Nodal and mesh analysis
引入| English | 释义 | 拼音 |
|---|---|---|
| mesh analysis/meʃ əˈnæləsɪs/ | 网孔法 | wǎng kǒng fǎ |
| nodal analysis/ˈnəʊdl əˈnæləsɪs/ | 节点法 | jié diǎn fǎ |
| reference node/ˈrefrəns nəʊd/ | 参考节点 | cān kǎo jié diǎn |
| supernode/ˈsuːpənəʊd/ | 超节点 | chāo jié diǎn |
| mesh current/meʃ ˈkʌrənt/ | 网孔电流 | wǎng kǒng diàn liú |
A recipe a computer could run
- A circuit with four nodes and six branches can be solved by clever guessing, once. A designer meets the same circuit a hundred times and needs a method that always works.
- Two such methods exist, nodal and mesh analysis. Both turn the circuit into a few simultaneous linear equations.
Nodal analysis: KCL does the walking
- Pick a reference node 参考节点 (ground), label the other node voltages, and write one KCL equation per node with each current written as a voltage difference over a resistance. Nodal analysis 节点法 is what every circuit simulator solves.
- 12 V feeds node A through 2 Ω; A has 6 Ω to ground and a 1 A source pushing current in. Currents leaving A: $\dfrac{V_{A} - 12}{2} + \dfrac{V_{A}}{6} - 1 = 0$.
Multiply by 6: $3V_{A} - 36 + V_{A} - 6 = 0$, so $V_{A} = 10.5\,\text{V}$. Check: the source delivers $\dfrac{12 - 10.5}{2} = 0.75\,\text{A}$, plus 1 A in, and the 6 Ω carries exactly $\dfrac{10.5}{6} = 1.75\,\text{A}$.
一个节点、四个滑动变阻器、一个方程
Solve the node live and watch the signed KCL sum stay at zero.
A 12 V source feeds a node through 4 Ω; a 6 Ω runs from the node to ground. What is the node voltage?
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The divider 12 × 6/(4+6) = 7.2 V equals the nodal answer; both branch currents then equal 1.2 A.
Put the steps of nodal analysis in order.
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The reference must come first, because every other node voltage is measured from it. The final check uses a route the solve never touched.
A floating voltage source: the supernode
- A voltage source tied to ground fixes one node outright. A source between two unknown nodes is "floating": its current is unknown, so KCL at either end alone cannot be written.
- Draw a supernode 超节点 around both ends: write one KCL for the whole region, and add the source's equation $V_{P} - V_{Q} = V_{s}$.
A 5 V source joins P (+) to Q. P has 10 Ω to ground, Q has 5 Ω to ground, and 2 A is injected into P. Supernode: $\dfrac{V_{P}}{10} + \dfrac{V_{Q}}{5} = 2$ with $V_{Q} = V_{P} - 5$, so $3V_{P} - 10 = 20$: $V_{P} = 10\,\text{V}$, $V_{Q} = 5\,\text{V}$. Check: 1 A + 1 A leave, 2 A came in.
A 4 V source joins P (+) to Q. Both P and Q have 4 Ω to ground, and 2 A is injected into P. What is V_Q, in volts?
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Supernode: $\dfrac{V_{P}}{4} + \dfrac{V_{Q}}{4} = 2$, so $V_{P} + V_{Q} = 8$; with $V_{P} - V_{Q} = 4$ this gives $V_{P} = 6\,\text{V}$ and $V_{Q} = 2\,\text{V}$.
Mesh analysis: KVL around each window
- Give every window a clockwise mesh current 网孔电流 and write KVL per window; a shared resistor appears as $R(I_{1} - I_{2})$. Mesh analysis 网孔法 suits planar circuits and current sources.
- 12 V through 1 Ω, sharing 2 Ω with a source-free loop that has its own 2 Ω: $3I_{1} - 2I_{2} = 12$ and $2(I_{2} - I_{1}) + 2I_{2} = 0$.
The second equation gives $I_{1} = 2I_{2}$, so $4I_{2} = 12$: $I_{2} = 3\,\text{A}$, $I_{1} = 6\,\text{A}$, and the shared resistor carries $6 - 3 = 3\,\text{A}$. Loop 1 check: $-12 + 6 + 6 = 0$.
Mesh 1: a 12 V source and 2 Ω. Mesh 2: 6 Ω and no source. The meshes share 3 Ω. What current flows in the shared resistor, in amperes?
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Mesh 2: $3(I_{2} - I_{1}) + 6I_{2} = 0$, so $I_{1} = 3I_{2}$. Mesh 1: $5I_{1} - 3I_{2} = 12$ gives $12I_{2} = 12$: $I_{2} = 1\,\text{A}$, $I_{1} = 3\,\text{A}$, shared current $3 - 1 = 2\,\text{A}$.
Two unknown nodes, one clean finish
- For a chain like 12 V → 2 Ω → X → 4 Ω → Y → 6 Ω → ground with 1 A into Y, write both equations and clear fractions before solving.
- Verify through a node or branch you did not use for the solve: the check is where errors are caught.
$3V_{X} - V_{Y} = 24$ and $-3V_{X} + 5V_{Y} = 12$ add to $4V_{Y} = 36$, so $V_{Y} = 9\,\text{V}$ and $V_{X} = 11\,\text{V}$. At Y: 0.5 A in from X, 1 A in from the source, $\dfrac{9}{6} = 1.5\,\text{A}$ out. Balanced.
In the X-Y chain (V_X = 11 V, V_Y = 9 V), what does the 6 Ω to ground carry?
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9/6 = 1.5 A out, which balances the 0.5 A from X plus the 1 A source — KCL closes at Y.
Choose the method by counting
- Nodal needs one equation per unknown node; each voltage source removes one. Mesh needs one equation per window; each current source removes one.
- Count both before writing anything, and pick the smaller set. Choosing the method is part of the answer.
A planar circuit has three loops and five nodes, with two grounded voltage sources. Which method gives fewer unknowns?
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Grounded sources fix two nodes outright, leaving 5 − 1 − 2 = 2 nodal unknowns versus 3 mesh currents.
Three common slips
- The shared resistor's current. In mesh analysis it is the difference of the two mesh currents, never one of them alone.
- Nodal for a circuit full of current sources. Each source becomes a constraint you already know; mesh would have been shorter.
- Ground inside a voltage source's pair of nodes. Put the reference at the source's negative terminal, so the source fixes one node directly.
In mesh analysis, a resistor shared by two meshes carries the sum of the two mesh currents.
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With both mesh currents drawn clockwise, they run in opposite directions through the shared resistor, so it carries their difference.
Recap
- Nodal: KCL at each unknown node, currents as voltage differences over resistances.
- Floating voltage source: one KCL around both ends, plus the source equation.
- Mesh: KVL per window, shared elements see differences; choose by counting equations.