Resonance and Power in AC Circuits
In an ideal series RLC circuit with fixed supply amplitude, cancelling reactances minimize impedance and maximize current. Tuned circuits can favor a chosen frequency. Average sinusoidal AC power includes the power factor; an ideal purely reactive load takes zero net energy over a cycle.
What a learner can do afterwards
- Finds the resonant frequency of a series RLC circuit and describes the current peak
- Calculates average power using root mean square values and the power factor
- Explains why a capacitor or inductor alone takes no net energy from the supply over a cycle
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Consider ideal series RLC components with R > 0 and fixed supply voltage amplitude. Current is largest when XL = Xc. Setting omega0 L = 1/(omega0 C) gives omega0^2 = 1/(LC), so omega0 = 1/sqrt(LC). This is angular frequency in rad/s. The corresponding frequency is f0 = omega0/(2 pi), in hertz. Resistance limits the current peak; the peak need not be sharp.
For sinusoidal voltage and current, Pavg = Irms Vrms cos phi. The product Irms Vrms is apparent power in volt-amperes (VA). The power factor cos phi is R/Z for an ideal series RLC circuit. Rms means square the instantaneous values, average over a cycle, then take the square root. A sinusoid has Irms = I0/sqrt(2) and Vrms = V0/sqrt(2). In this circuit, Pavg = Irms^2 R: net energy is dissipated in the resistor.
An ideal capacitor or inductor stores energy in its field and returns it to the circuit. In sinusoidal steady state, current and voltage differ by a quarter cycle. Instantaneous power alternates between positive and negative values, with equal energy taken and returned over a full cycle. Its average power is zero, even though current flows. Real components can dissipate energy through their losses.
Suppose Irms = 3 A, Vrms = 12 V and cos phi = 0.5. Average power is 3 times 12 times 0.5 = 18 W. The product Irms Vrms alone is 36 VA, not 36 W of real power. At resonance cos phi = 1, so Pavg = Irms Vrms. Use rms values with this formula; peak values would need an extra factor of one-half.
Compare circuits with the same L and C and the same fixed supply voltage amplitude. Smaller R gives a taller current peak because I0 = V0/R at resonance. Define bandwidth using the two frequencies where average power is half its peak value. For this ideal series model, angular bandwidth is R/L and Q = omega0 L/R. Smaller R gives narrower bandwidth and higher Q.
An LC-based tuner favors signals near its resonant frequency. Adjusting C at fixed L changes f0 = 1/(2 pi sqrt(LC)). If C is multiplied by k, f0 is divided by sqrt(k). Increasing C lowers the tuned frequency; decreasing C raises it. A narrow bandwidth helps separate nearby frequencies, but centering the resonance alone does not guarantee that another station disappears.
For fixed supply amplitude, series RLC current peaks when XL = Xc. Sinusoidal average power is Irms Vrms cos phi. Ideal L and C return their stored energy each cycle.
Original lesson, questions and study cards authored by Marble.
The rewritten readings were written for this dataset. Video references credit Khan Academy.
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