Source transformations and network theorems
Introduzido| English | Significado |
|---|---|
| Thévenin's theorem | Teorema de Thévenin |
| open-circuit voltage | open-circuit voltage |
| Norton's theorem/ˈnɔːtnz ˈθɪərəm/ | Teorema de Norton |
| source transformation/sɔːs trænsfɔːˈmeɪʃn/ | transformação de fonte |
| maximum power transfer/ˈmæksɪməm ˈpaʊə ˈtrænsfɜː/ | A load receives the most power when its resistance equals the source's Thévenin resistance. |
| superposition/ˌsuːpəpəˈzɪʃn/ | princípio da superposição |
What does the lamp see?
- You replace a sensor board and the lamp it feeds glows slightly brighter. To the lamp, the board was never a schematic: it was one source and one resistance.
- Three theorems make that replacement exact. They all say one thing: a linear network is known completely by its behaviour at two terminals.
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 开路电压; $R_{th}$ is the resistance seen with the sources zeroed: voltage sources shorted, current sources opened.
- Two measurements find it, and any load then sees only those two numbers.
10 V with 4 Ω in series, feeding a 6 Ω shunt: $V_{th} = 10 \times \dfrac{6}{10} = 6\,\text{V}$, and $R_{th} = 4 \parallel 6 = 2.4\,\Omega$.
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.
Norton, and swapping between the two
- Norton's theorem 诺顿定理 is the same port as a current source $I_{N} = \dfrac{V_{th}}{R_{th}}$ in parallel with the same $R_{th}$.
- Swapping one form for the other is a source transformation 电源变换. Use whichever makes the next step easier: series forms add in a loop, parallel forms add at a node.
The 6 V, 2.4 Ω equivalent above becomes $I_{N} = \dfrac{6}{2.4} = 2.5\,\text{A}$ in parallel with 2.4 Ω. Open-circuit check: $2.5 \times 2.4 = 6\,\text{V}$.
A network has V_th = 9 V and R_th = 3 Ω. What is its Norton current, in amperes?
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$I_{N} = \dfrac{V_{th}}{R_{th}} = \dfrac{9}{3} = 3\,\text{A}$, in parallel with the same 3 Ω.
Maximum power, and its 50% catch
- The load power is $P = \left(\dfrac{V_{th}}{R_{th}+R_{L}}\right)^{2} R_{L}$. It peaks when $R_{L} = R_{th}$, at $P_{max} = \dfrac{V_{th}^{2}}{4R_{th}}$. With 6 V and 2.4 Ω: $\dfrac{36}{9.6} = 3.75\,\text{W}$.
- At the match the source resistance burns as much as the load receives: 50% efficiency. This holds for a real series resistor; a Thévenin equivalent keeps terminal behaviour, not the original network's internal losses.
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.
A source has V_th = 12 V and R_th = 4 Ω. What is the largest power it can deliver to a load, in watts?
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Match the load, $R_{L} = 4\,\Omega$: $P_{max} = \dfrac{12^{2}}{4 \times 4} = 9\,\text{W}$. The source resistance also dissipates 9 W.
A signal rule, not an energy rule
- Maximum power transfer 最大功率传输 matters where the signal is weak and precious: audio amplifiers, antennas and radio receivers match resistance (impedance) on purpose.
- Power grids never match. At 50% efficiency every bill would double, so a supply is built with a source resistance far below the load's.
Power stations match their source resistance to the load so that homes receive the most power.
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Matching gives only 50% efficiency. Supplies are built with a source resistance far below the load's, so most of the energy reaches the load.
Superposition: add responses, not powers
- Superposition 叠加定理 adds the responses to each independent source acting alone, with the others zeroed. It works for voltages and currents because the circuit is linear.
- Power is not linear: it goes with the square. Find the total current or voltage first, then compute the power.
A node fed by 6 V through 3 Ω and 3 V through 3 Ω, with 3 Ω to ground. Alone, the sources give 2 V and 1 V; together 3 V. Adding the "alone" powers in the 3 Ω, $1.33 + 0.33 = 1.67\,\text{W}$, misses the true $\dfrac{3^{2}}{3} = 3\,\text{W}$.
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.
Three common slips
- $R_{th}$ with the sources still alive. Zero them first: short each voltage source, open each current source.
- Superposition applied to power. Add currents or voltages, then square.
- Treating 50% efficiency as a design target. Matching is for signals; supplies aim for a source resistance much smaller than the load.
Which steps are correct when finding R_th? Select all that apply.
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R_th is the resistance seen at the terminals with the independent sources zeroed and the load removed. Measuring with live sources gives V_th, not R_th.
Recap
- Thévenin: $V_{th}$ open-circuit, $R_{th}$ with sources zeroed. Norton: $I_{N} = \dfrac{V_{th}}{R_{th}}$ in parallel.
- Maximum power at $R_{L} = R_{th}$: $\dfrac{V_{th}^{2}}{4R_{th}}$, at 50% efficiency.
- Superposition adds responses, never powers.