Second-order RLC circuits
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| natural frequency/ˈnætʃərəl ˈfriːkwənsi/ | tần số riêng |
| damping/ˈdæmpɪŋ/ | Tắt dần |
| overdamped | Quá tắt dần |
| critically damped | Tắt dần tới hạn |
| underdamped | Chưa tắt dần |
| Q factor | Hệ số chất lượng |
Second order: two numbers decide everything
- A series RLC has natural frequency $ω_{0} = 1/\sqrt{LC}$ and damping $α = R/(2L)$; one comparison picks the regime — overdamped ($α > ω_{0}$, two real decays), critically damped ($α = ω_{0}$, fastest return without ringing), underdamped ($α < ω_{0}$, ringing at $ω_{d} = \sqrt{ω_{0}^{2} - α^{2}}$). The Q factor $ω_{0}/(2α)$ says how long it rings.
- Two energy stores give two constants, so two endpoint checks are free: initial value and initial slope (no instantaneous current through L, no instantaneous voltage jump on C).
$L = 0.2$ H, $C = 0.05$ F gives $ω_{0} = 10$ rad/s. $R = 4$ Ω: $α = 10$, critical. $R = 2$ Ω: $α = 5$, underdamped at $ω_{d} = \sqrt{100 - 25} = 8.66$ rad/s. $R = 8$ Ω: $α = 20$, overdamped with roots $-20 \pm \sqrt{300}$, i.e. $-2.68$ and $-37.32$ per second (machine-verified); released from 1 V, $A = 37.32/34.64 = 1.077$ and $B = -0.077$ — the slow root $-2.68$ owns the tail, the fast one only the first moments.

{exercise:uni-ee4x-second}
R = 4 Ω, L = 0.2 H, C = 0.05 F. The regime is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
ω₀ = 1/√(LC) = 10 and α = R/2L = 10 — exactly equal: critical. Change R to 2 Ω and α = 5 < 10 becomes underdamped with ω_d = 8.66.
Series RLC with R = 8, L = 0.2 H, C = 0.05 F: the two natural exponents are
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
alpha = 20 > omega-0 = 10, overdamped: s = −20 ± sqrt(300) = −2.68 and −37.32. The slow root −2.68 owns the tail (dominant pole).
Series RLC with alpha = omega-0 is
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The regime test is one comparison: alpha = omega-0 is the boundary; R = 4 ohms with L = 0.2 H, C = 0.05 F lands exactly on it (both 10).