Take the same coil, capacitor and resistor from the series RLC world, but wire them in PARALLEL and push a fixed AC current through them. Does the tank still 'prefer' ω₀ = 1/√(LC) — and does it respond the same way the series loop did, or the opposite?
▶ Launch the interactive simulationDrive a parallel L‖C‖R tank with a fixed-amplitude current source I₀cos ωt, so its admittance is Y = 1/R + j(ωC − 1/ωL) and its impedance |Z|(ω) = 1/√((1/R)² + (ωC−1/ωL)²). The lab sweeps the drive frequency through ω₀ and traces |Z|(ω), shading the half-power band, while the live coil (½LI_L²) and capacitor (q²/2C) hand the energy back and forth and a pulse races the circulating current round the L↔C loop. The same AM tank — L=250 µH, C=100 pF — is read as a parallel resonator with loss R_p≈250 kΩ and swept with ±1.5% reading noise.
It is the exact DUAL of the series loop. The impedance |Z| PEAKS sharply at ω₀ = 1/√(LC) (where it dipped before), so a fixed drive current produces a huge node voltage there and the tank looks like a bare resistor R — antiresonance. The sharpness flips to Q = R√(C/L) = ω₀R_pC, so a BIGGER R now sharpens the tank (it damped the series one). And at ω₀ the two branch currents are equal and opposite, nearly cancelling at the source: a current Q·I₀ circulates in the L↔C loop while the source draws only I₀ (current magnification). A high impedance at ω₀ BLOCKS that frequency in a line — the parallel tank is the band-stop 'trap' and the oscillator's frequency-setting resonator, the mirror of the series tank that PASSES ω₀. Reading the real tank recovers f₀ ≈ 1.01 MHz and Q ≈ 158 — and that same Q is the current magnification I_circ/I₀ ≈ 158×