Norton's Theorem

Norton current. Remove the 6 Ω load and short \(a\!-\!b\). The 12 Ω now has zero volts across it and carries nothing, so the entire short-circuit current comes through the 4 Ω:

Shorting a terminal often removes elements from the calculation — the reason \(I_{sc}\) is frequently easier to find than \(V_{oc}\).

Norton resistance. Deactivate the source. A dead voltage source is a short, placing the 4 Ω directly across the 12 Ω:

Reconnect the load — the equivalent is now a current divider between \(R_N\) and \(R_L\):

The resistance in the numerator is the other branch's — \(R_N\), not \(R_L\). Current takes the easier path, so the larger the load, the smaller its share.

Check via Thévenin. The same network has \(V_{TH} = 24 \times 12/16 = 18\ \text{V}\) and \(R_{TH} = 3\ \Omega\), so \(I_L = 18/(3+6) = 2\ \text{A}\;\checkmark\), and \(I_N = V_{TH}/R_{TH} = 18/3 = 6\ \text{A}\;\checkmark\)

Route 1 — from Thévenin. Set 9, Problem 2 established that any linear two-terminal network obeys

Rearranging for \(I\) is the whole proof:

which is exactly KCL at the terminals of a current source \(I_N = V_{TH}/R_{TH}\) in parallel with \(R_N = R_{TH}\).

Route 2 — from duality. Set 8, Problem 14 gave the dictionary: voltage source ↔ current source, series ↔ parallel, \(R \leftrightarrow G\). Applying it term by term to Thévenin's statement:

Norton's theorem is Thévenin's theorem applied to the dual network. It required no separate proof and never did.

The geometry. Both forms describe the same relation between terminal voltage and terminal current:

A straight line in the \((V, I)\) plane with intercepts \(V_{oc}\) and \(I_{sc}\) and slope \(-1/R_{TH}\). Thévenin names the voltage intercept, Norton names the current intercept, and both name the same slope.

A network is therefore completely characterised by any two points on that line. This is what makes Problem 12 possible: two measurements at arbitrary loads determine both parameters, without either terminal ever being opened or shorted.

Norton current. Short \(a\!-\!b\). The 6 A now divides between the 3 Ω shunt and the 6 Ω path leading to the short, and only the second part reaches the terminals:

Current division again, and again the opposite resistance sits on top: the 3 Ω shunt appears in the numerator of the current through the 6 Ω.

Norton resistance. Deactivating the current source opens it, leaving the 3 Ω and 6 Ω as the only path between the terminals — in series:

With the load:

Check directly: with the 9 Ω connected, the 6 A divides between the 3 Ω and the \(6+9 = 15\ \Omega\) branch, giving \(6 \times 3/18 = 1\ \text{A}\;\checkmark\)

Norton resistance. Short the 20 V source and open the 2 A source. Only the 24 Ω then joins \(a\) to \(b\):

Norton current. Short \(a\!-\!b\), so that \(V_a = 0\). The 20 V source then drives its full current through the 24 Ω into node \(a\), while the 2 A source removes 2 A from it:

Cross-check by open circuit. With the terminals open, KCL at \(a\) gives

Reading the sign. A negative \(I_N\) means the arrow was drawn the wrong way: 1.167 A actually flows from \(a\) to \(b\) inside the equivalent. Reversing the arrow and writing \(+1.167\) A describes the identical network.

Note also that the 2 A source overwhelms the 20 V one — it demands more current than the source can supply through 24 Ω, so the terminal voltage is dragged below zero. Nothing is wrong; it simply means node \(a\) sits 28 V below node \(b\).

Norton resistance. Short the 120 V source and open the 2 A source. The 20 Ω and 40 Ω are then in series, and that chain is in parallel with the 12 Ω:

Norton current. Short \(a\!-\!b\), which puts the 12 Ω out of action. Transform the 2 A source with its parallel 40 Ω into a voltage source:

The network is now a single loop containing 120 V, 20 Ω, 80 V and 40 Ω, closed by the short.

KVL round that loop, with the two sources opposing:

Independent check by open circuit. Leave the 12 Ω in place. The same single loop now runs through it as well, with total resistance \(20+40+12 = 72\ \Omega\) and net driving voltage \(120-80 = 40\ \text{V}\):

Norton resistance. Remove the load, short the 36 V source and open the 3 mA source. Looking back from the terminals: the 10 kΩ, then the 2 kΩ, then the two source-side resistors in parallel:

Norton current by superposition. Short the terminals and take one source at a time.

a36 V source alone, with the 3 mA source opened. Transform the 36 V–12 kΩ pair into a 3 mA source with a 12 kΩ shunt, which combines with the 24 kΩ to give \(12 \parallel 24 = 8\ \text{k}\Omega\). That 3 mA now divides between the 8 kΩ shunt and the \(2 + 10 = 12\ \text{k}\Omega\) path to the short:

b3 mA source alone, with the 36 V source shorted. From node \(Y\) there are now two routes to the reference: back through the 2 kΩ into \(12 \parallel 24 = 8\ \text{k}\Omega\), total 10 kΩ — and forward through the 10 kΩ to the short. Two equal paths, so the current splits evenly:

Negative because this contribution reaches the short in the direction opposite to \(I_1\).

Superposing:

Reconnect the 1 kΩ load to the Norton source by current division, then multiply by the load:

Replace the left network by its Norton equivalent. Deactivating its sources — the 2 A opened, the 12 V shorted — leaves the 6 Ω and 4 Ω as the only path between the terminals:

Short-circuit current. Shorting \(a\!-\!b\) reconnects the 6 Ω branch to the reference. KCL at node \(v\):

The short carries whatever the 6 Ω branch carries:

Now combine. Both the left equivalent and the right-hand 4 A source are current sources across the same terminals, so they simply add:

Current division into the 5 Ω:

Short-circuit current. Short \(a\!-\!b\) and write KCL at both nodes. At \(v_1\), with terminal \(a\) now at zero:

At \(v_2\), which receives \(i_1\) from the middle resistor and \(2i_1\) from the dependent source:

The second pair gives \(3(v_1-v_2)/2 = v_2/2\), so \(v_2 = \tfrac34 v_1\). Substituting into the first:

Expected: with no independent source the equations are homogeneous, so every node sits at zero — Set 9, Problem 15.

Norton resistance by test source. Both \(V_{oc}\) and \(I_{sc}\) vanish, so their ratio is useless. Apply \(v_0\) at \(a\!-\!b\). KCL at \(v_1\) now has the test source at the far end of the 2 Ω:

The relation \(v_2 = \tfrac34 v_1\) is unchanged, since it came from node \(v_2\) alone. Substituting:

Hence \(v_2 = \tfrac12 v_0\) and \(i_1 = (v_1-v_2)/2 = v_0/12\).

The current drawn from the test source is what enters through the 2 Ω plus what leaves through the dependent source:

With \(I_N = 0\) the equivalent collapses to a bare 3 Ω resistor — a zero-valued current source in parallel with \(R_N\) is simply an open circuit beside it.

Short-circuit current. Remove the 18 Ω and connect \(c\) to \(d\) with a wire. They become a single node \(m\), and the bridge becomes two pairs of parallel resistors between the 8 V rail and the reference. KCL at \(m\):

That is \(\tfrac14(8-V_m) = \tfrac{5}{18}V_m\), so

The current in the shorting wire is whatever arrives at \(c\) and does not leave through the 9 Ω:

Norton resistance is the \(R_{TH}\) of Set 9, Problem 7 — shorting the 8 V supply merges \(a\) and \(b\):

Consistency with Thévenin. Set 9 found \(V_{TH} = 32/15\ \text{V}\), so

Reconnecting the arm by current division:

The technique alternates between the two forms. A voltage source in series with a resistor becomes a current source in parallel with it, which can then absorb any shunt resistor; converting back lets the next series resistor be absorbed, and so on down the ladder.

Step 1. Transform the source and its first series resistor:

Step 2. The 15 Ω shunt is now in parallel with that 4 Ω, and merges with it:

Step 3. Convert back and absorb the second series 4 Ω:

Step 4. Absorb the 18 Ω shunt, convert back, and absorb the final 3 Ω:

These are precisely the \(V_{TH}\) and \(R_{TH}\) that Set 9, Problem 8 obtained by open-circuit voltage and deactivation — reached here without a single KCL equation.

Step 5. One last transformation gives the Norton form:

And the load current by division:

Convert both to Norton form, where parallel connection is trivial:

Current sources in parallel add, and so do their conductances:

Converting back gives Millman's theorem. In general, for \(n\) sources in parallel:

Here \(V_{TH} = 5.5 \times 2 = 11\ \text{V}\), or directly \((12 \times 6 + 9 \times 3)/(3+6) = 99/9 = 11\ \text{V}\;\checkmark\)

Reading the result. The combined voltage is a conductance-weighted average of the individual voltages, so it always lies between them — never outside. The stiffer source (smaller \(R\)) dominates: here the 3 Ω branch is weighted twice as heavily, pulling the result to 11 V rather than the arithmetic mean of 10.5 V.

The circulating current. Even with nothing attached, current flows from the stronger source into the weaker one:

Confirm at the terminals: the 12 V branch delivers \((12-11)/3 = 1/3\ \text{A}\) and the 9 V branch absorbs \((11-9)/6 = 1/3\ \text{A}\;\checkmark\)

Problem 2 established that the terminal behaviour is a straight line, so two points determine it completely. Convert each measurement to a \((V, I)\) pair:

The slope gives the resistance. Since \(I = I_N - V/R_N\):

The intercept gives the current. Substituting either point:

Check with the second point: \(1.5 + 12/4 = 4.5\;\checkmark\)

Verification. Predicting the original measurements from the equivalent:

In Thévenin terms the box is 18 V behind 4 Ω, and its short-circuit current would be 4.5 A — inferred, never measured.

Testing linearity. Two points always fit a line, so they prove nothing. Take a third measurement at some other load — say 4 Ω, where the model predicts \(4.5 \times 4/8 = 2.25\ \text{A}\). Agreement supports linearity; disagreement means no Norton equivalent exists and the two-point fit was meaningless.

Norton current. Short \(a\!-\!b\), so \(V_a = 0\). That kills the dependent source — its control variable is zero — and puts zero volts across the shunt 5 Ω. Everything the 20 V source delivers goes into the short:

Norton resistance by test source. Deactivate the independent 20 V source but not the dependent one. Applying \(V\) at the terminals, three paths draw current:

Without the dependent source the answer would be \(5 \parallel 5 = 2.5\ \Omega\); the VCCS draws extra current and stiffens the source.

Cross-check by open circuit. KCL at \(a\) with the terminals open:

With the load:

No Norton form: an ideal voltage source. Here \(R_{TH} = 0\), so

The terminal characteristic is the vertical line \(V = V_{TH}\): the voltage is fixed and the current takes any value the load demands. A Norton form would need an infinite source shunted by zero resistance, which is not a circuit.

No Thévenin form: an ideal current source. Here \(R_N = \infty\), so \(V_{TH} = I_N R_N\) is unbounded. The characteristic is the horizontal line \(I = I_N\).

Summarising by the slope of the terminal line:

Why this is not merely pedantic. Both extremes occur in practice as good approximations. A mains supply or a regulated bench supply has \(R_{TH}\) of milliohms and is treated as ideal; a transistor in saturation, or a current mirror, presents megohms and is treated as an ideal current source. In each case the missing equivalent is missing because the model has been idealised, not because the real device lacks one.

Note that Problem 8's network — \(I_N = 0\) with \(R_N = 3\ \Omega\) — has both forms, and they coincide: a zero current source in parallel with 3 Ω and a zero voltage source in series with 3 Ω are the same bare resistor.

Set 8, Problem 9 derived the nodal system as

Look at the right-hand side: it is a vector of currents injected at nodes. Nodal analysis has a slot for a current source and no slot at all for a voltage source.

A Norton source fits directly. A current \(I_N\) into node \(j\) and out of node \(k\) adds \(+I_N\) and \(-I_N\) to the source vector, and its parallel \(R_N\) stamps into \(\mathbf{Y}\) like any other resistor. Nothing else changes.

A Thévenin source does not. A voltage source between two nodes forces a supernode (Set 7) or an extra unknown in modified nodal analysis (Set 7, Problem 20) — either way, a structural complication. But converting it removes the difficulty entirely:

What this makes of the "by inspection" rules. Set 6's recipe — diagonal entries are sums of conductances, off-diagonals are minus the shared conductance, right-hand side is the net injected current — is now simply a description of what Norton sources and resistors stamp. The rules were never arbitrary.

The limit. A voltage source with no series resistance cannot be converted, by Problem 14. That single case is exactly what supernodes and modified nodal analysis exist to handle — and it is why they exist at all.

By current division, the load current and hence its power:

Differentiating with respect to \(R_L\):

which vanishes at \(R_L = R_N\) — the same condition as the Thévenin derivation, as it must be, since \(R_N = R_{TH}\).

The maximum:

Confirming directly: at \(R_L = 3\ \Omega\), \(I_L = 6 \times 3/6 = 3\ \text{A}\) and \(P_L = 9 \times 3 = 27\ \text{W}\;\checkmark\)

The two expressions side by side:

Identical, since \(V_{TH} = I_N R_N\). Here \(V_{TH} = 18\ \text{V}\) and \(18^2/(4 \times 3) = 27\ \text{W}\;\checkmark\)

The behaviour at the extremes is worth noting. As \(R_L \to 0\) the load carries the full \(I_N = 6\ \text{A}\) but drops no voltage, so \(P_L \to 0\). As \(R_L \to \infty\) it holds the full 18 V but carries nothing, so again \(P_L \to 0\). The peak sits between, and the current-source view makes the first limit the obvious one — the mirror of the Thévenin view, where the open circuit is obvious.

Work from the far end. At node 3 the shunt 2 Ω sits in parallel with the 2 Ω terminator:

Node 2 sees that 1 Ω through the series 1 Ω, giving 2 Ω to its right — which is in parallel with its own 2 Ω shunt:

Node 1 repeats the pattern exactly:

The structure is self-similar: from every node, the resistance looking right is \(2R\) and the resistance including the shunt is \(R\). That is the whole design principle of the R–2R network.

Now the currents. At node 1 the incoming 8 A meets two equal 2 Ω paths — the shunt, and everything to the right — so it divides exactly in half. The same happens at every node:

The four branch currents sum to \(4+2+1+1 = 8\ \text{A}\;\checkmark\)

Why the halving is exact. Each node's Norton equivalent, seen by the remainder of the ladder, has \(R_N = 2\ \Omega\) facing a shunt of \(2\ \Omega\). Equal resistances split current equally, and the equality is preserved at every stage because the reduction returns the same value — which is what "R–2R" is chosen to achieve.

Current division gives the load current in each case:

The first delivers 99% of its nominal current; the second delivers 9%. The source's rating is the same in both cases — the difference lies entirely in \(R_N\).

The design rule. A good current source needs \(R_N \gg R_L\), so that the shunt steals a negligible share:

The fractional error is roughly \(R_L/R_N\): with \(R_N = 100R_L\) the output is within 1%.

The dual rule for a voltage source is \(R_{TH} \ll R_L\), so that little voltage is lost internally. Both say the same thing — the internal resistance should be at the opposite extreme from the load:

Note the middle row. A good current source is deliberately inefficient: with \(R_N = 100\text{k}\) and \(R_L = 1\text{k}\), the shunt dissipates a hundred times less than... in fact the load takes 99% of the current but the source must sustain the terminal voltage across the whole shunt. Precision, not efficiency, is what is being bought.

Stage 1. Transform the source: \(20\ \text{V} + 4\ \Omega \to 5\ \text{A} \parallel 4\ \Omega\). The rest of the ladder, seen from node 1, presents

The first shunt and the onward branch then follow immediately:

Stage 2. That 1.2 A now drives the 18 Ω shunt in parallel with the remaining \(3+6 = 9\ \Omega\):

Stage 3. The final 0.8 A passes through the 3 Ω into the 6 Ω load:

Every figure matches the earlier sets exactly:

What this shows about equivalence. Problem 10 reduced the same ladder to a single Norton equivalent, which correctly gave the load current but discarded everything internal. Reducing in stages keeps a Norton equivalent at each node, and so recovers the entire solution. The equivalence is exact at whichever terminal pair you choose to stop at — and stopping at several gives the whole circuit.

Choosing which quantity to compute. The decision is made before the form is chosen, by looking at what each operation does to the topology:

Whichever you compute, the other follows from \(V_{oc} = I_{sc}R_{TH}\) — so this choice costs nothing and can halve the work.

Choosing which form to present.

What the two theorems accomplish. They replace "solve this network" with "characterise it at two terminals", reducing an arbitrary linear network to two numbers. Every load question then becomes one division. The saving grows without limit as the load varies, and it is what makes modular design possible: a subsystem can be specified by \(V_{TH}\) and \(R_{TH}\) and connected to anything.

What they do not accomplish. Four things, each already met:

Set 9, Problem 20 measured the first three — 38 W internal against 2 W in \(R_{TH}\). The fourth is the dangerous one, because the arithmetic completes and the answer means nothing.

And what they need. Only linearity of the network, by Set 9, Problem 2. Not planarity, not independence of the sources, and not linearity of the load.