RC and RL transients, impedance and resonance
| English | 中文 | Pinyin |
|---|---|---|
| impedance | 阻抗 | zǔ kàng |
| time constant/taɪm ˈkɒnstənt/ | 时间常数 | shí jiān cháng shù |
A decision before an answer
- A capacitor voltage and an inductor current cannot change instantly under finite ordinary circuit drives, but their steady-state behaviour is different.
- Your goal: Solve RC charging and discharging with stated initial conditions.
Solve the capacitor response
- For a resistor R in series with a capacitor C and a DC source V_s, Kirchhoff’s voltage law is R dq/dt+q/C=V_s. Define V_C=q/C and τ=RC. After a switch to a constant source, V_C(t)=V_s+(V_C(0)−V_s)exp(−t/τ). An initially uncharged capacitor therefore has V_C=V_s(1−exp(−t/τ)) and charging current I=(V_s/R)exp(−t/τ).
- With the source removed and the resistor connected across it, V_C=V_C(0)exp(−t/τ). Capacitor voltage is continuous across the switch if there is no impulsive current; the resistor current may change instantly.
After one time constant, an initially uncharged RC capacitor driven by a constant source reaches approximately:
V_C/V_s=1−exp(−1)≈0.632; exp(−1) is the remaining difference.
Track inductor current and energy
- In a series RL circuit, Kirchhoff’s law gives L dI/dt+RI=V_s and τ=L/R. The current after a constant drive is I(t)=V_s/R+(I(0)−V_s/R)exp(−t/τ). Inductor current is continuous for a finite voltage; its voltage may change abruptly as switching changes dI/dt.
- At late times an ideal inductor in a DC circuit acts as a zero-voltage connection, while an ideal capacitor has zero DC current. Stored energies are LI²/2 and CV_C²/2. During decay into a resistor, the initially stored energy becomes resistor heat rather than disappearing when the source is disconnected.
Using exp(iωt), a series circuit with X=ωL−1/(ωC)>0 has:
Z has a positive phase; I=V/Z has a negative phase relative to voltage.
Use the phasor convention
- Use the declared phasor convention exp(iωt). A resistor has impedance R, an inductor iωL and a capacitor 1/(iωC)=−i/(ωC). A series RLC circuit therefore has Z=R+iX with X=ωL−1/(ωC). Divide the source voltage phasor by Z to obtain current.
- Its magnitude is V_rms/sqrt(R²+X²) when rms quantities are used. The impedance phase φ satisfies tanφ=X/R; current lags voltage for X>0 and leads it for X<0. Adding the scalar magnitudes R, ωL and 1/(ωC) loses the vector phase information.
For R=2000 Ω, C=0.001 F, V_s=12 V and V_C(0)=0, τ=RC=2 s. At t=τ, V_C=12(1−e^(−1))=7.585 V and I=(12/2000)e^(−1)=2.207 mA. A separate RL circuit with L=0.4 H, R=2 Ω and V_s=10 V has τ=0.2 s and I(τ)=5(1−e^(−1))=3.161 A. For series R=10 Ω, L=0.1 H, C=0.001 F and V_rms=20 V, ω₀=100 rad/s; resonance gives I_rms=2 A and mean power 40 W.
An RL circuit has L=0.6 H and R=3 Ω. Its time constant is ____ s.
τ=L/R=0.6/3=0.2 s.
Separate resonance from decay
- Series resonance occurs at ω₀=1/sqrt(LC) when X=0. Current is maximal for a fixed voltage in this ideal series model, and the voltage/current phase difference is zero. Mean real power is V_rms I_rms cosφ=I_rms²R; ideal L and C exchange energy but dissipate no average power.
- This differs from transient natural frequency: a damped series circuit can oscillate at sqrt(1/(LC)−(R/(2L))²) when underdamped. Large resistance removes such free oscillation but does not change the condition X=0 of the ideal driven series impedance. State whether a question asks for a step response, free decay or sinusoidal steady state.
RC has τ=RC, while RL has τ=L/R. Do not confuse current amplitude with rms current, or resonance of a driven impedance with damped free-oscillation frequency.
Which answer fits this case?
Solve RC charging and discharging with stated initial conditions
An ideal inductor stores energy but dissipates no average real power in sinusoidal steady state.
Its ideal voltage/current phase is a quarter cycle; average power is zero while LI²/2 varies.
Keep the distinctions
- time constant 时间常数 — The exponential response scale, RC for an RC circuit and L/R for an RL circuit.
- impedance 阻抗 — The complex voltage-to-current phasor ratio in sinusoidal steady state.
- Solve RC charging and discharging with stated initial conditions.
- Solve RL current response and account for stored energy.
- Use complex impedance, phase and resonance in sinusoidal circuits.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.