Visual walkthrough — Use Thevenin equivalent circuits

See also the parent: Thevenin topic note.

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Step 1 — What "two terminals" even means

WHAT. Draw a big scribbly box full of resistors and batteries. Cover all of it with your hand except two wires poking out — call them the terminals, top one $a$, bottom one $b$.

WHY. The whole theorem is about what you can observe from outside. You are only allowed to touch $a$ and $b$. So the only two things you can ever measure are:

• the voltage between $a$ and $b$ — call it $V$ (like a height between two points),
• the current $I$ that flows out of $a$, through whatever you attach, and back into $b$.

PICTURE. In the figure, the tangle is hidden behind a violet curtain. Only $a$, $b$, $V$ and $I$ escape. That pair $(I,V)$ is all the information the outside world has.

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Step 2 — Poke the box and collect points

WHAT. Attach different external resistors, one at a time. Each one lets a different amount of current flow, and each gives you a reading. Plot every reading as a dot on a graph: horizontal axis $I$, vertical axis $V$.

WHY. We do not yet know how the box behaves. So we experiment. If the box is made only of resistors and sources that obey $V=IR$ (that is what linear means — no bending, no squaring), then something magical is about to show up in the dots.

PICTURE. The magenta dots — from many different loads — all fall on one straight line. Not a curve. A line.

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Step 3 — Read off the first number: the intercept $V_{th}$

WHAT. Follow the line left until $I=0$ (nothing drawn from the box, terminals open). Read the height there.

WHY. $I=0$ is the easiest experiment of all — attach nothing. The voltage you read is called the open-circuit voltage, $V_{oc}$. On the graph it is where the line crosses the vertical axis: the intercept.

PICTURE. The orange dot sits on the $V$-axis. Its height is the intercept. We name it $V_{th}$.

$V{|}_{I=0}=\underset{\text{the intercept}}{\underbrace{V_{th}}}$

• $V{|}_{I=0}$ — the terminal voltage when no current flows.
• $V_{th}$ — the name we give that height. Nothing more mysterious than "where the line hits the axis."

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Step 4 — Read off the second number: the slope $-R_{th}$

WHAT. Now measure how much the line drops in height for each extra amp you pull. Take any two points, divide the fall in $V$ by the rise in $I$. That ratio is the slope.

WHY. The line tilts downward: the more current you draw, the more the terminal voltage sags. That sag is caused by an internal resistance. A slope has units of volts-per-amp — which is ohms. So the slope literally is a resistance (with a minus sign, because it droops).

PICTURE. The violet right-triangle on the line shows a horizontal run $\Delta I$ and a downward drop $\Delta V$. Their ratio is the tilt.

$\text{slope}=\frac{\Delta V}{\Delta I}=-R_{th}$

• $\Delta V$ — how far the voltage fell (a negative number, the line goes down).
• $\Delta I$ — how much more current we pulled (positive).
• $-R_{th}$ — the negative tells us "voltage sags as current rises"; the size $R_{th}$ is the internal resistance in ohms.

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Step 5 — Assemble the equation from the two numbers

WHAT. A straight line is "intercept plus slope times the input." Plug in our two numbers:

WHY. We now have both pieces the line needs. Writing them together is the terminal law — no new physics, just naming what the graph already shows.

PICTURE. The full line, with the intercept labelled $V_{th}$ and the tilt labelled $-R_{th}$, and the equation written along it.

$\underset{\text{what we read}}{\underbrace{V}}=\underset{\text{intercept, from Step 3}}{\underbrace{V_{th}}}\underset{\text{slope}\times \text{current, from Step 4}}{\underbrace{-R_{th}I}}$

• $V$ — terminal voltage, the output.
• $V_{th}$ — where the line starts (the open-circuit height).
• $-R_{th}I$ — how much that height drops once current $I$ flows.

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Step 6 — Build the physical box that makes this line

WHAT. Ask: what is the simplest real circuit whose terminal graph is this exact line? Answer: one battery $V_{th}$ in series with one resistor $R_{th}$.

WHY. Walk current $I$ through that little circuit. The battery lifts you to $V_{th}$; the resistor drops you by $I{R}_{th}$. Terminal voltage $=V_{th}-I{R}_{th}$ — the same line. Since the outside world only sees the line, it cannot tell this tiny circuit apart from the original tangle.

PICTURE. Left: the violet tangle. Right: battery $V_{th}$ stacked on resistor $R_{th}$. An "=" sign between them, because their $(I,V)$ graphs are identical.

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Step 7 — The edge cases: read them off the line

Every scenario is just a special spot on our line. Nothing new to compute.

WHAT / WHY / PICTURE (one figure, three marked points):

1. Open circuit ($I=0$): the intercept. $V=V_{th}$. Nothing attached ⇒ full open-circuit voltage. (orange dot on the $V$-axis)
2. Short circuit ($V=0$): where the line crosses the horizontal axis. Set $0=V_{th}-I_{sc}{R}_{th}$, so $I_{sc}=\frac{V_{th}}{R_{th}}⟹R_{th}=\frac{V_{th}}{I_{sc}}.$ This is the third route to $R_{th}$ — cross-check with Nodal and Mesh Analysis. (magenta dot on the $I$-axis)
3. Degenerate $R_{th}=0$ (ideal source): the line is flat — voltage never sags, no matter the current. A perfect battery.
4. Degenerate $V_{th}=0$ (no internal sources): the line passes through the origin — the box is just a plain resistor $R_{th}$, giving $V=-I{R}_{th}$ (i.e. it only ever absorbs).

• $I_{sc}$ — short-circuit current, the current when you clip the terminals together.
• The four marked spots cover every possibility: any load you attach lands somewhere on this one line.

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The one-picture summary

One figure holds the entire derivation: the hidden tangle on the left, its measured straight line in the middle (intercept $V_{th}$, slope $-R_{th}$, marked open- and short-circuit points), and the two-part equivalent box on the right.

Recall Feynman retelling — the whole walkthrough in plain words

You have a sealed machine with two plug holes. You cannot open it. So you experiment: plug in different resistors and, each time, write down two things — how hard it pushes (voltage $V$) and how much flows (current $I$). Dot by dot, you graph those pairs.

Surprise: all the dots line up straight. A straight line has just two facts to it — where it starts (its height with nothing plugged in, that's $V_{th}$) and how fast it droops as you pull more current (its steepness in ohms, that's $R_{th}$). Two facts, two dots' worth of information.

So you write the line as $V=V_{th}-I{R}_{th}$: start at $V_{th}$, sag by $I{R}_{th}$. Then you notice the simplest real gadget that makes exactly this line — one battery ($V_{th}$) sitting on top of one resistor ($R_{th}$). From the outside it's a perfect twin of the sealed machine. That twin is the Thevenin equivalent, and the two special dots — one on each axis — are the open-circuit voltage and short-circuit current you can always use to double-check your two numbers.

Recall Quick self-check

The $V$–$I$ line crosses the vertical axis at ::: $V_{th}$ (open circuit, $I=0$). The $V$–$I$ line crosses the horizontal axis at ::: $I_{sc}=V_{th}/R_{th}$ (short circuit, $V=0$). The slope of the line equals ::: $-R_{th}$ (volts dropped per amp drawn). Why must the graph be a straight line? ::: Because the network is linear — only linearity forbids curves, leaving $V=V_{th}-I{R}_{th}$. If $R_{th}=0$ the line looks like ::: a flat horizontal line — an ideal voltage source that never sags.

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Connections

• Use Thevenin equivalent circuits — the parent topic this page unpacks visually.
• Superposition Theorem — the reason we may split the line into "intercept part" + "slope part".
• Norton Equivalent Circuits — same line, read as a current source; $I_N=I_{sc}$, $R_N=R_{th}$.
• Source Transformation — swap freely between the battery+resistor and current-source forms.
• Maximum Power Transfer Theorem — sliding the load along this same line to the sweet spot $R_L=R_{th}$.
• Voltage and Current Dividers · Nodal and Mesh Analysis — tools that compute the two numbers $V_{th}$, $R_{th}$.