Parallel Tank: \(C \parallel (R+L)\)

The satisfying part: anti-resonance (impedance peak), circulating reactive currents, and clean pulse intuition

This topology is a “voltage-driven parallel tank”: the source voltage \(v(t)\) is applied across both branches and the source current splits. Its signature behavior is not a current peak (series RLC) — it is an impedance peak (anti-resonance).

Topology definition. A capacitor \(C\) is in parallel with a series branch \(R+L\). The same voltage appears across both: \[ v_C(t)=v(t),\qquad v_{RL}(t)=v(t) \] Branch currents: \[ i_{src}(t)=i_{RL}(t)+i_C(t) \]

1) One sentence intuition (the correct “mental picture”)

Why this feels “different” than series RLC. Series resonance gives a current maximum. Parallel anti-resonance gives a source current minimum (impedance maximum) — while still allowing large internal circulating reactive currents.

2) Sine excitation (the right tool is admittance)

Parallel networks are easiest in admittance: \[ I_{src}=V\cdot Y \]

Branch admittances

\[ Z_{RL}=R+j\omega L,\qquad Y_{RL}=\frac{1}{Z_{RL}}=\frac{R-j\omega L}{R^2+(\omega L)^2} \] \[ Y_C=j\omega C \] \[ Y=Y_{RL}+Y_C \]

Anti-resonance condition: \(\operatorname{Im}(Y)=0\)

The “magic point” is when net susceptance is zero:

\[ \operatorname{Im}(Y)=0 \] \[ \Rightarrow\ \omega C-\frac{\omega L}{R^2+(\omega L)^2}=0 \] \[ \Rightarrow\ \omega^2=\frac{1}{LC}-\frac{R^2}{L^2} \]
What this means physically. At this frequency, the capacitor draws a leading reactive current while the RL branch provides a lagging reactive component. They cancel in the vector sum \(I_{src}=I_{RL}+I_C\), so the source sees (almost) pure conductance.

Impedance peak (why it’s satisfying)

At anti-resonance, the total admittance becomes (approximately) real: \[ Y(\omega_{ar}) \approx G(\omega_{ar})=\frac{R}{R^2+(\omega_{ar}L)^2} \] so the equivalent impedance peak is: \[ |Z_{eq}(\omega_{ar})|\approx \frac{1}{G}=\frac{R^2+(\omega_{ar}L)^2}{R} \] That can be much larger than \(R\), which is why the source current becomes small even though branch currents can be large.

What you should look for in the simulator (sine)

3) Pulse excitation (edges + slow branch)

In pulse mode, the capacitor voltage is clamped by the source: \[ v_C(t)=v(t) \] Therefore the capacitor current is entirely determined by the voltage edge shape: \[ i_C(t)=C\frac{dv}{dt} \]

Do not use ideal steps for intuition. An ideal step has \(dv/dt\to\infty\) (impulse current). Real edges have finite rise time \(t_r\). A good lab estimate is: \[ i_{C,peak}\approx C\frac{\Delta V}{t_r} \]

RL branch under pulses

The RL branch follows the same first-order dynamic as the pure RL case, driven by the same \(v(t)\): \[ L\frac{di_{RL}}{dt}+R\,i_{RL}=v(t),\qquad \tau=\frac{L}{R} \] So you get:

What you should look for in the simulator (pulse)

4) Worked examples (replicable in the simulator)

The simulator does not expose a separate “edge rise time” parameter, so the worked examples below avoid any \(t_r\)-based peak-current claims and instead focus on things you can set and verify directly: frequency, L/C/R totals, phase angle, P/Q/S/PF, and the impedance peak (“anti-resonance”) behavior.

Example A — Sine anti-resonance (impedance peak + PF improvement)

Simulator setup

Prediction (frequency)

Parallel anti-resonance (net susceptance zero) occurs when: \[ \omega_{ar}^2=\frac{1}{LC}-\frac{R^2}{L^2} \] For \(L=10\,\text{mH}\), \(C=10\,\mu\text{F}\), \(R=1\,\Omega\): \[ \omega_{ar}=\sqrt{10^7-10^4}=3160.70\,\text{rad/s} \qquad\Rightarrow\qquad f_{ar}=\frac{\omega_{ar}}{2\pi}=503.04\,\text{Hz} \]

What to verify in the simulator

Why this is the “parallel wow moment”. Series resonance gives a current maximum. Parallel anti-resonance gives a source current minimum. So the correct “success criterion” here is a dip in source current + PF improvement near \(f_{ar}\).

Example B — Sine detuning: sign flip of reactance around the peak

Simulator setup: keep the same values as Example A.

Do this

What should happen (matches the simulator)

Sign convention note (important). Use the simulator’s Q sign as the authoritative indicator: \(Q>0\) → net inductive, \(Q<0\) → net capacitive. The displayed phase \(\phi\) follows the same convention shown in the UI.
Replicable check. You do not need internal branch-current probes to validate this topology. The sign flip of \(Q\) and the PF peak around \(f_{ar}\) is an external, instrument-style validation that the admittance cancellation is happening.

Example C — Pulse train: isolate the RL time constant inside the parallel network

Pulse mode in this topology can look “less clean” than series RLC because the capacitor branch reacts instantly to changes in \(v(t)\). Instead of chasing edge peaks, the replicable check is: the slow envelope in current is still governed by \(\tau=L/R\).

Simulator setup

Prediction

The RL branch time constant is: \[ \tau=\frac{L}{R}=\frac{10\,\text{mH}}{1\,\Omega}=10\,\text{ms} \] Since \(T=5\,\text{ms} < \tau\), the “slow” part of the waveform cannot fully rise or decay within one cycle. So you should observe a gentle exponential envelope (limited change per period).

What to verify in the simulator

Key idea. In parallel \(C \parallel (R+L)\), the capacitor branch explains “fast” behavior and the RL branch explains “slow” behavior. If your simulator shows the correct \(\tau\)-scaling when you vary \(L\) or \(R\), the model is behaving correctly.

5) Why this model is “safe” in the RLC Analyzer

Practical takeaway. If you can confirm (1) a source current dip near \(f_{ar}\), (2) PF peak / \(\phi\to 0^\circ\) there, and (3) the correct sign flip of \(Q\) across \(f_{ar}\), then you have robust parallel-tank intuition and a strong reason to trust the simulator here.

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