RCおよびRL過渡現象、インピーダンスおよび共振
導入済み| English | 意味 |
|---|---|
| impedance/ɪmˈpiːdəns/ | インピーダンス |
| time constant/taɪm ˈkɒnstənt/ | 時間定数 |
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:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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:
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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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
このケースに当てはまる答えはどれか？
RC回路の充電および放電を、指定された初期条件下で解く
An ideal inductor stores energy but dissipates no average real power in sinusoidal steady state.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.