Thevenin's and Norton's Theorems

Reducing a whole network to two numbers

Thevenin's theorem: any linear circuit, viewed from just two terminals, behaves exactly like a single voltage source $V_{th}$ in series with a single resistor $R_{th}$ , no matter how complicated the actual network behind those terminals is.

Norton's theorem is the same idea with a current source instead: a current source $I_N$ in parallel with $R_N$ . The two are directly related, same resistance, and:

R_N = R_{th}, \qquad I_N = \dfrac{V_{th}}{R_{th}}

$V_{th}$ is the open-circuit voltage across the terminals (nothing connected). $R_{th}$ is the resistance seen from the terminals with every independent source turned off (voltage sources shorted, current sources opened).

Finding the equivalent, step by step

Pick the two terminals and disconnect whatever load you want to analyse from them.
Find $V_{th}$ : the voltage across the open terminals, usually by a voltage divider, nodal analysis or KVL.
Find $R_{th}$ : turn off every independent source (short voltage sources, open current sources) and work out the resistance looking back into the terminals.
Or find $I_N$ directly: short the terminals and calculate the current through the short. Then $R_{th} = \dfrac{V_{th}}{I_N}$ , which is the method to use when the circuit has dependent sources.
Reconnect the load to the simple equivalent and solve that instead of the original network.

Worked example

A 12V source feeds a 4Ω resistor in series, which then splits into a 6Ω resistor to the return path. Find the Thevenin and Norton equivalents seen from the two ends of that 6Ω resistor, as a load connected across it would see the circuit.

The worked example's circuit. A and B are the two ends of the 6Ω resistor, where a load connects.

$V_{th}$ : with nothing connected across the terminals, the only current flows through the 4Ω and 6Ω in series, so the terminal voltage is just the voltage-divider result across the 6Ω:

V_{th} = 12 \times \dfrac{6}{4+6} = 12 \times 0.6 = 7.2\text{ V}

$R_{th}$ : shorting the 12V source and looking back into the terminals, the 4Ω and 6Ω are now simply in parallel with each other:

R_{th} = \dfrac{4\times6}{4+6} = \dfrac{24}{10} = 2.4\,\Omega

Norton equivalent:

I_N = \dfrac{V_{th}}{R_{th}} = \dfrac{7.2}{2.4} = 3\text{ A}, \qquad R_N = 2.4\,\Omega

What a load across A and B sees: 7.2 V behind 2.4Ω (Thevenin), or equivalently 3 A in parallel with 2.4Ω (Norton).

Cross-check: the Norton current should also equal the short-circuit current at the terminals. Shorting the terminals shorts out the 6Ω branch entirely, leaving only the 4Ω resistor across the 12V source:

I_{sc} = \dfrac{12}{4} = 3\text{ A}

Matches $I_N$ exactly, confirming both equivalents independently.

Worked example: connecting a load

A 2.4Ω load is connected across the same terminals. Find the load current and voltage.

Solution: With the Thevenin equivalent, this is one series loop of $V_{th}$ , $R_{th}$ and the load:

I_L = \dfrac{V_{th}}{R_{th} + R_L} = \dfrac{7.2}{2.4 + 2.4} = 1.5\text{ A}, \qquad V_L = I_L R_L = 1.5 \times 2.4 = 3.6\text{ V}

Check against the original circuit: the 6Ω and the 2.4Ω load are in parallel, $\dfrac{6 \times 2.4}{6 + 2.4} = \dfrac{12}{7}\,\Omega$ . In series with the 4Ω, the source sees $\dfrac{40}{7}\,\Omega$ , so it supplies $12 \div \dfrac{40}{7} = 2.1$ A, and the voltage across the parallel pair is $2.1 \times \dfrac{12}{7} = 3.6$ V. Same answer, with a lot more arithmetic, which is why the equivalent is worth finding when the load changes.

Maximum power transfer

A load draws the most power from a circuit when its resistance equals the Thevenin resistance, $R_L = R_{th}$ . The power it then receives is:

P_{max} = \dfrac{V_{th}^2}{4R_{th}}

The 2.4Ω load above is exactly that case: $P_{max} = \dfrac{7.2^2}{4 \times 2.4} = \dfrac{51.84}{9.6} = 5.4$ W, which matches $I_L^2 R_L = 1.5^2 \times 2.4 = 5.4$ W. Inside the Thevenin equivalent the efficiency at this point is 50%, since $R_{th}$ dissipates as much as the load. The real network can do worse: here the source supplies $12 imes 2.1 = 25.2$ W, so the load gets only about 21% of it.

Common mistakes

Leaving the load in when finding $V_{th}$ . $V_{th}$ is an open-circuit voltage. Decide which element is the load, take it out, then measure across the gap.
Turning off dependent sources. Only independent sources are deactivated. A dependent source stays in the circuit, and you find $R_{th}$ from $\dfrac{V_{oc}}{I_{sc}}$ or with a test source.
Shorting a current source. A turned-off current source is an open circuit (zero current), not a short.
Getting the Norton direction wrong. $I_N$ points so that it drives current out of the terminal that is positive in the Thevenin circuit.

Where this comes up in exams

Thevenin's and Norton's theorems are listed in the GATE 2027 EC (Networks, Signals and Systems), EE (Electric circuits) and IN (Electrical Circuits and Machines) syllabi, and all three also name maximum power transfer. They are not in the JEE Main 2026, JEE Advanced 2026 or NEET (UG) 2026 syllabi, which cover Kirchhoff's laws but not network theorems.

Checked against GATE 2027 EC syllabus, GATE 2027 EE syllabus, GATE 2027 IN syllabus, JEE (Main) 2026 syllabus (NTA), JEE (Advanced) 2026 syllabus and NEET (UG) 2026 syllabus (NMC, published by NTA). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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