03 · Intermediate

Worked resistor networks

From the familiar divider to a multi-node network where Thévenin is genuinely useful.

45–60 minIntermediate

In this lesson

Recognize the output resistance of a loaded divider.
Use nodal analysis to find VTH in a network that cannot be reduced by one divider formula.
Reduce the source-deactivated network to find RTH.
Predict any load voltage without re-solving the whole original circuit.

Starting question

If a circuit has two unknown node voltages, is Thévenin still useful, or is it merely extra work after nodal analysis?

The answer is the reason equivalent circuits matter in engineering: nodal analysis solves the internal network once; the Thévenin model stores the part of that solution that every future load needs. The payoff appears when the load changes, when tolerances are swept, or when the “load” is an instrument or transistor input.

solve the source network oncereuse the port model many times

03.1Why the familiar divider shortcut stops working

Consider the network below. A 12 V source feeds R₁ = 2 kΩ into node V_A. From that node, R₂ = 3 kΩ goes to ground and R₃ = 4 kΩ continues to the output. Finally R₄ = 6 kΩ connects the output to ground.

Figure 3.1 — The network seen before the load is connected

The load port is at VOUT. R2 already loads node VA, so the right-hand branch cannot be treated as an isolated divider.

two-node topology verified

find V_TH at VOUTthen look back for R_TH

A tempting mistake is to call R₃ and R₄ a voltage divider driven by an independently known V_A. But V_A itself depends on the current flowing through R₃ and R₄. The two nodes are coupled. That is precisely the situation nodal analysis was designed for.

03.2Complete worked example — find the open-circuit port voltage

The future external load is removed, but R₄ remains part of the source network. Let the two unknown node voltages be V_A and V_B, with V_B = V_TH because node B is the open output port.

node A: (V_A − 12)/2 kΩ + V_A/3 kΩ + (V_A − V_B)/4 kΩ = 0

node B: (V_B − V_A)/4 kΩ + V_B/6 kΩ = 0

Clear the denominators rather than jumping directly to a calculator. Multiplying both equations by 12 kΩ gives a small linear system whose structure remains visible.

13V_A − 3V_B = 72

−3V_A + 5V_B = 0

therefore V_A = 45/7 V = 6.429 V

V_TH = V_B = 27/7 V = 3.857 V

03.3Now find the resistance seen from the port

For R_TH, the 12 V independent source is set to 0 V and therefore replaced by a short. Seen from node A, R₁ and R₂ both go to ground.

R₁ ∥ R₂ = (2 kΩ × 3 kΩ)/(2 kΩ + 3 kΩ) = 1.2 kΩ

R₃ + (R₁ ∥ R₂) = 4.0 kΩ + 1.2 kΩ = 5.2 kΩ

R_TH = 6.0 kΩ ∥ 5.2 kΩ = 2.786 kΩ

Notice that the algebra is now easier than the open-circuit voltage calculation. Thévenin extraction does not require the same method for the intercept and the slope: this is the distinction established in chapter 02.

03.4The payoff — change the load without re-solving the source network

Figure 3.3 — The same port problem after reduction

Rounded equivalent values are shown. Once VTH and RTH are known, changing the load no longer requires solving the two-node internal network.

equivalent topology verified

Connect R_L = 4.7 kΩ. The entire five-component source network is now represented at the port by V_TH = 3.857 V and R_TH = 2.786 kΩ.

V_L = V_TH · R_L/(R_TH + R_L)

= 3.857 × 4.7/(2.786 + 4.7) = 2.422 V

I_L = 2.422 V / 4.7 kΩ = 0.515 mA

LoadPredicted port voltageInterpretation

∞3.857 VOpen circuit: no loading.

100 kΩ3.753 VSmall loading because RL ≫ RTH.

10 kΩ3.017 VLoading is already visible.

4.7 kΩ2.422 VThe worked load case.

1 kΩ1.019 VThe source is heavily loaded.

This table is the practical reason for doing the reduction. Without Thévenin, each row would invite another two-node solve. With the equivalent, a load sweep becomes one divider equation.

03.5Independent verification — solve the loaded original once

Figure 3.2 — Reconnect a 4.7 kΩ load

The original network is now loaded. We will solve it once more only as an independent check of the Thévenin prediction.

loaded topology verified

prediction first: V_L = 2.422 Vdirect nodal solution second

A trustworthy worked example should not verify itself by repeating the same algebra. Add the 4.7 kΩ load to the original node-B KCL equation and solve the original network directly.

node B: (V_B − V_A)/4 kΩ + V_B/6 kΩ + V_B/4.7 kΩ = 0

direct solution: V_B = 1269/524 V = 2.4218 V

The direct result agrees with the 2.422 V equivalent-circuit prediction to rounding. The agreement is stronger evidence than the fact that the reduced drawing “looks reasonable”: equivalence means equal terminal behaviour.

03.6Faded support — change one component and do more of the work yourself

Keep the same network, but change R₄ from 6 kΩ to 12 kΩ. Before opening the hints, predict qualitatively what should happen to the open-circuit output voltage and to R_TH.

Hint 1 — set up, do not solve yet

The node-A KCL is unchanged in form. In node-B KCL, replace VB/6 kΩ by VB/12 kΩ.

Hint 2 — resistance path

After deactivating the source, R1 ∥ R2 = 1.2 kΩ still holds. The second path seen from the output is therefore R3 + 1.2 kΩ = 5.2 kΩ.

Check your result

VTH = 216/43 V = 5.023 V and RTH = 12 kΩ ∥ 5.2 kΩ = 3.628 kΩ. Both increased, which matches the qualitative prediction: the output is less strongly shunted to ground.

03.7Three mistakes worth learning to diagnose

MistakeWhy it failsDiagnostic question

Treat R3–R4 as an isolated dividerVA is not an ideal voltage source; the right branch loads node A.Does current drawn by the branch change VA?

Remove R4 when the external load is openedR4 belongs to the source network, not to the external load.Which component was explicitly defined as the load?

Deactivate the 12 V source while finding VTHThat destroys the intercept we are trying to calculate.Am I finding the intercept or the slope?

03.8Retrieval check and practice bridge

Answer without rereading the derivation

Why does the open-circuit condition remove the external load but not R4?
Why can R1 and R2 be placed in parallel only after the independent voltage source is zeroed?
What is the main computational benefit of the Thévenin model when RL changes repeatedly?
If RL becomes much larger than RTH, what should VL approach?

Answers. R4 is internal to the source network; zeroing the ideal source ties the left end of R1 to ground; the internal network is solved once and each new load becomes a divider calculation; VL approaches VTH.

If the extraction procedure itself is not yet automatic, rehearse it on the simpler verified divider in E-002, then return to the multi-node variation above.

Practice E-002