A source has V_Th = 10\ V and R_Th = 5.0\. Find the matched load resistance. [1 mark]

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How does Thevenin's theorem reduce a complicated network to a single source and resistance, and when is power transfer greatest?

Thevenin's theorem: replacing a linear network by an equivalent electromotive force and series resistance, finding the Thevenin voltage and resistance, and the maximum power transfer condition.

An Eduqas A-Level Electronics answer on Thevenin's theorem: how to replace any linear two-terminal network by a single equivalent electromotive force in series with a resistance, how to find the Thevenin voltage and Thevenin resistance, and the maximum power transfer theorem with its impedance-matching consequence.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Thevenin's theorem says any linear network with two output terminals behaves, as far as anything connected to those terminals is concerned, like a single electromotive force $V_\text{Th}$ in series with a single resistance $R_\text{Th}$ . The Thevenin voltage is the open-circuit voltage at the terminals (nothing connected). The Thevenin resistance is the resistance looking back into the terminals with every independent source removed (voltage sources shorted, current sources opened). Once you have the Thevenin model, the current into any load $R_L$ is $I = \frac{V_\text{Th}}{R_\text{Th} + R_L}$ . The maximum power transfer theorem says the load receives most power when $R_L = R_\text{Th}$ , and that maximum is $P_\text{max} = \frac{V_\text{Th}^2}{4 R_\text{Th}}$ .

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What this dot point is asking

Eduqas wants you to apply Thevenin's theorem to replace a linear two-terminal network by a single electromotive force in series with a resistance, find the Thevenin voltage and Thevenin resistance, and state and use the maximum power transfer theorem. These tools turn a messy network into one a single load can be analysed against.

The answer

Thevenin's theorem

Finding the Thevenin voltage and resistance

Maximum power transfer

Reducing a loaded network to its Thevenin model

A $20\ \text{V}$ supply feeds a $100\ \Omega$ resistor to a node, and a $150\ \Omega$ resistor runs from that node to the $0\ \text{V}$ rail. Find the Thevenin equivalent at the node, then the voltage across a $300\ \Omega$ load connected there.

step 1: Thevenin voltage (open circuit)

With nothing connected at the node, the two resistors form a divider: $V_\text{Th} = 20 \times \frac{150}{100 + 150} = 20 \times 0.60 = 12\ \text{V}$ .

step 2: Thevenin resistance (source shorted)

Short the $20\ \text{V}$ supply; the $100\ \Omega$ and $150\ \Omega$ resistors are then in parallel: $R_\text{Th} = \frac{100 \times 150}{100 + 150} = \frac{15000}{250} = 60\ \Omega$ .

step 3: Add the load

The $300\ \Omega$ load with the $60\ \Omega$ Thevenin resistance forms a divider on $V_\text{Th}$ : $V_L = 12 \times \frac{300}{60 + 300} = 12 \times \frac{300}{360} = 10\ \text{V}$ .

Examples in context

Thevenin's theorem is how engineers model a sensor's output stage, an amplifier's output, or a power supply rail as "a voltage behind a resistance", so the effect of any load is a one-line calculation. Maximum power transfer governs audio output matching and radio-frequency aerial design, where the aim is to push the most signal power into the load. By contrast, a mains power supply is built with a very low output (Thevenin) resistance so its output voltage barely sags under load, the opposite design goal.

Try this

Q1. State how you find the Thevenin resistance of a network. [2 marks]

Cue. Remove the load, replace voltage sources by shorts and current sources by opens, then find the resistance looking into the terminals.

Q2. A source has $V_\text{Th} = 10\ \text{V}$ and $R_\text{Th} = 5.0\ \Omega$ . Find the matched load resistance. [1 mark]

Cue. $R_L = R_\text{Th} = 5.0\ \Omega$ .

Q3. For the source in Q2, find the maximum power delivered to a matched load. [2 marks]

Cue. $P_\text{max} = \frac{V_\text{Th}^2}{4 R_\text{Th}} = \frac{100}{20} = 5.0\ \text{W}$ .

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20216 marksA potential divider has a

6.0\ \text{k}\Omega

and a

3.0\ \text{k}\Omega

resistor in series across a

9.0\ \text{V}

supply, with the output taken across the

3.0\ \text{k}\Omega

resistor. Find the Thevenin equivalent (the Thevenin voltage and Thevenin resistance) seen at the output terminals.

Show worked answer →

Thevenin voltage (up to 3 marks): this is the open-circuit output, the unloaded divider value: $V_\text{Th} = 9.0 \times \dfrac{3.0}{6.0 + 3.0} = 9.0 \times \dfrac{1}{3} = 3.0\ \text{V}$ .

Thevenin resistance (up to 3 marks): replace the supply by a short circuit, leaving the two resistors in parallel as seen from the output terminals: $R_\text{Th} = \dfrac{6.0 \times 3.0}{6.0 + 3.0} = \dfrac{18}{9} = 2.0\ \text{k}\Omega$ .

Markers reward $V_\text{Th} = 3.0\ \text{V}$ (open-circuit voltage), shorting the source to find $R_\text{Th}$ , and the parallel combination $R_\text{Th} = 2.0\ \text{k}\Omega$ .

Eduqas 20204 marksState the maximum power transfer theorem, and calculate the load resistance and the power delivered to it when a source of Thevenin voltage

12\ \text{V}

and Thevenin resistance

8.0\ \Omega

is matched.

Show worked answer →

Theorem (up to 1 mark): maximum power is transferred to the load when the load resistance equals the source (Thevenin) resistance, $R_L = R_\text{Th}$ .

Matched load (up to 1 mark): $R_L = R_\text{Th} = 8.0\ \Omega$ .

Power delivered (up to 2 marks): at match the source voltage splits equally, so the load voltage is $6.0\ \text{V}$ and $P = \dfrac{V_L^2}{R_L} = \dfrac{6.0^2}{8.0} = \dfrac{36}{8.0} = 4.5\ \text{W}$ (equivalently $P_\text{max} = \frac{V_\text{Th}^2}{4 R_\text{Th}} = \frac{144}{32} = 4.5\ \text{W}$ ).

Markers reward the matching condition $R_L = R_\text{Th}$ , the value $8.0\ \Omega$ , and the maximum power $4.5\ \text{W}$ .

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)