Source transformations and network theorems
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| Thévenin's theorem | Định lý Thévenin |
| Norton's theorem/ˈnɔːtnz ˈθɪərəm/ | Định lý Norton |
| source transformation/sɔːs trænsfɔːˈmeɪʃn/ | Biến đổi nguồn |
| superposition/ˌsuːpəpəˈzɪʃn/ | định lý chồng chập |
Thévenin: one source, one resistor, same terminals
- Thévenin's theorem 戴维南定理 replaces any linear two-terminal network by $V_{th}$ in series with $R_{th}$. $V_{th}$ is the open-circuit voltage; $R_{th}$ is the resistance seen with sources zeroed — voltage sources shorted, current sources opened.
- Norton's theorem 诺顿定理 is the parallel form $I_{N} = V_{th}/R_{th}$; swapping is the source transformation 电源变换.
10 V with 4 Ω series feeding a 6 Ω shunt: $V_{th} = 10 \times 6/10 = 6 V$, $R_{th} = 4 ∥ 6 = 2.4 Ω$. Any load now sees just those two numbers.
{exercise:uni-ee14-thevenin}
A 10 V source feeds 5 Ω in series with a 20 Ω shunt. What are V_th and R_th at the shunt terminals?
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V_th = 10 × 20/25 = 8 V; with the source shorted, R_th = 5 ∥ 20 = 4 Ω. Adding the series resistor to R_th is the classic slip.
Maximum power, and its 50% catch
- A load draws the most power when $R_{L} = R_{th}$, and then $P_{max} = V_{th}^{2}/(4R_{th})$. With the panel's 6 V and 2.4 Ω:
36/9.6 = 3.75 W. - For a supply with a physical series resistor, the match gives equal resistor and load power — 50% efficiency. A Thévenin equivalent preserves terminal behaviour, not the original network's internal losses.
{exercise:uni-ee14-maxpower}
An ideal voltage source has a physical series resistor R_s. With R_L = R_s, what is the power-delivery efficiency?
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Equal resistances in series share the voltage equally, so they dissipate equal powers — the source supplies twice the load's power.
Superposition: add responses, not powers
- Superposition 叠加定理 adds the responses to each independent source alone (others zeroed). 9 V through 2 Ω gives 0.9 A in the 8 Ω; a 1 A source pulling out of the node contributes 0.2 A up; total 0.7 A down — matching the nodal answer
5.6/8. - Power is quadratic: adding sub-powers (6.48 + 0.32 W) misses the true 3.92 W.
Two 3 Ω-fed sources (6 V and 3 V) into a 3 Ω ground: alone they give 2 V and 1 V; together 3 V. But alone-dissipations 1.33 W + 0.33 W ≠ the true $3^{2}/3 = 3 W$.
{exercise:uni-ee14-superposition}
Why does superposition fail for power?
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p = i²R; (i₁ + i₂)² contains the cross term 2i₁i₂ that the sum of separate powers omits.