How Do You Find Thevenin Resistance When a Circuit Has No Independent Source?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•7 min read

Connect a test source across the two terminals, keep every dependent source switched on, and divide the test voltage by the current the test source delivers: Rth = Vt ÷ It. You need this because a circuit with no independent source has an open-circuit voltage of zero and a short-circuit current of zero, so the usual Voc ÷ Isc shortcut gives 0 ÷ 0 and tells you nothing.

What you'll learn

  • Why Voc ÷ Isc fails when there is nothing in the circuit to drive a current
  • How the test-source method works, and whether to use a test voltage or a test current
  • A fully worked example with a 6 Ω, a 2 Ω and a 4·Ix dependent source, where Rth comes out as 3 Ω
  • Why switching the dependent source off gives the wrong answer (1.5 Ω here)
  • How a dependent source can make Thevenin resistance negative or infinite

Why does Voc ÷ Isc stop working?

Thevenin's theorem replaces a linear two-terminal network with a voltage Vth in series with a resistance Rth. The quick route to Rth is to find Vth as the open-circuit voltage, find the short-circuit current Isc, and divide. That route relies on something in the network pushing current. A dependent source cannot do that on its own: its value is a multiple of some other voltage or current in the same circuit, so if every independent source is absent, the only consistent solution with the terminals open or shorted is every voltage and current equal to zero. Vth is 0 V, Isc is 0 A, and the ratio is undefined. The network still has a definite resistance seen from its terminals; you just have to excite it from outside to measure it.

How does the test-source method work?

The test-source method for Thevenin resistanceA box labelled network with independent sources turned off and dependent sources kept has terminals a and b. A test voltage source Vt is connected across a and b and delivers a current It into terminal a. The Thevenin resistance is Vt divided by It.Network:independent sourcesoff, dependentsources keptab+−VtItRth = Vt ÷ It
The test-source method: with independent sources turned off and dependent sources left in, apply Vt across a and b, find the current It that the source pushes into terminal a, and take Rth = Vt ÷ It.
  • Turn off every independent source: voltage sources become short circuits, current sources become open circuits. (In this post's circuits there are none to turn off.)
  • Leave every dependent source in place, with its controlling variable defined exactly as before.
  • Connect a test source across the terminals. Convention used here: It is the current leaving the test source's positive terminal and entering terminal a, so Rth = Vt ÷ It is positive for an ordinary passive network.
  • Solve the circuit by nodal or mesh analysis. The controlling variable is no longer zero, because the test source now drives the circuit.
  • Divide. Because the network is linear, the ratio does not depend on the test value you picked.

Should the test source be a voltage or a current?

Either gives the same Rth. A 1 V test voltage is convenient when the controlling variable is a current through an element tied to terminal a, because that current then follows directly from Ohm's law. A 1 A test current is convenient when you plan to use nodal analysis, because it enters the KCL equation at terminal a as a known number. Pick whichever makes the controlling variable easiest to write down.

Worked example: a 6 Ω, a 2 Ω and a 4·Ix source

Worked example circuit with a 1 volt test sourceA current-controlled voltage source of value 4 times Ix, positive at the top, sits on the left. Its top connects through a 2 ohm resistor to node a. From node a a 6 ohm resistor carries current Ix down to the bottom wire. A 1 volt test source across terminals a and b delivers It equal to one third of an amp into a. Ix is one sixth of an amp.+−4·Ix2 Ω6 ΩIx =1/6 Aab+−Vt =1 VIt = 1/3 A
Terminal a connects to a 6 Ω resistor down to b, with Ix the current flowing down through it. From a, a 2 Ω resistor leads to a current-controlled voltage source of 4·Ix volts (positive at the top) whose bottom is tied to b. A 1 V test source is connected across a and b.
  • Apply Vt = 1 V across a and b. The 6 Ω resistor now has 1 V across it, so Ix = 1 ÷ 6 = 1/6 A, flowing down.
  • The dependent source sets its top node to 4·Ix = 4 × 1/6 = 2/3 V above b.
  • The 2 Ω resistor has 1 − 2/3 = 1/3 V across it, so it carries (1/3) ÷ 2 = 1/6 A from a towards the dependent source.
  • KCL at a: the test current splits between the two branches, so It = 1/6 + 1/6 = 1/3 A.
  • Rth = Vt ÷ It = 1 ÷ (1/3) = 3 Ω.

Check the answer with a 1 A test current

Drive 1 A into terminal a instead and call the terminal voltage Vt. Then Ix = Vt ÷ 6, the dependent source sits at 4Vt ÷ 6, and the 2 Ω branch carries (Vt − 4Vt ÷ 6) ÷ 2 = Vt ÷ 6. KCL gives 1 = Vt ÷ 6 + Vt ÷ 6 = Vt ÷ 3, so Vt = 3 V and Rth = 3 V ÷ 1 A = 3 Ω, the same answer. Writing the gain as a general k instead of 4 gives It ÷ Vt = 1/6 + (1 − k/6) ÷ 2, which simplifies to Rth = 12 ÷ (8 − k). Setting k = 4 returns 3 Ω.

Why is shorting the dependent source wrong?

If you treat the 4·Ix source like an independent source and replace it with a short, the 2 Ω resistor drops straight to b and you get 6 Ω in parallel with 2 Ω, which is 1.5 Ω. That is half the true value. The dependent source is part of how the network responds to a voltage at its terminals: when the test source raises a, Ix rises, the dependent source rises with it, and less current flows through the 2 Ω branch than would flow into a short. Shorting it throws that feedback away. Only independent sources are turned off; dependent sources always stay.

Can Thevenin resistance be negative?

Thevenin resistance against the dependent-source gain kComputed curve of Rth equals 12 divided by 8 minus k for k from 0 to 7 ohms. Rth starts at 1.5 ohms when k is 0, which is 6 ohms in parallel with 2 ohms, reaches 3 ohms at k equals 4, the worked example, and rises to 12 ohms at k equals 7.01234567036912gain k (Ω)Rth (Ω)k = 4: 3 Ωk = 0: 1.5 Ω
Rth = 12 ÷ (8 − k) for the example circuit, computed for k from 0 to 7 Ω. At k = 0 the source is just a wire and Rth is 6 ∥ 2 = 1.5 Ω; at k = 4 it is 3 Ω.

Yes. Because the dependent source can supply energy, the formula Rth = 12 ÷ (8 − k) is not limited to positive values. As k approaches 8 Ω the dependent source holds its node almost exactly at the terminal voltage, the 2 Ω branch carries almost nothing, and only the 6 Ω path is left: at exactly k = 8 the two branch currents cancel, the test source delivers no current at all and terminal a looks like an open circuit. Push k past 8 and the 2 Ω branch drives current back out through terminal a. With k = 18, a 1 V test source sees It = −5/6 A, so Rth = −1.2 Ω. A negative Thevenin resistance is a real result for active circuits, and the same idea underlies negative-resistance oscillator analysis. It is not a sign error, as long as you kept the It convention fixed.

Common mistakes

  • Switching off a dependent source while finding Rth. Only independent sources are turned off.
  • Trying Voc ÷ Isc on a circuit with no independent source and getting 0 ÷ 0, then guessing.
  • Assuming the controlling variable is zero during the test. It is not: the test source drives the whole circuit, Ix included.
  • Mixing current directions, for example taking It as the current leaving terminal a. That flips the sign of Rth.
  • Rejecting a negative or infinite Rth as impossible. With a dependent source either can be the correct answer.

Check your understanding

  • Q1. In the same circuit, the gain is changed from 4 Ω to 2 Ω. What is Rth? Answer: 12 ÷ (8 − 2) = 2 Ω.
  • Q2. With k = 4 and a 1 V test source, what current flows in the 2 Ω resistor? Answer: 1/6 A, from a towards the dependent source, since 1 − 2/3 = 1/3 V sits across 2 Ω.
  • Q3. For what value of k does the test source deliver no current, so that terminal a looks open? Answer: k = 8 Ω, where It = 1/6 + (1 − 8/6) ÷ 2 = 0.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists Thevenin's theorem under Section 2, Networks, Signals and Systems (checked against the official PDF from IIT Madras on 10 October 2026). Every number in the example was recomputed by exact nodal analysis in code.

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