RLC Current Waveform Away from Resonance
1 · Predict
This RLC circuit is driven off-resonance (at half its resonant frequency), so its current lags or leads the driving voltage by a fixed phase angle. Does the instantaneous current still oscillate sinusoidally at the same frequency as the drive?
- Yes — the current oscillates at the exact drive frequency, just phase-shifted relative to the voltage.
- No — the current oscillates at a different frequency than the drive.
- The current isn't sinusoidal at all in steady state.
2 · Set Up
- Open the rlc-off-resonance preset and press Reset. R = 20 Ω, L = 1 H, C = 0.01 F, drive ω = 5 rad/s (half the resonant ω₀ = 10 rad/s), V₀ = 10 V.
- Enable the instantaneous-current readout.
- Read the instantaneous current at each listed time.
3 · Collect Data
| Time t (s) | Instantaneous current i(t) (A) |
|---|---|
| 0.1 | |
| 0.3 | |
| 0.6 |
Plot current i(t) (y-axis) against time t (x-axis) for your three readings, and sketch a full cycle to show the phase lag relative to a sine wave in phase with the drive voltage.
4 · Analyze
- First compute Z = √(R² + X²) and φ = atan2(X, R) with X = ωL − 1/(ωC), using R = 20 Ω, L = 1 H, C = 0.01 F, ω = 5 rad/s (giving Z = 25 Ω, φ ≈ −0.64 rad). Then compute i(t) = (V₀/Z)sin(ωt − φ) using V₀ = 10 V. Compare to the table.
- The phase angle here is negative, meaning the circuit is capacitor-dominated at this frequency (below resonance). Explain what a negative φ means for whether the current leads or lags the driving voltage.
5 · Extend
- The current amplitude here (V₀/Z = 10/25 = 0.4 A) is smaller than it would be exactly at resonance (V₀/R = 10/20 = 0.5 A). Explain why driving an RLC circuit off-resonance always results in a smaller current amplitude than driving it at resonance.
- Because off-resonance frequencies produce smaller current amplitudes, an RLC circuit acts as a natural frequency filter. Explain why this circuit would respond weakly to a very high-frequency or very low-frequency signal mixed in with its resonant frequency.