Use Norton equivalent circuits: Visual walkthrough

Step 1 — What is a "two-terminal network"?

WHAT. A two-terminal network is any tangle of resistors and sources with exactly two wires poking out — a top wire and a bottom wire. We call these two exposed wires the terminals. Everything you are ever allowed to do to the box is: connect something between those two terminals and measure.

WHY start here. The whole Norton idea is about what the box looks like from outside. So the first thing to nail down is: the outside world only ever touches those two dots. Nothing inside is observable except through them.

PICTURE. The grey cloud is the mystery circuit. Only the two lavender dots — the terminals — are exposed. We name the top dot $a$ and the bottom dot $b$ .

Step 2 — The box speaks only one language: a straight line

WHAT. For a linear box (resistors + independent sources, no weird nonlinear parts), the relationship between $V$ and $I$ is always a straight line. Sweep $V$ from small to large and plot the $I$ you measure — the dots fall on a line: $I = a V + b$

$I$ ::: current leaving terminal $a$
$V$ ::: voltage across the terminals
$a$ ::: the slope — how much $I$ changes per volt (negative: more voltage ⇒ less current out)
$b$ ::: the intercept — the current when $V = 0$

WHY a straight line and not a curve? Because of Superposition: in a linear circuit every response adds up proportionally. Doubling a driving voltage doubles its contribution to $I$ . An "add-up-proportionally" rule is exactly the recipe for a straight line. No squares, no bends.

PICTURE. The coral line is the box's entire personality. A line needs only two numbers to pin it down — where it crosses ( $b$ ) and how steep it is ( $a$ ). Remember that "two numbers" — it is the seed of the whole equivalent.

Step 3 — Read off the intercept: short the terminals to get $I_{N}$

WHAT. Take the box and connect the two terminals directly with a plain wire — a short circuit. A wire has (ideally) zero resistance, so it forces $V = 0$ . Measure the current racing through that wire; call it $I_{N}$ , the Norton current.

WHY this move. Look at the line $I = aV + b$ . Setting $V = 0$ kills the slope term entirely: $I = a \cdot 0 + b = b .$ So the short-circuit current is the intercept $b$ . The short reads the intercept off the graph directly — no algebra needed. We rename $b \to I_{N}$ .

PICTURE. The wire across $a$ – $b$ collapses the terminal voltage to zero (green marker sits at $V = 0$ ), and the current through the wire is the height where the line meets the vertical axis.

Step 4 — Read off the slope: kill the sources to get $R_{N}$

WHAT. Now we want the slope $a$ . Trick: turn off every independent source inside the box — replace each voltage source with a wire (a $0 V$ source is a wire), and each current source with a gap (a $0 A$ source is an open). What remains is a pure pile of resistors. Measure the resistance between $a$ and $b$ ; call it $R_{N}$ , the Norton resistance.

WHY kill the sources. The slope tells you how current sags as voltage rises — that is a purely resistive effect. Sources only shift the line up or down (they move the intercept); they never tilt it. So to isolate the tilt, we must remove the shift. Deactivating the sources drops the whole line down to pass through the origin, leaving only its slope to measure.

Relating slope to $R_{N}$ . With sources off, the box is just a resistance $R_{N}$ . By Ohm's Law the current out is $I = - V / R_{N}$ (out-current falls as you raise $V$ against it). Comparing to $I = aV$ : $a = - \frac{1}{R _{N}} ⟹ R_{N} = - \frac{1}{a} .$ The steeper (more negative) the slope, the smaller $R_{N}$ .

PICTURE. Top: sources alive, line sits high (intercept $I_{N}$ ). Bottom: sources killed, the same line slides down to the origin — same steepness, and that steepness is $1/ R_{N}$ .

Step 5 — Rebuild the line from the two numbers

WHAT. We now have the intercept $I_{N}$ and the slope $- 1/ R_{N}$ . Plug them into $I = aV + b$ : $I = slope a - \frac{1}{R _{N}} V + intercept b I_{N} = I_{N} - \frac{V}{R _{N}} .$

Every term is now earned:

$I$ ::: current delivered to whatever load you attach (we now call it $I_{L}$ )
$I_{N}$ ::: the full current the source pushes in (from Step 3)
$\frac{V}{R _{N}}$ ::: the piece of that current that leaks back through $R_{N}$ (Ohm's Law: current = voltage ÷ resistance)
$V$ ::: the terminal voltage the load develops

WHY it is the leak. This is Kirchhoff's Current Law at the top node. The source shoves $I_{N}$ in. $R_{N}$ sits across the terminals feeling voltage $V$ , so it siphons $V / R_{N}$ back down. What survives, $I_{N} - V / R_{N}$ , must exit to the load — current in equals current out.

PICTURE. The lavender arrow $I_{N}$ enters the node. A mint arrow $V / R_{N}$ peels off down through $R_{N}$ . The coral arrow — the remainder — leaves to the load. That is the whole Norton circuit, and the equation is just bookkeeping at the dot.

Step 6 — Every case on the load line (nothing left unshown)

WHAT. The one line $I_{L} = I_{N} - V / R_{N}$ must cover all possible loads. Let us walk the line end to end and make sure no scenario surprises the reader.

Short circuit ( $V = 0$ , load = wire): $I_{L} = I_{N}$ . Maximum current out. Left end of the line.
Open circuit ( $I_{L} = 0$ , load = gap): solve $0 = I_{N} - V / R_{N} \Rightarrow V = I_{N} R_{N}$ . Maximum voltage, zero current. Right end of the line — and this $V_{oc} = I_{N} R_{N}$ is exactly the Thévenin voltage.
Matched load ( $R_{L} = R_{N}$ ): current splits equally, $I_{L} = I_{N} /2$ , $V = I_{N} R_{N} /2$ . This is the Maximum power transfer point — the midpoint of the line.
In between ( $0 < V < I_{N} R_{N}$ ): ordinary operation, current-divider $I_{L} = I_{N} \cdot R_{N} / (R_{N} + R_{L})$ .

WHY these edges matter. A real load can be anything from a dead short to nothing at all. The line already answers all of them; a reader who only saw the "middle" case would panic at the ends. Here are the ends, drawn.

PICTURE. The same coral load line, now with three marked points: short-circuit (left, on the $I$ -axis), matched load (middle), open-circuit (right, on the $V$ -axis).

Step 7 — Numbers on the line (worked, from the parent)

WHAT. Parent Worked Example: $12 V$ source, $R_{1} = 4 Ω$ in series, $R_{2} = 6 Ω$ across the output.

Step 3 in action ( $I_{N}$ ): short the terminals ⇒ $R_{2}$ shorted out ⇒ $I_{N} = 12/4 = 3 A$ .
Step 4 in action ( $R_{N}$ ): kill the $12 V$ (→ wire) ⇒ $R_{1} ∥ R_{2} = \frac{4 \cdot 6}{4 + 6} = 2.4 Ω$ .
Line: $I_{L} = 3 - V /2.4$ .
Attach $R_{L} = 2.4 Ω$ : $I_{L} = 3 \cdot \frac{2.4}{2.4 + 2.4} = 1.5 A$ , and $V = 1.5 \times 2.4 = 3.6 V$ .

WHY it lands on our line. The point $(V, I_{L}) = (3.6, 1.5)$ must satisfy $I_{L} = 3 - V /2.4$ : check $3 - 3.6/2.4 = 3 - 1.5 = 1.5$ . ✓ The abstract line and the concrete numbers are the same object.

PICTURE. The line $I_{L} = 3 - V /2.4$ with the load-resistor line $I_{L} = V /2.4$ crossing it. The crossing dot is the operating point $(3.6 V, 1.5 A)$ .

The one-picture summary

Everything above in one frame: the mystery box collapses to a line, the line's intercept is $I_{N}$ (short the terminals), its slope is $- 1/ R_{N}$ (kill the sources), and rebuilt they give the current source ∥ resistor that obeys $I_{L} = I_{N} - V / R_{N}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Picture a sealed box with two wires sticking out. You are never allowed inside; you can only push voltage across the wires and watch the current. When you do, the current-vs-voltage points always land on a straight line — that is what "linear" buys you. A straight line is just two facts: where it starts and how steeply it drops. To get where it starts, short the wires (voltage = 0) and read the current — that's $I_{N}$ , the biggest current the box can give. To get the steepness, switch off the internal sources (batteries become plain wire, current pumps become gaps) and measure the leftover resistance — that's $R_{N}$ . Now you can throw the box away: a single current source $I_{N}$ sitting beside a single resistor $R_{N}$ reproduces the exact same line. Attach any load, and the current it gets is "everything the source pushes, minus whatever leaks back through $R_{N}$ " — $I_{L} = I_{N} - V / R_{N}$ . Short load? All the current comes out. No load? All the voltage builds up. Matched load? Split it in half. One box, two numbers, every case covered.

Recall Self-test

Why is the $I$ – $V$ relationship a straight line? ::: Linearity/superposition makes $I$ an affine function of $V$ . What does shorting the terminals read off? ::: The intercept $I_{N}$ (current at $V = 0$ ). What does killing the sources isolate? ::: The slope, i.e. $R_{N} = - 1/ a$ . What is the open-circuit terminal voltage of the Norton model? ::: $V_{oc} = I_{N} R_{N}$ (the Thévenin voltage). What happens to the line when $R_{N} \to \infty$ ? ::: It goes flat — a pure ideal current source.

Connections

Use Norton equivalent circuits — parent topic; this page is its picture-derivation.
Superposition — why the terminal law is a straight line.
Ohm's Law — supplies the $V / R_{N}$ leak term and the slope.
Kirchhoff's Current Law — bookkeeping at the parallel node.
Thévenin equivalent circuits — the open-circuit end of the same line.
Source transformation — converts between the two descriptions.
Maximum power transfer — the matched-load midpoint of the line.