Thévenin & Norton: practice problems with step-by-step solutions

Also called: Thévenin’s theorem · Norton’s theorem · Thévenin equivalent circuit

Free Thévenin & Norton practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: divider network, with series resistor and load, with current source and load.

Free practice problemEasy
R1​ = 6 Ω+−V1​18 VR2​12 ΩAab

Find the Norton current IN: the current that flows from a to b if the terminals are shorted.

V1 = 18 VR1 = 6 ΩR2 = 12 Ω
A
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What you practise

  1. Easydivider network
  2. Mediumwith series resistor and load
  3. Hardwith current source and load

How to solve Thévenin & Norton problems

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

Common mistakes

  • Leaving the load connected while finding Rth.
  • Adding resistors in series that are actually in parallel when seen from the terminals.
  • Forgetting that maximum power is delivered when Rload = Rth.

Key formulas

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Learn Thévenin & Norton from scratch

Before this: Series & parallel resistors

1What a Thévenin equivalent is

Often you only care about what happens at two terminals of a circuit, for example where a load is connected.

Thévenin’s theorem says: seen from two terminals a and b, any circuit of sources and resistors behaves exactly like one voltage source in series with one resistor.

+−Vth​Rth​ab

A battery is a good picture: inside it is a chemical cell (the source) with some internal resistance. From the outside, those two things are all you notice.

This saves a lot of work. Once you have Vth and Rth, you can try many different loads with one line of maths each.

Self-check What does a Thévenin equivalent replace a circuit with?
  1. One current source in series with one resistor
  2. One resistor only
  3. One voltage source in series with one resistor
  4. Two voltage sources in parallel

Answer: One voltage source in series with one resistor. Vth in series with Rth. The Norton form uses a current source, but in parallel with Rth, not in series.

2Finding Vth and Rth

You find the two numbers with two separate small calculations.

  • Vth is the open-circuit voltage: the voltage between a and b with nothing connected to them.
  • Rth is the resistance you see looking into a and b with every independent source switched off.

Switching off works as in superposition: a voltage source becomes a wire, a current source becomes a gap.

The Norton equivalent is the same circuit in another form: a current source in parallel with the same Rth.

IN​Rth​ab
IN = Vth / Rth
Self-check A circuit has Vth = 12 V and Rth = 4 Ω. What is its Norton current?
  1. 0.33 A
  2. 16 A
  3. 3 A
  4. 48 A

Answer: 3 A. IN = Vth/Rth = 12/4 = 3 A. It is also the current that would flow if you shorted a to b.

3The method step by step

  1. Remove the load (if there is one) and mark the terminals a and b.
  2. Find the open-circuit voltage between a and b. This is Vth.
  3. Switch off all independent sources. Find the resistance between a and b with series and parallel rules. This is Rth.
  4. Draw the equivalent: Vth in series with Rth, then put the load back.
  5. The load current is IL = Vth / (Rth + RL).

Useful fact for step 2: with a and b open, no current flows in a resistor that leads only to a. So it has no voltage across it.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A load RL = 4 Ω is connected to a–b. Find the Thévenin equivalent and the load current IL.

1

Look at the circuit

R1​ = 6 ΩR2​3 ΩR3​ = 2 Ω+−Vs​24 VabRL​4 Ω

We replace everything to the left of a–b by Vth and Rth.

2

Remove the load and find Vth

R1​ = 6 ΩR2​3 ΩR3​ = 2 Ω+−Vs​24 VabVth​ ?+−

With a–b open, no current flows through R3, so a sits at the same voltage as the node between R1 and R2.

That node is a voltage divider:

Vth = 24 · 3 / (6 + 3) = 8 V
3

Switch the source off and find Rth

R1​ = 6 ΩR2​3 ΩR3​ = 2 Ω0 VabRth​ ?

The source becomes a wire. Looking in from a: R3 in series with (R1 ∥ R2).

R1 ∥ R2 = 6·3 / (6 + 3) = 2 Ω
Rth = 2 + 2 = 4 Ω
4

Put the load back

+−8 V4 Ωab
IL = Vth / (Rth + RL) = 8 / (4 + 4) = 1 A
5

The Norton form

IN = 8 / 4 = 2 A

Same circuit, other form: a 2 A source in parallel with 4 Ω gives the same 1 A in the load.

Answer

Vth = 8 V, Rth = 4 Ω and IL = 1 A.

5Maximum power to the load

A classic follow-up question: which load gets the most power?

The answer is the load equal to the Thévenin resistance:

RL = Rth ⇒ Pmax = Vth² / (4·Rth)

In the example, a 4 Ω load gets 8² / (4·4) = 4 W. That is the 1 A we found: P = 1² · 4 = 4 W.

Self-check A circuit has Vth = 10 V and Rth = 5 Ω. Which load resistance receives the most power?
  1. 0 Ω
  2. 10 Ω
  3. 2.5 Ω
  4. 5 Ω

Answer: 5 Ω. Maximum power transfer happens when RL = Rth = 5 Ω. It then receives 10²/(4·5) = 5 W.

6Common mistakes

  • Leaving the load connected while finding Vth or Rth. Remove it first.
  • Forgetting to switch the sources off for Rth, or switching a voltage source off as a gap instead of a wire.
  • Adding resistors in series that are actually in parallel when seen from a and b. Redraw the circuit from the terminals’ point of view.
  • Counting a voltage drop across a resistor that carries no current in the open circuit.

7How exam questions look

  • Easy: a source with a voltage divider. You find Vth or Rth.
  • Medium: the same, with an extra series resistor to the terminal, as in the example.
  • Hard: a current source as well, and a load whose current you have to find. The switched-off current source becomes a gap.

Write Vth and Rth down clearly before you use them. Many questions ask for one of them on the way.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
R1​ = 10 Ω+−V1​20 VR2​8 ΩAab

Find the Thévenin voltage Vth seen from terminals a–b.

V1 = 20 VR1 = 10 ΩR2 = 8 Ω

Find Vth (V)

Show hints
  1. You need two numbers: Rth (switch off every source and look into a–b) and Vth (the voltage across a–b with nothing connected). Norton current and load values follow from those two.
  2. With V₁ shorted, R₁ and R₂ are in parallel as seen from a–b.
  3. Vth = V1·R2/(R1+R2) = 20·8/(10+8).
Show the answer and the step-by-step solution

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 10 ΩSHORTR2​8 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

Rth = R1 ∥ R2 = 10·8/(10+8) = 4.444 Ω
R1​ = 10 Ω+−V1​20 VR2​8 ΩAab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, no current leaves at the terminals, so R₁ and R₂ form a plain voltage divider.

Vth = V1·R2/(R1+R2) = 20·8/(10+8) = 8.889 V
Example 2Medium
R1​ = 2 ΩR3​ = 3 Ω+−V1​9 VR2​12 ΩAab

Find the Thévenin resistance Rth seen from terminals a–b.

V1 = 9 VR1 = 2 ΩR2 = 12 ΩR3 = 3 Ω

Find Rth (Ω)

Show hints
  1. You need two numbers: Rth (switch off every source and look into a–b) and Vth (the voltage across a–b with nothing connected). Norton current and load values follow from those two.
  2. With V₁ shorted, R₁ ∥ R₂ is in series with R₃ as seen from a–b. With a–b open, no current flows in R₃.
  3. R1 ∥ R2 = 1.714 Ω.
Show the answer and the step-by-step solution

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 2 ΩR3​ = 3 ΩSHORTR2​12 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 2·12/(2+12) = 1.714 Ω
Rth = (R1 ∥ R2) + R3 = 1.714 + 3 = 4.714 Ω
Example 3Hard
R1​ = 20 ΩR3​ = 8 Ω+−V1​9 VR2​4 ΩAIs​0.5 ABRL​2 ΩIL​

A load RL = 2 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.

V1 = 9 VIs = 0.5 AR1 = 20 ΩR2 = 4 ΩR3 = 8 ΩRL = 2 Ω

Find PL (W)

Show hints
  1. You need two numbers: Rth (switch off every source and look into a–b) and Vth (the voltage across a–b with nothing connected). Norton current and load values follow from those two.
  2. With V₁ shorted and Iₛ open, Rth = (R₁ ∥ R₂) + R₃. For Vth, all of Iₛ must flow through R₃ into A.
  3. Rth = 11.33 Ω, Vth = 7.167 V.
Show the answer and the step-by-step solution

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 20 ΩR3​ = 8 ΩSHORTR2​4 ΩAOPENBabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 20·4/(20+4) = 3.333 Ω
Rth = (R1 ∥ R2) + R3 = 3.333 + 8 = 11.33 Ω
R1​ = 20 ΩR3​ = 8 Ω+−V1​9 VR2​4 ΩAIs​0.5 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 1.5 + 0.5·3.333 = 3.167 V
Vth = VB = VA + Is·R3 = 3.167 + 0.5·8 = 7.167 V
Rth​ = 11.33 Ω+−Vth​7.167 VRL​2 ΩIL​

Connect the load to the Thévenin equivalent

The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.

IL = Vth/(Rth + RL) = 7.167/(11.33 + 2) = 0.5375 A
PL = IL²·RL = 0.5375²·2 = 0.5778 W

Questions and answers

What is Thévenin & Norton?

Often you only care about what happens at two terminals of a circuit, for example where a load is connected.

Thévenin’s theorem says: seen from two terminals a and b, any circuit of sources and resistors behaves exactly like one voltage source in series with one resistor.

How do I solve Thévenin & Norton problems step by step?
  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.
What are common mistakes in Thévenin & Norton?
  • Leaving the load connected while finding Rth.
  • Adding resistors in series that are actually in parallel when seen from the terminals.
  • Forgetting that maximum power is delivered when Rload = Rth.
What do the Thévenin & Norton exercises cover?
  • Easy: divider network
  • Medium: with series resistor and load
  • Hard: with current source and load

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

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Series & parallel resistors

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Wheatstone bridge & strain gauges

Calculators

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Circuit Analysis 1Circuit Analysis 2: AC circuitsElectrical Engineering FundamentalsMeasurement and InstrumentationFE Electrical and Computer exam practice

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Printable worksheet

Thévenin & Norton worksheet with answers
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