Parallel Resonance

Sweep a current-driven parallel RLC tank. Capacitor and inductor currents cancel at resonance, maximizing tank impedance and voltage. L1 is 10 mH and C1 is approximately 2.533 µF, setting the resonant frequency to 1 kHz. R1 = 1 kΩ represents parallel loss and makes the tank voltage finite.

Resonance condition

$$\begin{aligned} f_0&=\frac{1}{2\pi\sqrt{LC}}=1\,\mathrm{kHz}\\ X_L(f_0)&=|X_C(f_0)|\approx62.83\,\Omega \end{aligned}$$

All three branches share the same voltage. Inductor current lags by 90° and capacitor current leads by 90°. At resonance these two branch currents cancel, leaving the source to supply the resistor loss.

$$\begin{aligned} Y&=\frac{1}{R}+j\left(\omega C-\frac{1}{\omega L}\right)\\ V(f_0)&=(1\,\mathrm{mA})(1\,\mathrm{k}\Omega)=1\,\mathrm V\\ |I_L(f_0)|&=|I_C(f_0)|\approx15.92\,\mathrm{mA}\\ Q&=\frac{R}{\omega_0L}\approx15.92 \end{aligned}$$

Read the circuit and response

I1 supplies AC analysis magnitude = 1 mA upward into the tank. Green Vtank measures tank voltage. Blue IL and purple IC measure the reactive branch currents downward. Their amplitudes exceed the source current near resonance because energy circulates between L1 and C1.

The example opens in AC sweep, from 100 Hz to 10 kHz, with 100 points per decade and a logarithmic frequency axis. The chart shows magnitude and phase. Magnitude in dB is $20\log_{10}(|V|/1\,\mathrm V)$ for voltage or $20\log_{10}(|I|/1\,\mathrm A)$ for current; a 1 V voltage peak is 0 dB. These current dB values are referenced to one ampere, not voltage gain. Open Table to inspect numeric frequency samples, including 1 kHz.

At resonance Vtank is 1 V and is in phase with the source current. IL and IC have equal magnitudes and opposite phases. Away from resonance their cancellation is incomplete and the tank voltage falls. AC sweep uses the source’s AC analysis magnitude/phase fields. The transient waveform Frequency and Phase fields do not control this sweep.

Try two changes

1. Double C1 → Capacitance to approximately 5.066 µF (5.0661 µF). Resonance moves down to $f_0/\sqrt{2}\approx707.1\,\mathrm{Hz}$. Restore C1 to 2.533 µF.
2. Double R1 → Resistance to 2 kΩ. The voltage peak doubles to 2 V, Q doubles and the response narrows. The reactive branch-current peaks increase. Restore R1 to 1 kΩ afterward.

Model assumptions

This is small-signal sinusoidal steady-state analysis of ideal linear L and C with explicit resistance. It does not show switch-on ringing, saturation, breakdown, winding/ESR frequency dependence or parasitic resonances. Parallel resonance maximizes driving-point impedance; the current-source drive makes that visible as a voltage peak.