mu\text{F}$ are in series with a $10\ \text{V}$ rms source. Find the resonant frequency and the current and power there.
- $f_0 = \dfrac{1}{2\pi\sqrt{(0.10)(100 \times 10^{-6})}} = 50.3\ \text{Hz}$.
- $Z = R$, so $I_{\text{rms}} = 10/20 = 0.50\ \text{A}$ and $\bar{P} = I^2R = 5.0\ \text{W}$.
Answer: $f_0 \approx 50.3\ \text{Hz}$; $0.50\ \text{A}$; $5.0\ \text{W}$
Worked example: Off resonance
The same circuit is driven by $120\ \text{V}$ rms at $60\ \text{Hz}$. Find the power factor and the average power.
- $X_L = 37.7\ \text{Ω}$, $X_C = 26.5\ \text{Ω}$, $Z = \sqrt{20^2 + (11.2)^2} = 22.9\ \text{Ω}$.
- $I_{\text{rms}} = 120/22.9 = 5.24\ \text{A}$ and $\cos\phi = R/Z = 0.873$.
- $\bar{P} = I_{\text{rms}}V_{\text{rms}}\cos\phi = (5.24)(120)(0.873) = 549\ \text{W}$, the same as $I^2R$. Not $I_{\text{rms}}V_{\text{rms}} = 629\ \text{W}$.
Answer: $\cos\phi \approx 0.873$, $\bar{P} \approx 549\ \text{W}$
Worked example: How sharp is the peak?
With the resistance lowered to $2.0\ \Omega$, find $Q$.
- $Q = \dfrac{\omega_0L}{R} = \dfrac{(2\pi)(50.3)(0.10)}{2.0} = 15.8$.
- The peak's width is about $f_0/Q \approx 3.18\ \text{Hz}$: a low-resistance circuit responds to a narrow band of frequencies, which is how a radio picks one station out of many.
Answer: $Q \approx 15.8$
Common mistake
"The power is always $I_{\text{rms}}V_{\text{rms}}$." That is true only when current and voltage are in step (a resistor, or any circuit at resonance). Otherwise part of the current only shuttles energy into the inductor or capacitor and back. In the second example, $IV$ overstates the power by about 15%. Check: the average power must equal $I_{\text{rms}}^2R$, the heat in the resistance.