Thevenin and Norton Theorems for GATE, with Solved Examples

October 7, 2026•Updated October 21, 2026•Last verified October 7, 2026•8 min read

Thevenin's and Norton's theorems are listed in the GATE 2027 syllabi for EC (under Networks, Signals and Systems, with maximum power transfer), EE (under Electric circuits, with maximum power transfer) and IN. GATE questions rarely stop at a single source and two resistors. They add a second source, a dependent source or a load to optimise. This post works through two such circuits, one with two independent sources and one bridge with a load to optimise, each solved more than one way so you can check your own answers. Circuits with dependent sources have their own post, linked below.

The method in four lines

  • Thevenin voltage Vth: the open-circuit voltage across the two terminals
  • Norton current In: the current through a short placed across the terminals
  • Thevenin resistance Rth = Vth ÷ In. With only independent sources you can also turn them off (voltage sources shorted, current sources opened) and combine resistors
  • With a dependent source, never turn it off. Use Vth ÷ In, or turn off only the independent sources and apply a test source

Example 1: two independent sources

Circuit: a 10 volt source, positive at the top, feeds a 2 ohm resistor in series to node A. A 3 ohm resistor runs from node A down to the return wire, and a 2 amp current source, arrow pointing up, also connects the return wire to node A. Terminals A (top) and B (bottom) are at the right.+10 V2 Ω3 Ω2 AAB
Example 1: a 10 V source with 2 Ω in series feeds node A. From A to the return wire there is a 3 Ω resistor, and a 2 A source pushes current up into A.
  • Vth by nodal analysis at A: (Va − 10) ÷ 2 + Va ÷ 3 − 2 = 0. Multiply by 6: 3Va − 30 + 2Va − 12 = 0, so 5Va = 42 and Vth = 8.4 V
  • Rth with both sources turned off (10 V shorted, 2 A opened): 2 Ω in parallel with 3 Ω = 6 ÷ 5 = 1.2 Ω
  • In from Vth ÷ Rth: 8.4 ÷ 1.2 = 7 A
  • Check In directly: shorting A to B puts 0 V across the 3 Ω, so the 10 V source drives 10 ÷ 2 = 5 A into the short and the current source adds 2 A, giving 7 A. Both routes agree
  • Maximum power transfer: a 1.2 Ω load takes Vth² ÷ (4Rth) = 8.4² ÷ 4.8 = 70.56 ÷ 4.8 = 14.7 W

Example 2: a bridge with a load to optimise

A 12 V source feeds two branches in parallel. In the left branch, 4 Ω sits above node A and 12 Ω below it. In the right branch, 6 Ω sits above node B and 6 Ω below it. The bottom ends of both branches return to the source's negative terminal. A load RL connects A to B. Find the Thevenin equivalent seen by the load, the RL that takes maximum power, and that power.

  • Vth from two voltage dividers: VA = 12 × 12 ÷ (4 + 12) = 9 V and VB = 12 × 6 ÷ (6 + 6) = 6 V, so Vth = VA − VB = 3 V, with A positive
  • Rth with the 12 V source shorted: the top and bottom rails become one node, so A sees 4 Ω in parallel with 12 Ω (3 Ω) and B sees 6 Ω in parallel with 6 Ω (3 Ω). In series between A and B, Rth = 3 + 3 = 6 Ω
  • Check with In: short A to B and call the joined node C. Nodal analysis gives (12 − VC) ÷ 4 + (12 − VC) ÷ 6 = VC ÷ 12 + VC ÷ 6, so VC = 7.5 V. The current from A through the short is (12 − 7.5) ÷ 4 − 7.5 ÷ 12 = 1.125 − 0.625 = 0.5 A, and Vth ÷ In = 3 ÷ 0.5 = 6 Ω. Both routes agree
  • Maximum power transfer: RL = Rth = 6 Ω takes Vth² ÷ (4Rth) = 9 ÷ 24 = 0.375 W. As a check, 3 V across 6 + 6 Ω drives 0.25 A, and 0.25² × 6 = 0.375 W
  • A 3 Ω load would take (3 ÷ 9)² × 3 ≈ 0.333 W, less than the matched load, as the theorem says it must
  • If both branches had the same ratio (say 6 Ω over 6 Ω on the left too), VA would equal VB at 6 V, Vth would be 0 and no current would flow in the load: the balanced bridge

Circuits with dependent sources

The method in four lines above covers the one rule that changes: never turn a dependent source off while finding Rth. Use Vth ÷ In, or turn off only the independent sources and apply a test source. When a circuit has no independent source at all, Vth and In are both zero and only the test-source method works. That case, with a worked example, has its own post: how to find Thevenin resistance when a circuit has no independent source.

Maximum power transfer in AC circuits

In sinusoidal steady state the same idea uses impedances: the load that takes maximum average power is the complex conjugate of the Thevenin impedance, ZL = Zth*. The reactances cancel, and the power is |Vth|² ÷ (4Rth) with Vth as an RMS value. If only a pure resistance is allowed as the load, the best choice is RL = |Zth|, which gives less power than the conjugate match.

Common mistakes

  • Turning off a dependent source while finding Rth
  • Replacing a current source with a short instead of an open circuit when turning it off
  • Getting the sign of In wrong: In points the same way as the current Vth would push through an external short
  • Using peak values in |Vth|² ÷ (4Rth) when the question gives RMS, or the other way round
  • Forgetting to check: Vth ÷ In and the turned-off-source method must give the same Rth whenever both apply

Go deeper

Official sources

Plan with Studyloaf

Sources and last verified date

Every GATE 2027 fact in this post was checked against the official GATE 2027 website, information brochure and syllabus PDFs from IIT Madras on 7 October 2026. Official dates and rules can change, so confirm anything you plan around on the official site.

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