Thévenin's Theorem

Step 1 — remove the load and find the open-circuit voltage. With nothing connected at \(a\!-\!b\), no current is drawn from the terminals, so the 18 V source drives a simple series pair and the 3 Ω acts as the lower arm of a divider:

Step 2 — deactivate the independent source and look back into the terminals. A dead voltage source is a short, which places the 6 Ω directly across the 3 Ω:

Step 3 — reconnect the load to the equivalent, which is now a single loop:

Check against the original circuit. With the 4 Ω connected, the 3 Ω and 4 Ω are in parallel at \(12/7\ \Omega\), so the source current is \(18/(6 + 12/7) = 2.333\ \text{A}\) and the terminal voltage is \(2.333 \times 12/7 = 4\ \text{V}\). The load then carries \(4/4 = 1\ \text{A}\;\checkmark\)

Let the network \(N\) be linear, with terminals \(a\!-\!b\) carrying current \(I\) into a load and terminal voltage \(V\). By the substitution principle, replacing the load by an ideal current source of exactly \(I\) changes nothing inside \(N\) — the same current is drawn at the same voltage.

Now the network contains its own internal sources plus this one external source. Because \(N\) is linear, superposition applies. Split the terminal voltage into two contributions:

The first term. With the external source dead — that is, opened, so \(I = 0\) — the terminal voltage is by definition the open-circuit voltage:

The second term. With all internal independent sources deactivated, what remains is a passive network — possibly containing dependent sources — presenting some resistance \(R_{TH}\) at its terminals. Driving it with \(I\) flowing out of the terminal gives

Adding:

This is precisely the terminal relation of a source \(V_{TH}\) in series with \(R_{TH}\). Since \(I\) was arbitrary, the two networks are indistinguishable at the terminals for every load.

What the proof required. Exactly two things:

Nothing about planarity, topology, source type, or the number of elements. The load itself need not be linear — a diode load is fine, since only \(N\) was superposed.

Open-circuit voltage. With the terminals open, no current can flow in the 2 Ω series resistor, so it drops nothing and the whole 3 A passes through the 6 Ω:

The 2 Ω is invisible to \(V_{TH}\) — but not, as the next step shows, to \(R_{TH}\).

Thévenin resistance. A dead current source is an open circuit, not a short. That leaves the 6 Ω and 2 Ω in series along the only remaining path:

With the load:

Check directly: with the 4 Ω connected, the 3 A divides between the 6 Ω and the \(2 + 4 = 6\ \Omega\) branch. Equal resistances share equally, so 1.5 A flows to the load \(\checkmark\)

With the load removed the network is two meshes. Taking both clockwise and writing the equations by inspection (Set 4, Problem 4):

Mesh 1 totals \(2+1 = 3\ \Omega\), mesh 2 totals \(1+1+2 = 4\ \Omega\), and they share the 1 Ω. The 5 V source opposes mesh 2's clockwise direction, hence the negative entry.

The determinant is \(12 - 1 = 11\), so by Cramer's rule:

Open-circuit voltage. Walk from the reference up through the 5 V source and back along the 2 Ω that carries \(i_2\):

Thévenin resistance. Short both sources. Looking in from \(a\) there are two paths to the reference: the 2 Ω directly, and the 1 Ω to \(c\) followed by \(1 \parallel 2 = \tfrac23\ \Omega\):

Reconnecting the 2 Ω load:

The elevenths cancel exactly, which is the usual sign that \(V_{TH}\) and \(R_{TH}\) came from the same determinant.

Remove the 15 Ω. Convert the 2 A source with its parallel 5 Ω into a voltage source in series with the same resistance (Set 12's method, used here in advance):

The network is now a single loop containing 40 V, 20 Ω, 5 Ω and 10 V.

KVL round that loop, with the two sources opposing:

Open-circuit voltage at the terminals, taken across the transformed branch:

Thévenin resistance. Deactivate both sources — the 40 V shorted, the 2 A opened, which leaves its 5 Ω in place. The two resistors are then in parallel across the terminals:

With the load:

First the open-circuit voltage, by KCL at \(a\) with the terminals open:

Route 1 — deactivation. Short the 12 V source and open the 2 A source. The 6 Ω and 3 Ω are then in parallel across the terminals:

Valid here because every source is independent. This is always the quickest route when it is available.

Route 2 — open-circuit / short-circuit. Shorting \(a\!-\!b\) puts \(V_a = 0\), so the 3 Ω carries nothing and both sources drive straight into the short:

Route 3 — test source. Deactivate the independent sources, apply \(V_x = 1\ \text{V}\) at the terminals, and find the current it draws:

When each route is the right one:

With the 18 Ω removed, the bridge falls apart into two independent dividers across the 8 V source — the whole reason this circuit is worth doing by Thévenin:

The open-circuit voltage is their difference:

A balanced bridge is exactly the case \(V_{TH} = 0\), which is why its arm carries nothing whatever its resistance.

Thévenin resistance. Short the 8 V source, which merges \(a\) and \(b\) into a single node. Node \(c\) then reaches it by two parallel paths, and so does \(d\):

These two are in series along the only route from \(c\) to \(d\), since the direct arm has been removed:

Reconnecting the arm:

Thévenin resistance first, because it is the easier of the two here. Short the 20 V source and reduce from the source end outwards:

Note the direction of travel: for \(R_{TH}\) the reduction runs from the shorted source towards the load, the opposite of the usual ladder reduction.

Open-circuit voltage. With the 6 Ω removed the 3 Ω carries no current, so \(V_{oc}\) is simply the voltage at the 18 Ω node. Reducing from the open end: the 18 Ω sees \(4 + (15 \parallel 4)\) back towards the source, and nodal analysis gives

The load current:

The 239ths cancel and leave a clean 0.8 A — exactly the current Set 2 obtained by ladder reduction, Set 4 by mesh analysis and Set 6 by nodal analysis.

When this is worth doing. For a single 6 Ω load it is not — direct ladder reduction is quicker. It becomes worthwhile the moment the load varies: an attenuator whose final resistor is switched, or a line whose termination is being matched, needs \(V_{TH}\) and \(R_{TH}\) once and nothing thereafter.

Deactivation cannot be used: the dependent source must stay active, so there is nothing to reduce. Apply a test source \(V\) across \(P\!-\!Q\) and find the current it delivers.

The test current is whatever flows through the 5 Ω, which is \(i_0\) itself:

KCL at node \(x\). Current arrives as \(i_0\) through the 5 Ω and as \(2i_0\) from the dependent source; all of it leaves through the 10 Ω:

Substituting back into the first relation:

The terminal current is \(i_0\), so

Compare the wrong answer. Deactivating the dependent source would leave \(5 + 10 = 15\ \Omega\) — less than half the truth. The dependent source triples the current in the 10 Ω without adding any to the terminals, so the network drops far more voltage than its resistors alone suggest.

Open-circuit voltage. Remove the 4 Ω. KCL at \(A\), with the current leaving through the 2 Ω balanced by the dependent source:

Expanding:

A degenerate but perfectly legitimate result: at \(V_{oc} = 10\ \text{V}\) the controlling drop \(v_s\) is zero, so the dependent source is off and no current flows anywhere. Both sides of the equation are zero.

Short-circuit current. Now short \(A\!-\!B\) — but with the 4 Ω still in place inside the network, since it is not the load being removed in this step. KCL at the node:

Multiplying by 4:

Hence

Reconnecting the 4 Ω load:

KVL on the input side, with the feedback branch carrying \(V_0/10^4\):

On the output side the dependent source drives the whole 30 kΩ:

Substituting (ii) into (i):

Solving exactly rather than in rounded decimals — multiply (i) by \(2.25 \times 10^{6}\) and substitute \(I = -V_0/(2.25 \times 10^{6})\):

Short-circuit current. Shorting the output forces \(V_0 = 0\), so the feedback branch carries nothing and (i) reduces to a single term:

Hence

Apply a test source \(V_x = 1\ \text{V}\) at \(C\!-\!D\). The controlling variable is the test voltage itself, so the dependent source delivers a known \(10^{-4}\ \text{A}\) into node \(A\).

KCL at \(A\) — current arrives from the test source through one 5 kΩ and from the dependent source, and leaves through the other 5 kΩ:

Multiplying by 5000:

The test current is whatever flows into the network through the first resistor:

Hence

Without the dependent source the answer would be \(5\text{k} + 5\text{k} = 10\ \text{k}\Omega\). The injected current partly supplies the second resistor, so the terminals see half the current and twice the resistance.

Short-circuit current first, because shorting the terminals kills the dependent source and makes the circuit trivial. With \(a\!-\!b\) shorted, \(v_1 = 0\), so \(v_1/100 = 0\):

Open-circuit voltage. Now \(v_1 = v_{oc}\) and KCL at the node gives

Multiplying by 100:

Hence

What the negative sign means. The open-circuit voltage is 125 V with terminal \(b\) positive with respect to \(a\) — a real, measurable voltage, simply of opposite polarity to the reference direction chosen at the start. Reversing the labels would give \(+125\) V and the identical circuit.

Note also that \(V_{oc}\) exceeds the 100 V supply. There is nothing wrong: the dependent source contributes energy of its own, and a network containing one is not bounded by its independent sources.

Apply a test voltage \(V\) at the terminals. The resistor current follows Ohm's law:

KCL at \(a\). The terminal supplies \(I\) and the dependent source supplies \(k\,i_x\); together they must equal the resistor current:

Hence

The behaviour across the range:

At \(k = 1\) the dependent source exactly supplies the resistor's demand, so no current is drawn at the terminals whatever the voltage — the network looks like an open circuit, and \(V_{oc}/I_{sc}\) would give a division by zero.

At \(k = 2\) the source supplies twice what the resistor needs, so the excess flows back out of terminal \(a\). Applying a positive voltage produces a negative current: the network delivers power to whatever is connected, which is exactly what \(R_{TH} = -4\ \Omega\) encodes.

Why \(V_{TH} = 0\). With the terminals open, the network's equations are homogeneous: every element law is of the form \(v = Ri\) or \(v = \alpha v_c\) or \(i = \beta i_c\), all with zero on the right when no independent source drives them. In matrix form,

and provided \(\mathbf{Y}\) is non-singular, the only solution is \(\mathbf{V}_n = \mathbf{0}\). Every node sits at zero and so do the terminals.

What the equivalent becomes. A source of zero volts is a short, so the Thévenin equivalent collapses to a single resistance:

Which method works. Both \(V_{oc}\) and \(I_{sc}\) are zero, so

The open-circuit/short-circuit route fails completely. Deactivation is also unavailable, since dependent sources may not be killed. Only the test source remains.

This is not an exotic case. It is exactly the situation in Problems 9, 12 and 14, and it is what every small-signal input or output resistance calculation amounts to: the bias sources are dead, the controlled sources are alive, and a test signal is applied precisely because there is nothing else to work with.

Every entry is one division into the same two numbers:

The sweep:

The two extremes bracket everything. The short-circuit current \(V_{TH}/R_{TH} = 3\ \text{A}\) is the largest current the network can deliver; the open-circuit voltage 6 V is the largest voltage. No load produces more of either.

The power peaks at \(R_L = 2\ \Omega = R_{TH}\). Differentiating:

which vanishes at \(R_L = R_{TH}\), giving \(P_{\max} = V_{TH}^2/4R_{TH} = 36/8 = 4.5\ \text{W}\). That is the maximum power transfer theorem, and Set 13 develops it.

Count the work. Six load values, six divisions. Solving the original circuit six times would mean six parallel combinations, six source currents and six terminal voltages — and the saving grows without limit as the number of loads increases.

The Norton equivalent is a current source \(I_N\) in parallel with a resistance \(R_N\). Matching the short-circuit currents of the two forms:

Verifying the equivalence. The Thévenin form has terminal relation \(V = V_{TH} - IR_{TH}\). For the Norton form, KCL at the terminals gives

Identical. The two are the same straight line in the \((V, I)\) plane, with intercepts \(V_{oc} = 6\ \text{V}\) and \(I_{sc} = 3\ \text{A}\) and slope \(-1/R_{TH}\).

Checking against the sweep of Problem 16 at \(R_L = 4\ \Omega\). By current division:

Which form to use:

The one case where conversion fails. An ideal voltage source has \(R_{TH} = 0\), so \(I_N = V_{TH}/0\) is undefined — no Norton form exists. Dually, an ideal current source has \(R_N = \infty\) and has no Thévenin form. Every network with a finite non-zero \(R_{TH}\) has both.

Measuring \(V_{TH}\). Connect a voltmeter with nothing else attached. A good voltmeter has input resistance far above \(R_{TH}\), so it draws almost no current and reads the open-circuit voltage directly.

The half-voltage method. Connect the variable resistor and adjust it until the terminal voltage falls to exactly half its open-circuit value. From the divider relation,

Reading the resistor's setting gives \(R_{TH}\) without knowing any current, and without trusting the ammeter's own resistance.

Why not just short the terminals? The relation \(R_{TH} = V_{oc}/I_{sc}\) is exact in theory and dangerous in practice:

The half-voltage method never draws more than \(I_{sc}/2\), and at that point the load absorbs the maximum power the source can give — itself a useful confirmation.

What could go wrong anyway. If the box is not linear, no single \(R_{TH}\) exists, and the resistor setting will depend on where you measure. Testing at a third point — say quarter voltage, which should need \(R_L = R_{TH}/3\) — checks linearity in one extra measurement.

Case 1: \(V_{oc} = 0\) and \(I_{sc} = 0\). A network with no independent sources, as in Problem 15. The ratio is \(0/0\).

This is the normal situation in small-signal analysis, where every input and output resistance is computed this way.

Case 2: \(R_{TH} = 0\). The network is an ideal voltage source. Then \(I_{sc} \to \infty\), the ratio is \(V_{oc}/\infty\), and shorting the terminals is physically impossible to model.

Case 3: \(R_{TH} = \infty\). The dependent source exactly supplies the internal demand, as at \(k = 1\) in Problem 14. Then \(I_{sc} = 0\) while \(V_{oc} \ne 0\), giving division by zero.

Case 4: the network is not linear. No \(V_{TH}\) and \(R_{TH}\) exist at all, since the terminal relation is not a straight line. Both measurements can still be made, and their ratio will be some number, but it describes nothing.

This is the dangerous case, because the calculation succeeds and the answer is meaningless. Problem 18's third measurement is the test.

Case 5: a singular network. If the dependent-source gain makes \(\mathbf{Y}\) singular — Set 4, Problem 20's condition — the circuit has no unique solution and neither \(V_{oc}\) nor \(I_{sc}\) is defined.

The original network. With the 4 Ω connected, the 3 Ω and 4 Ω are in parallel at \(12/7\ \Omega\), so the source current is

Element by element:

The load figures agree exactly, as they must. Everything else differs by a factor of seven — the original network burns 38 W internally while the equivalent burns 2 W.

Why this is not a contradiction. The proof of Problem 2 established one thing only: that the terminal relation \(V = V_{TH} - IR_{TH}\) is identical for both. It said nothing whatever about internal currents, and the substitution step deliberately discarded them.

The practical consequence. Efficiency questions cannot be answered from the equivalent. Here the original delivers 4 W of 42 W — under 10% efficient — while the equivalent appears to deliver 4 W of 6 W, or 67%. Anyone sizing a power supply, or asking whether a resistor will overheat, must go back to the real circuit.

The same warning covers component ratings, fault currents in internal branches, and the effect of a component failing inside the network. All of these are invisible to a two-terminal model, by construction rather than by oversight.