AC resonance and active power

In a series circuit containing $R$, $L$ and $C$, the impedance is $Z=\sqrt{R^2+(X_L-X_C)^2}$, where $X_L=\omega L$ and $X_C=\dfrac{1}{\omega C}$.

Resonance is the regime in which $X_L=X_C$: the reactances cancel each other, $Z=R$ is at its minimum, and the current reaches its maximum $I=U/R$. The resonant frequency follows from $\omega_0 L=\dfrac{1}{\omega_0 C}$, so $\omega_0=\dfrac{1}{\sqrt{LC}}$ and $f_0=\dfrac{1}{2\pi\sqrt{LC}}$.

The phase shift between voltage and current is set by $\cos\varphi=\dfrac{R}{Z}$: for $f<f_0$ the circuit is capacitive ($X_C>X_L$), for $f>f_0$ it is inductive ($X_L>X_C$), and at $f=f_0$ the current is in phase with the voltage ($\varphi=0$).

Active power is the average power over one period; it is dissipated only in the resistor: $P=UI\cos\varphi=I^2R$. Apparent power $S=UI$ and reactive power $Q=UI\sin\varphi$ satisfy $S^2=P^2+Q^2$; at resonance $Q=0$, $\cos\varphi=1$, and $P=S=UI$ is maximal for a given source voltage.

The sharpness of the resonance is described by the quality factor $Q_L=\dfrac{\omega_0 L}{R}=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$, which shows how many times the voltage across the inductor and capacitor at resonance exceeds the source voltage.

A series circuit has $R=10\ \Omega$, $L=100$ mH, $C=10\ \mu$F. The source rms voltage is $U=20$ V. Find the resonant frequency $f_0$, the current and the active power at resonance.

The same circuit ($R=10\ \Omega$, $L=100$ mH, $C=10\ \mu$F, $U=20$ V) operates at $\omega=2000$ rad/s. Determine the nature of the circuit, the impedance, the current, $\cos\varphi$ and the active power. Compare with the resonant regime.

A single-phase load (equivalent: $R=8\ \Omega$ and $X_L=6\ \Omega$ in series) is connected to a $U=220$ V, $f=50$ Hz supply. Find the current, the active and reactive power. What capacitance must be connected in parallel to make $\cos\varphi$ equal to one, and how does the line current change?

Simulating resonance (safe, no 220 V mains).

1. Open an AC circuit simulation (for example PhET "Circuit Construction Kit: AC") or a spreadsheet and build a series circuit with $R=10\ \Omega$, $L=100$ mH, $C=10\ \mu$F and a $U=20$ V source.
2. Sweeping the frequency from 50 Hz to 400 Hz in 25 Hz steps, record the current $I$ and the active power $P=I^2R$.
3. Plot $I(f)$ and locate the frequency of the maximum; compare it with the prediction $f_0=\dfrac{1}{2\pi\sqrt{LC}}\approx 159$ Hz.
4. Verify that at $f_0$ the condition $\cos\varphi=1$ holds (voltage and current are in phase), while for $f<f_0$ and $f>f_0$ the phase shifts in opposite directions.
5. Varying $R$ (for example 5, 10, 20 $\Omega$), describe how the height and width of the resonance curve change, and relate this to the quality factor $Q_L$.

Important: never build this circuit on household 220 V mains — resonant overvoltages across $L$ and $C$ are dangerous. Work only in simulation or with a low-voltage source.

To show that resonance is the condition $X_L=X_C$ at which the impedance is minimal, the current is maximal, and the active power reaches its largest value with $\cos\varphi=1$; at the same time to relate active, reactive and apparent power.

Graph. The horizontal axis is frequency $f$ (Hz); the vertical axis carries two curves: $|Z(f)|=\sqrt{R^2+(\omega L-1/(\omega C))^2}$ (falling monotonically to the minimum $Z=R$ at $f_0$, then rising) and $I(f)=U/|Z(f)|$ (a symmetric peak at $f_0=\dfrac{1}{2\pi\sqrt{LC}}$). To the left of $f_0$ the curves are labelled "capacitive region, $X_C>X_L$, $\varphi<0$", to the right "inductive region, $X_L>X_C$, $\varphi>0$", and at the peak "resonance: $\varphi=0$, $\cos\varphi=1$, $P=P_{\max}$".

Inset diagram. On the right, a power triangle: the horizontal leg is $P=I^2R$, the vertical leg is $Q=UI\sin\varphi$, and the hypotenuse is $S=UI$ with the angle $\varphi$ between $S$ and $P$. A dashed version shows the resonant case, where the triangle collapses into a segment: $Q=0$, $S=P$.

A phone with an FM receiver and an old radio pick up many stations at once. The input circuit is a coil $L = 10\ \mu\text{H}$ and a variable capacitor tuned from $10\ \text{pF}$ to $100\ \text{pF}$. At $C = 100\ \text{pF}$ the resonant frequency of the circuit is $f_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{10^{-5}\cdot 10^{-10}}} \approx 5.0\ \text{MHz}$, and at $C = 10\ \text{pF}$ it is $\approx 16\ \text{MHz}$: turning the knob tunes the receiver across the band.

The scenario exercises the resonance condition $X_L = X_C$, that is $\omega_0 L = \dfrac{1}{\omega_0 C}$, hence $\omega_0 = \dfrac{1}{\sqrt{LC}}$. At this frequency the term $(X_L - X_C)$ in the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ vanishes, $Z$ drops to $R$, the current is maximal, and the station at $f_0$ produces the largest output voltage. Shifting the capacitance moves $\omega$ away from $\omega_0$ and sharply reduces the response.

The meter in the hallway counts active energy in kW$\cdot$h. A fridge compressor is a motor with windings; at $U = 230\ \text{V}$ its active power is $P = 150\ \text{W}$ at $\cos\varphi = 0.6$. Then the apparent power is $S = P/\cos\varphi = 250\ \text{VA}$, and the wire current is $I = S/U = 250/230 \approx 1.09\ \text{A}$ instead of $0.65\ \text{A}$ for a purely resistive load of the same power.

The scenario separates active from apparent power: $P = UI\cos\varphi = I^2R$ is the power irreversibly converted into heat and mechanical work, while $Q = \sqrt{S^2 - P^2} = \sqrt{250^2 - 150^2} = 200\ \text{var}$ oscillates between the grid and the magnetic field of the windings and is not charged by the meter. The factor $\cos\varphi = R/Z$ is the same phase shift between current and voltage that the impedance sets.

The charging pad and the phone contain identical coils $L = 20\ \mu\text{H}$ with capacitors $C = 56\ \text{nF}$, tuned to $f_0 = \dfrac{1}{2\pi\sqrt{LC}} \approx 150\ \text{kHz}$. With a coil resistance $R = 0.4\ \Omega$ the quality factor is $Q = \dfrac{\omega_0 L}{R} = \dfrac{2\pi\cdot 150000\cdot 2\cdot 10^{-5}}{0.4} \approx 47$, and an alternating current $I = 2\ \text{A}$ builds up a voltage $I\omega_0 L \approx 38\ \text{V}$ across the coil.

Both coils are tuned to the same $\omega_0$, so at this frequency $X_L = X_C$ and each loop's impedance is only its $R$; the current is in phase with the voltage, and $P = I^2R = 2^2\cdot 0.4 = 1.6\ \text{W}$ is the active power lost as heat in the coil, while the rest of the energy sloshes between the coil's magnetic field and the capacitor. Because $Q \approx 47$ is large, a small drift of frequency away from $f_0$ breaks the matching of the resonances and the transferred power drops sharply.