Electrical Circuits and Sensors · Module 3 of 10
Nodal, Mesh, and Thevenin Analysis
When a network is too tangled to reduce by inspection, two systematic methods write the equations for you, and one powerful theorem replaces a whole circuit with a single source and resistor.
Readiness check
This module systematises the laws of Module 2. Tick only what you can do closed-notes.
- Write KCL at a node and KVL around a loop.
- Combine series and parallel resistors.
- Solve two simultaneous linear equations.
- Use the voltage and current divider rules.
- Identify the two terminals of a load in a circuit.
The core idea
Nodal analysis solves for node voltages with KCL; mesh analysis solves for loop currents with KVL. The Thevenin theorem then collapses everything outside a load into one source and one resistance.
nodal: KCL at each node → node voltagesmesh: KVL around each loop → loop currentsThevenin: VTh in series with RThNodal analysis picks a reference node, assigns an unknown voltage to every other node, and writes KCL at each, giving as many equations as unknown voltages. Mesh analysis assigns a circulating current to each independent loop and writes KVL around each, giving as many equations as loops. Either method turns any network into linear algebra. The Thevenin theorem goes further: any linear two-terminal network, however complex, behaves at its terminals exactly like a single voltage source VTh in series with a single resistance RTh. That equivalent makes load calculations trivial and underlies the idea of a sensor's output impedance.
The skills, taught in order
Five skills cover the two systematic methods, the Thevenin and Norton equivalents, and the maximum power transfer that follows.
3.1 Nodal analysis
Choose a reference (ground) node and label the others with unknown voltages. Write KCL at each non-reference node, expressing every branch current as a node-voltage difference over a resistance. Solve the resulting linear system for the node voltages, from which every current follows.
3.2 Mesh analysis
Assign a circulating current to each independent loop (mesh) and write KVL around each, summing the IR drops. A resistor shared by two meshes carries the difference of their currents. Solve for the mesh currents; branch currents are sums or differences of them.
3.3 The Thevenin equivalent
To find the equivalent seen at a load port, remove the load. The open-circuit voltage at the terminals is VTh. The Thevenin resistance RTh is the resistance looking back into the port with all independent sources zeroed (voltage sources shorted, current sources opened).
| Method | Law used | Unknowns | Best when |
|---|---|---|---|
| Nodal | KCL | node voltages | few nodes, many loops |
| Mesh | KVL | loop currents | few loops, planar circuit |
| Thevenin | both | VTh, RTh | one load to study repeatedly |
Choosing the method with the fewest unknowns saves the most work.
3.4 The Norton equivalent
The Norton equivalent is the dual: a current source IN in parallel with RN. The two are related by IN = VTh/RTh and RN = RTh, so you can convert freely between them depending on which form the rest of the circuit needs.
3.5 Maximum power transfer
A load draws the most power from a source when its resistance equals the Thevenin resistance, RL = RTh. At that match the load receives VTh2/(4RTh), and the source and load each dissipate half the total. This sets the impedance match in measurement and power circuits.
Engineering connection: a sensor and its conditioning amplifier are a Thevenin source feeding a load; matching their resistances is a recurring design choice.
Worked example 1: a Thevenin equivalent
A 32 V source feeds R1 = 4 Ω in series, then R2 = 12 Ω to ground, with the load port taken across R2. Find the Thevenin voltage and resistance seen at that port.
- ProblemFind VTh and RTh at the load port in Figure 1.
- Given / findV = 32 V, R1 = 4 Ω, R2 = 12 Ω, port across R2. Find VTh and RTh.
- AssumptionsLinear network; load removed for the open-circuit voltage.
- ModelVTh is the divider voltage across R2; RTh is R1 parallel R2 with the source shorted.
- EquationsVTh = V · R2/(R1 + R2)RTh = R1R2/(R1 + R2)
- SolveVTh = 32 × 12/(4 + 12) = 32 × 12/16 = 24 V. RTh = (4 × 12)/16 = 48/16 = 3 Ω.
- CheckThe open-circuit current through the series pair is 32/16 = 2 A, and 2 × 12 = 24 V across R2, confirming VTh.
- ConclusionAny load placed at this port sees a 24 V source behind 3 Ω, no matter how the original network was drawn.
Worked example 2: loading the Thevenin source
Connect a load RL = 6 Ω to the Thevenin equivalent from Example 1 (VTh = 24 V, RTh = 3 Ω). Find the load current and the load voltage.
- ProblemFind the load current and load voltage for the circuit in Figure 2.
- Given / findVTh = 24 V, RTh = 3 Ω, RL = 6 Ω. Find IL and VL.
- AssumptionsThe Thevenin equivalent fully represents the original network at this port.
- ModelSeries loop: the current is VTh over the total resistance; VL = ILRL.
- EquationsIL = VTh/(RTh + RL)VL = IL RL
- SolveIL = 24/(3 + 6) = 24/9 = 2.67 A. VL = 2.67 × 6 = 16 V.
- CheckThe drop across RTh is 2.67 × 3 = 8 V, and 8 + 16 = 24 V satisfies KVL. The divider gives VL = 24 × 6/9 = 16 V directly.
- ConclusionThe Thevenin model turns a loaded network into a one-line calculation, the payoff for finding the equivalent once.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| No reference node | Nodal equations have no solution | "Which node did I set to zero?" | Always fix one node as ground first. |
| Sources left in for RTh | RTh comes out too large | "Did I zero the independent sources?" | Short voltage sources, open current sources. |
| Shared resistor mishandled | Mesh equation double counts a drop | "Whose currents cross this resistor?" | Use the difference of the two mesh currents. |
| Wrong match for power transfer | Load chosen by largest current | "Does RL equal RTh?" | Maximum power needs RL = RTh, not the smallest RL. |
Practice ladder
A 10 V source feeds 2 Ω in series with 8 Ω to ground, port across the 8 Ω. Find VTh.
Show answer
VTh = 10 × 8/(2 + 8) = 8 V.
For the Level 1 circuit, find RTh and the current into a 4 Ω load.
Show answer
RTh = (2 × 8)/10 = 1.6 Ω. IL = 8/(1.6 + 4) = 8/5.6 = 1.43 A.
For the Example 1 source (VTh = 24 V, RTh = 3 Ω), what load gives maximum power, and how much power is that?
Show answer
RL = RTh = 3 Ω. Pmax = VTh2/(4RTh) = 576/12 = 48 W.
A sensor has an output resistance of 1 kΩ. Explain why a high-impedance amplifier input reads its voltage more accurately than a low one.
What good work looks like
The sensor is a Thevenin source; a load near 1 kΩ would halve the reading, so an input far above 1 kΩ draws little current and the measured voltage stays close to the true open-circuit value.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take a multi-resistor network with a defined load port, find its Thevenin equivalent, and verify it by computing the load current both from the full circuit and from the equivalent.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What does nodal analysis solve for?
The node voltages, by writing KCL at each non-reference node.
2. What does mesh analysis solve for?
The loop currents, by writing KVL around each mesh.
3. How do you find VTh and RTh?
VTh is the open-circuit terminal voltage; RTh is the resistance looking in with sources zeroed.
4. How are Thevenin and Norton related?
IN = VTh/RTh and RN = RTh.
5. State the maximum power transfer condition.
RL = RTh, giving Pmax = VTh2/(4RTh).
Textbook mapping
This module follows Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Nodal and mesh analysis | Alexander & Sadiku, Chapter 3 |
| Thevenin and Norton theorems | Alexander & Sadiku, Chapter 4 |
| Maximum power transfer | Alexander & Sadiku, Chapter 4 |
Chapter numbers refer to the 4th edition. The theorems are standard, so any recent edition will align closely.