Thevenin and Norton Equivalents

Section Overview

You now have the full toolkit — Ohm's law plus Kirchhoff's two laws — to solve any resistive network. This section gives you a shortcut so powerful it feels like cheating. Thevenin's theorem says that no matter how complicated a two-terminal network of sources and resistors is, everything the outside world can see at those two terminals behaves exactly like a single voltage source in series with a single resistance. Norton's theorem says the same network is equally well described by a single current source in parallel with a resistance. Learn to find these equivalents and a page of circuitry collapses into two numbers you can reason about instantly — which is exactly how professionals think about power supplies, sensor outputs, and any "source" feeding a load.

Why This Matters

Real sources aren't ideal. A battery, a power-supply rail, a sensor, an amplifier output — each behaves like some voltage behind some resistance, and that hidden resistance is why the output sags when you load it. The Thevenin equivalent is the exact model of that behavior: V_th is the unloaded voltage, and R_th is the internal resistance that causes the sag. Once you can reduce a source to V_th and R_th, you can predict — without re-solving the whole circuit — how much its voltage will drop under any load, how much current a load will draw, and whether a replacement part will be over- or under-driven. It is the single most practical analysis tool for reasoning about how one part of a circuit affects another.

Required Prerequisites

• Ohm's Law — The Foundation of Circuit Analysis — used at every step to relate the equivalent's voltage, current, and resistance.
• Series Circuits — the Thevenin form is a source in series with a resistance.
• Parallel Circuits — the Norton form is a source in parallel with a resistance, and finding R_th uses parallel combination.
• Kirchhoff's Voltage Law — underlies finding the open-circuit voltage.
• Kirchhoff's Current Law — underlies finding the short-circuit current.

Recommended Consumables

No consumables required. The exercises are pen-and-paper network reduction plus optional measurement on a single-battery divider circuit.

Recommended Practice Hardware

• A single battery (AA or 9 V) and holder
• Two or three resistors of different values to build a voltage divider, plus one more to act as a "load"
• Clip leads or a breadboard, and a multimeter if you have one (from Section 1.7) to measure the open-circuit output and the loaded output

Everything here runs from a single low-voltage battery — the same safe territory as the previous sections. No mains, no high current.

Real-World Applications

Every datasheet that lists an "output impedance," every power supply with a "load regulation" spec, and every rule of thumb like "don't load a divider with less than ten times its resistance" is Thevenin thinking. Technicians use the equivalent to answer questions like: will this sensor still read correctly when I connect my meter across it (does the meter load it)? How much will this rail droop when the circuit draws its full current? Why does this reference voltage shift when I add a stage? Reducing the source to V_th and R_th turns all of these into a one-line calculation. It's also the mental model behind a battery's "internal resistance" — the reason a tired battery reads fine unloaded but collapses the moment it has to deliver current.

Common Challenges

• Getting the source-zeroing rule backwards. To find R_th you turn the sources off — but "off" means shorting a voltage source (0 V across it is a wire) and opening a current source (0 A through it is a break). Reversing these is the most common Thevenin mistake, and this section drills it.
• Confusing V_th with the loaded voltage. V_th is the open-circuit voltage — what the terminals read with nothing connected. Once a load is attached, the terminal voltage drops below V_th; that drop is the whole point of R_th.
• Thinking the equivalent is only valid for one load. The Thevenin equivalent is a property of the source network, not of any particular load. Once you have V_th and R_th, they work for any load you connect — that's what makes them so useful.

Safety Notes

Risk Level: Low. This section is analysis and optional low-voltage measurement on a single-battery divider — the same safe territory as the previous sections. No mains, no high-energy sources.

Professional Tips Before Starting

• Always identify the two terminals first — the equivalent only means anything "as seen from" a specific pair of terminals. Mark them before you reduce anything.
• Find V_th and R_th by two independent routes when you can (open-circuit voltage for V_th; source-zeroed resistance for R_th; short-circuit current as a cross-check via I_sc = V_th ÷ R_th). If the numbers agree, you've almost certainly got it right.
• Think of R_th as the source's internal resistance. The instant you frame it that way, "why does the output sag under load?" becomes obvious — it's a voltage divider between R_th and your load.

Reducing a Network to One Source

Thevenin's Theorem: One Voltage, One Resistance

Thevenin's theorem states that any linear network of voltage sources, current sources, and resistors, viewed from two chosen terminals, behaves identically to a single voltage source V_th in series with a single resistance R_th. "Linear" here just means the ordinary resistors and ideal sources you've been working with; "viewed from two terminals" means that any load you connect across those terminals cannot tell the difference between the real network and its two-component equivalent. The Thevenin equivalent is that pair — V_th and R_th — and finding it is a two-step recipe.

Step 1 — Find V_th, the Open-Circuit Voltage

Disconnect whatever load will hang on the terminals, leaving them open, and find the voltage across them. That open-circuit voltage is V_th, because with no load drawing current there's no drop across R_th, so the full internal source voltage appears at the terminals.

Take a concrete network: a 12 V source with a 3 kΩ resistor (R₁) feeding a node, and a 6 kΩ resistor (R₂) from that node to ground — a voltage divider, with the two terminals taken across R₂. Open-circuit, no load, this is just the divider from Section 2.1:

V_th = 12 × R₂ ÷ (R₁ + R₂) = 12 × 6 ÷ (3 + 6) = 12 × 6 ÷ 9 = 8 V

Step 2 — Find R_th by Zeroing the Sources

Now find the resistance the terminals present looking back into the network, but with every independent source turned off. Turning a source off means replacing it with its zero-value version:

• A voltage source goes to 0 V, which is a short (a plain wire).
• A current source goes to 0 A, which is an open (a break).

Get this right and the rest is just series-parallel reduction. In our divider, short the 12 V source. Now, looking in from the terminals, R₁ and R₂ both run between the output node and the (now shorted) supply rail — they are in parallel:

R_th = R₁ ‖ R₂ = (R₁ × R₂) ÷ (R₁ + R₂) = (3 × 6) ÷ (3 + 6) = 18 ÷ 9 = 2 kΩ

So the entire divider, as seen from those two terminals, is exactly an 8 V source in series with 2 kΩ. That's the Thevenin equivalent — two numbers standing in for the whole thing.

Norton's Theorem and the Source Transformation

Norton's theorem is the same idea in current form: the Norton equivalent describes the network as a single current source I_n in parallel with a resistance R_n. The Norton resistance equals the Thevenin resistance (R_n = R_th), and the Norton current is the network's short-circuit current — what flows if you connect the two terminals directly together.

The two equivalents convert into each other by the source transformation:

V_th = I_n × R_n and R_th = R_n

For our example, R_n = R_th = 2 kΩ, and:

I_n = V_th ÷ R_th = 8 ÷ 2000 = 0.004 A = 4 mA

Cross-check with the definition of I_n as the short-circuit current: short the two terminals (across R₂), and R₂ carries no voltage, so all the source current flows through R₁: I_sc = 12 V ÷ 3 kΩ = 4 mA. The two routes agree — a 4 mA Norton source in parallel with 2 kΩ is the same network as an 8 V Thevenin source in series with 2 kΩ. Either form is correct; pick whichever makes the next step easier.

The Payoff: Loading Becomes One Line

Here's why this matters. Connect a 2 kΩ load across the terminals. With the Thevenin equivalent, the load and R_th simply form a series divider:

I_load = V_th ÷ (R_th + R_load) = 8 ÷ (2 kΩ + 2 kΩ) = 8 ÷ 4000 = 2 mA V_load = I_load × R_load = 2 mA × 2 kΩ = 4 V

The unloaded 8 V sags to 4 V under a 2 kΩ load — because that load is comparable to the 2 kΩ internal resistance. You found that in one line, without touching the original divider. And it's exactly right: solve the full circuit (R₂ ‖ R_load = 6 kΩ ‖ 2 kΩ = 1.5 kΩ, output = 12 × 1.5 ÷ (3 + 1.5) = 4 V) and you get the same 4 V. The equivalent isn't an approximation — it's an exact stand-in that made the loading calculation trivial.

That sag is the loading effect made concrete: a source with a high R_th relative to its load droops badly (a weak divider, a tired battery), while one with a low R_th holds its voltage (a stiff supply). Reading R_th as the internal resistance tells you at a glance how a source will behave when you actually ask it to deliver current.

Common Mistakes

• Zeroing sources the wrong way. Voltage source → short; current source → open. Swapping these gives a wrong R_th every time. Remember it by the value: a 0 V source is a wire, a 0 A source is a break.
• Using the loaded voltage as V_th. V_th is strictly the open-circuit voltage. If a load is connected while you measure, you're reading the divided-down output, not V_th.
• Forgetting R_n = R_th. The Norton and Thevenin resistances are identical; only the source form (voltage-in-series vs. current-in-parallel) differs.
• Reducing from the wrong terminals. The equivalent is defined relative to a specific terminal pair. Choose the terminals the load will actually connect to before you start.

Troubleshooting Guidance

Thevenin thinking is how you reason about loading faults — the ones where a circuit works until something is connected to it. If a reference voltage or sensor output reads correctly unloaded but drops when the next stage or your meter is attached, you're seeing R_th: the source has more internal resistance than the load can tolerate, so estimate V_th (the unloaded reading) and R_th (from how far it sags under a known load) and the numbers will tell you whether the source is too weak or the load too heavy. A battery that measures its rated voltage with no load but collapses under load has a high internal resistance — a classic Thevenin diagnosis, and often the real meaning of a "dead" battery that still shows voltage. Conversely, if adding a load doesn't sag the output at all, R_th is low and the source is stiff — look elsewhere for the fault.

Verification & Testing Methods

Check your understanding before moving on:

• [ ] State Thevenin's theorem and explain what "equivalent as seen from two terminals" means.
• [ ] Find V_th as the open-circuit voltage of a simple divider, and R_th by zeroing the source correctly.
• [ ] Convert a Thevenin equivalent to its Norton form using V_th = I_n × R_n.
• [ ] Use an equivalent to predict the load voltage and current when a given load is connected, and explain the sag in terms of R_th.

Then try the practice exercises below — pen-and-paper reduction, with optional measurement on a single-battery divider.

Practice Exercises

1. Find the Thevenin equivalent (10 minutes, pen and paper). A 10 V source feeds a divider: R₁ = 4 kΩ on top, R₂ = 4 kΩ to ground, terminals across R₂. Find V_th (open-circuit voltage) and R_th (source shorted, R₁ ‖ R₂). State the equivalent as "___ V in series with ___ kΩ."
2. Convert to Norton (5 minutes, pen and paper). Using your V_th and R_th from Exercise 1, find the Norton equivalent: I_n = V_th ÷ R_th and R_n = R_th. State it as "___ mA in parallel with ___ kΩ." As a check, confirm V_th = I_n × R_n.
3. Predict the loading (10 minutes, pen and paper). Connect a 4 kΩ load across the terminals of your Exercise 1 equivalent. Find the load current and load voltage using the series divider I_load = V_th ÷ (R_th + R_load). By how much did the output sag from V_th, and why?
4. Measure it (10 minutes, optional hardware). Build a divider from a single battery and two equal resistors. Measure the open-circuit output (V_th). Connect a load resistor equal to one divider resistor, measure the loaded output, and use the drop to estimate R_th. Compare your measured R_th to the calculated R₁ ‖ R₂.

These core ideas — the two-terminal equivalent, finding V_th and R_th, the Thevenin–Norton transformation, and predicting loading — are tested in the Chapter Quiz at the end of this chapter, where a score of 80% is required to continue.

Key Takeaways

• Thevenin's theorem: any linear two-terminal network reduces, as seen from those terminals, to one voltage source V_th in series with one resistance R_th — the Thevenin equivalent.
• V_th is the open-circuit voltage at the terminals; R_th (the Thevenin resistance) is the resistance looking back in with the sources zeroed.
• Zeroing sources: a voltage source becomes a short (0 V = wire); a current source becomes an open (0 A = break).
• Norton's theorem gives the same network as a current source I_n in parallel with R_n; R_n = R_th, and the source transformation relates them by V_th = I_n × R_n, with I_n the short-circuit current.
• R_th is the source's internal resistance — it is why a real source sags under load, and the sag is just a divider between R_th and the load.
• The equivalent belongs to the source network, not the load, so once found it predicts the behavior with any load in a single line.

Skills Learned

• You can now find the Thevenin equivalent (V_th and R_th) of a simple resistive network such as a loaded divider.
• You can now zero sources correctly — shorting voltage sources and opening current sources — to compute R_th.
• You can now convert between Thevenin and Norton forms with V_th = I_n × R_n.
• You can now predict how a source's output voltage and a load's current change when the load is connected, and explain the sag as an R_th effect.
• You can now recognize internal resistance and loading effects in real sources — batteries, dividers, supply rails — as Thevenin behavior.

Glossary Additions

• Thevenin equivalent — the representation of any linear two-terminal network as a single voltage source (V_th, the open-circuit voltage) in series with a single resistance (R_th), behaving identically to the original network as seen from those terminals.
• Norton equivalent — the representation of the same network as a single current source (I_n, the short-circuit current) in parallel with a resistance (R_n = R_th).
• Thevenin resistance — the resistance a two-terminal network presents looking back into it with all independent sources zeroed (voltage sources shorted, current sources opened); it equals the Norton resistance and models a source's internal resistance.
• source transformation — the conversion between a voltage source in series with a resistance and a current source in parallel with the same resistance, related by V_th = I_n × R.

Suggested Next Sections

Must read next:

• Superposition Principle — the last analysis tool of the chapter: how to handle a circuit with more than one source by considering each one at a time and adding the results.

Recommended:

• Parallel Circuits — the parallel combination you used to find R_th, and the form a Norton source takes.
• Ohm's Law — The Foundation of Circuit Analysis — every step of finding and using an equivalent is Ohm's law on the reduced circuit.