RLC Current Amplitude Across Frequencies (Linked to Circuits)
1 · Predict
This experiment uses the exact same R, L, C values as the rlc-series circuits experiment, viewed here as a phasor/AC problem instead of a switch-on transient. As you sweep the drive frequency away from resonance in either direction, does the current amplitude always decrease?
- Yes — moving away from resonance in either direction (higher or lower ω) decreases the current amplitude, since impedance rises on both sides.
- Current amplitude only decreases for frequencies above resonance.
- Current amplitude only decreases for frequencies below resonance.
2 · Set Up
- Open the rlc-in-circuits preset and press Reset. R = 20 Ω, L = 1 H, C = 0.01 F, V₀ = 10 V — identical to the rlc-series circuits experiment's components.
- Enable the current-amplitude readout.
- Set the drive angular frequency for each trial (as a multiple of ω₀ = 10 rad/s) and record the peak current amplitude.
3 · Collect Data
| Drive frequency ω (rad/s) | Peak current amplitude I_peak (A) |
|---|---|
| 5 | |
| 10 | |
| 20 |
Plot peak current I_peak (y-axis) against drive frequency ω (x-axis) for your three trials. Does the peak sit exactly at ω₀?
4 · Analyze
- For one trial, compute Z = √(R² + (ωL − 1/(ωC))²) using R = 20 Ω, L = 1 H, C = 0.01 F, then I_peak = V₀/Z using V₀ = 10 V. Compare to the table.
- Confirm your two off-resonance trials (below and above ω₀) give equal current amplitude when equally spaced from resonance in this specific case. Explain why current amplitude peaks exactly at ω₀ and falls off symmetrically on a log-frequency scale.
5 · Extend
- The rlc-series circuits experiment shows this same hardware's step response (switch-on transient) to a constant 10 V battery, not a sinusoidal drive. Explain why 'impedance' and 'resonance' are AC-steady-state concepts that don't directly apply to that DC switch-on scenario.
- How sharply current amplitude drops off away from resonance is described by a circuit's 'quality factor' Q = ω₀L/R. Using this circuit's R = 20 Ω, L = 1 H, ω₀ = 10 rad/s, compute Q and explain whether this circuit has a sharp or broad resonance peak.