Resonance
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| resonance/ˈrezənəns/ | cộng hưởng |
| bandwidth/ˈbændwɪdθ/ | băng thông |
Resonance: the imaginary parts cancel
- At resonance $ω_{0} = 1/\sqrt{LC}$ the series impedance drops to pure $R$ (its minimum) and current peaks; the bandwidth between the half-power points is $BW = R/L = ω_{0}/Q$, and each reactive element holds Q times the source voltage — check the magnification on both elements.
- The parallel tank is the dual: maximum impedance, circulating branch currents Q × the line current that cancel at the terminals.
$R = 5$ Ω, $L = 0.1$ H, $C = 10$ µF: $ω_{0} = 1000$ rad/s, $Q = ω_{0}L/R = 20$. A 10 V source drives 2 A, so $V_{L} = 200$ V and $V_{C} = 2/0.01 = 200$ V — equal, 180° apart. $BW = 1000/20 = 50$ rad/s with half-power frequencies 975.3 and 1025.3 rad/s (difference 50.0, machine-verified). One bandwidth away at 900 rad/s the current is $10/|5 + j(90 - 111)| = 0.45$ A — under a quarter of the peak: that selectivity is what a tuner sells.

{exercise:uni-ee4x-resonance}
A series RLC with Q = 20 at resonance on a 10 V source has V_C equal to —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
I = 10/R flows; V_C = I/(ω₀C) = Q × V_source = 200 V. V_L is also 200 V — equal and opposite, cancelling each other while each alone is twenty times the source.
A series RLC has omega-0 = 1000 rad/s and Q = 20. The half-power frequencies are
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
BW = omega-0/Q = 50 rad/s; the exact half-power points are omega-0(-1/2Q ± sqrt(1 + 1/4Q^2)) = 975.3 and 1025.3 - difference exactly 50.
At resonance with Q = 20 and a 10 V source, the inductor voltage is
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Current is V/R = 2 A and V_L = omega-0 L I = Q x source = 200 V; the capacitor also reads 200 V, 180 degrees apart - check both.