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CR · DC analysis

The Thévenin Equivalent

However complicated the inside of a box, seen from its two terminals it reduces to one voltage source and one resistor. This unit varies the load to confirm that the terminal V-I relation is a single straight line, and shows how that line can stand in for the whole circuit.

Sweep the load: the terminals trace one line

The straight line in the graph is the terminal voltage-current relation, and the dot on it is the operating point at the present load. Change the load resistance R_L and the operating point slides along the line; look for the spot where the terminal voltage is exactly half the open-circuit voltage. The R_L there is the box's internal resistance R_th.

Load resistance R_LR_L = 18.0 Ω
Terminal voltage V_L
V_L = 9.0 V
75% of the open-circuit voltage (V_th = 12 V)
Far from half

Thévenin voltage = open-circuit voltage

Disconnect the load completely and the current drops to 0. Then the voltage across the internal resistance is also 0, so the voltage inside the box appears unchanged at the terminals, and this is the largest voltage the box can deliver. That value is the Thevenin voltage V_th, also called the open-circuit voltage.

Thévenin resistance = slope of the line

Lowering the load resistance raises the current and drops the terminal voltage. How much the terminal voltage falls for each 1 ampere of extra current, that is, the slope of the line, is the internal resistance R_th. Short the terminals by setting the load to 0, and the terminal voltage becomes 0 while the current rises to the short-circuit current I_sc = V_th/R_th.

Half voltage points to R_th

The moment the load resistance equals the internal resistance, the two split the voltage evenly, so the terminal voltage is exactly half the open-circuit voltage. Find the load at which the terminal voltage falls to half, then, and that value is R_th. The box in the plot has V_th = 12 V and R_th = 6 Ω, so setting R_L to 6 Ω brings the terminal voltage to 6 V. In effect, the internal resistance has been measured without ever opening the box.

ObserveVL = Vth RL(RL + Rth)
The box behaves like a voltage divider.
ChooseIsc = Vth / ?
The short-circuit current is Vth divided by Rth.
Fill inVL = Vth2 → RL = ?
The load that gives half voltage is Rth.
On your ownRth = Voc / ?
Open-circuit voltage over short-circuit current is Rth.

Folding the box into one line

Know just one open-circuit voltage V_th and one internal resistance R_th, and from outside it is impossible to tell whether the box holds ten resistors or several sources. The single V-I line those two values define stands in for the entire box. As the figure in this section shows, redrawing the question-mark box as one voltage source and one resistor looks exactly the same from terminals a and b. When analyzing one part of a large circuit, folding the rest into this one line makes the problem far simpler.

Back to the first screen

While you changed R_L, the operating point always slid along the same straight line. With the load removed, the line pointed to the open-circuit voltage V_th on the horizontal axis; with the terminals shorted, it pointed to the short-circuit current I_sc on the vertical axis. Where the terminal voltage reached half, R_L was exactly R_th. Without a single look inside the box, one straight line stood in for everything in it.

The Thevenin equivalent replaces any linear two-terminal circuit with one voltage source and one series resistor. Vth is the open-circuit voltage with the load removed, and Rth is the internal resistance set by the slope of the terminal line, with Rth = Vth/Isc. The single terminal V-I line stands in for the whole box, and the load at which the terminal voltage is half the open-circuit voltage is Rth itself.