Thevenin equivalents

YOU WILL BE ABLE TO

State Thevenin's theorem and what it claims.
Find V_Th as the open-circuit voltage.
Find R_Th by killing sources, or from V_oc/I_sc.
Use a Thevenin equivalent to analyse any load quickly.

Here is the most useful theorem in circuit analysis. Any linear two-terminal network — no matter how many sources and resistors it contains — behaves, at its two terminals, exactly like one ideal voltage source V_Th in series with one resistor R_Th. Not approximately: exactly, for every possible load. The outside world cannot tell the difference.

Thevenin's claim: at terminals a–b, the whole network on the left is indistinguishable from the two-element equivalent on the right.

Finding the two numbers takes two experiments — on paper or on a bench. V_Th is the open-circuit voltage: leave the terminals unconnected and compute (or measure) the voltage across them. With no load current, nothing drops across R_Th, so the terminal voltage IS V_Th. For R_Th, when the network contains only independent sources: kill them all (shorts and opens, as in Lesson 1) and compute the resistance looking into the terminals.

V_{Th} = v_{oc}

R_{Th} = \frac{v_{oc}}{i_{sc}}

the two experiments — i_sc is the current with the terminals shorted

The second formula is the alternative route: compute both the open-circuit voltage AND the short-circuit current, and their ratio is R_Th. On a bench this is often the only route, and it is also why a real battery sags under load: a AA cell is a Thevenin source with V_Th ≈ 1.5 V and R_Th of a fraction of an ohm — draw an ampere and you lose tenths of a volt inside the cell. The "ideal source with internal resistance" promised back in CA-1 is precisely a Thevenin equivalent.

The worked example: find the Thevenin equivalent seen at terminals a–b.

WORKED EXAMPLE

Thevenin, start to finish

Find the equivalent at a–b in the figure: 12 V source, R_1 = 4 kΩ, R_2 = 12 kΩ.
V_Th = open-circuit voltage at a–b. With nothing connected, R_1–R_2 is a plain divider: V_Th = 12 × 12/(4+12) = 9 V.
R_Th: kill the 12 V source (short it). Looking into a–b you see R_2 in parallel with R_1: 4 ∥ 12 = 3 kΩ.
Cross-check by the second route: short a–b. That shorts R_2, so i_sc = 12/4 = 3 mA, and v_oc/i_sc = 9/3 = 3 kΩ. ✓ Two independent methods, one answer.

V_Th = 9 V, R_Th = 3 kΩ

WORKED EXAMPLE

Why bother: instant load analysis

Attach a 6 kΩ load to a–b. Without Thevenin: redraw, re-solve the whole network. With Thevenin: one loop.
i_L = V_Th/(R_Th + R_L) = 9/(3+6) = 1 mA; v_L = 6 V.
Now a 1 kΩ load: i_L = 9/4 = 2.25 mA. A 100 kΩ load: v_L = 9 × 100/103 ≈ 8.74 V. Each new load costs one line.
This is why data sheets summarise entire circuits as "output impedance" — they are handing you R_Th.

One equivalent, every load solved by a divider.

PRACTICE

Find the Thevenin equivalent of a 15 V source in series with 6 kΩ, feeding terminals taken across a 3 kΩ resistor to ground.

Using practice problem 1's equivalent, find the load voltage when a 3 kΩ load is attached.

On a bench you measure a mystery network: open-circuit voltage 8 V, short-circuit current 4 mA. Give its Thevenin equivalent, and predict the terminal voltage with a 6 kΩ load.

Explain in a sentence or two why the open-circuit terminal voltage equals V_Th exactly.