The black box that fits in your pocket
In rung 3 you learned to grind through nodal and mesh analysis — write the equations, solve the system, get every voltage and current. That always works, but it's brute force. Often you only care about one part of a circuit: the two terminals where you'll connect a speaker, a sensor, or another stage. Everything behind those two terminals is just "the source" as far as your component cares. Wouldn't it be nice to replace that whole mess with the simplest thing that behaves identically?
That is exactly what Thévenin's theorem promises. It says: *any* linear two-terminal network — any combination of resistors and independent sources, however baroque — is electrically indistinguishable, at those two terminals, from a single ideal voltage source V_th in series with a single resistor R_th. A discovery published by Léon Charles Thévenin in 1883, it is the single most-used shortcut in all of electronics. The dual statement, Norton's theorem, says the same box also equals a single current source I_N in parallel with that same resistor. Two pictures of the same truth.
Messy original Thévenin equivalent Norton equivalent
network (series source) (parallel source)
o---[R1]--+--[R2]--o a R_th o a o a
| | | ---/\/--+ +--+--+
(V1) [R3] (load) | | | |
| | | (V_th) (load) (I_N) R_th (load)
o---------+--------o b | | | |
o-------o b +--+--+--o b
any linear 2-terminal box = V_th in series R_th = I_N in parallel R_thFinding V_th and R_th: two measurements, two thought experiments
The recipe is wonderfully physical, and you can run it either on a real bench or on paper. To pin down the equivalent you need just two numbers. V_th is the open-circuit voltage: disconnect whatever load was on terminals a–b, and measure (or compute) the voltage across the gap. With nothing drawing current, no drop appears across R_th, so the terminal voltage equals the internal source — that's V_th by definition.
R_th is the resistance you'd feel looking back into the terminals — but with a crucial preparation: kill all the independent sources first. Killing a voltage source means replacing it with a wire (a short, because an ideal voltage source forced to deliver 0 V is a 0-Ω connection). Killing a current source means replacing it with a gap (an open, because an ideal current source forced to deliver 0 A passes nothing). Then collapse the leftover resistor network — series and parallel — into one number seen from a–b.
- Open the load. Remove whatever was connected across terminals a–b.
- Find V_th = the open-circuit voltage across a–b, using any rung-3 method (divider, nodal, mesh).
- Zero the independent sources: shorts for voltage sources, opens for current sources.
- Find R_th = the equivalent resistance looking into a–b of that source-free network.
- Get I_N for free: by Ohm's law the Norton current is I_N = V_th / R_th, and R_th is shared by both equivalents.
WORKED EXAMPLE — find the Thévenin equivalent at a-b
o---[ 6 kΩ ]---+-------o a Given: 12 V source, R1 = 6 k�, R2 = 3 kΩ.
| | Load (between a-b) is removed.
(12 V) [ 3 kΩ ]
| |
o-------------+-------o b
STEP 1 V_th = open-circuit voltage across the 3 kΩ
It's just a voltage divider:
V_th = 12 V x 3k / (6k + 3k) = 12 x 1/3 = 4.00 V
STEP 2 Kill the 12 V source -> replace it with a SHORT (a wire).
Now 6 kΩ and 3 kΩ both run from a-b to the same node:
they are in PARALLEL.
R_th = (6k x 3k)/(6k + 3k) = 18k/9k = 2.00 k�
STEP 3 Norton: I_N = V_th / R_th = 4 V / 2 kΩ = 2.00 mA
RESULT Thévenin: 4 V in series with 2 kΩ
Norton: 2 mA in parallel with 2 kΩ (identical box!)Superposition: solve one source at a time
What if your circuit has several batteries and current sources all pushing at once, and you just want the current through one branch? Trying to reason about all of them simultaneously is dizzying. The superposition theorem offers a cleaner mental move, and it works for the same deep reason Thévenin does: linear circuits add. The response (any voltage or current) caused by several independent sources acting together equals the sum of the responses caused by each source acting alone, with all the others turned off.
"Turning off" means exactly the same zeroing you learned for R_th: short the voltage sources, open the current sources. You then solve a sequence of one-source circuits — each one trivially simple — and add the partial answers, keeping track of sign. It trades one hard problem for several easy ones. Think of it like noise-cancelling headphones predicting the room: model each sound source on its own, then sum the waves.
SUPERPOSITION — current through the middle 4 kΩ with TWO sources
o--[2kΩ]--+--[3kΩ]--o Sources: 10 V on the left,
| | | 5 mA into the top node.
(10V) [4kΩ] (5mA)
| | |
o---------+---------o
PART A: keep 10 V, OPEN the 5 mA source
(current source off = gap, so the 3 kΩ branch carries no source)
I_A through 4 kΩ ... solve -> +1.00 mA (downward)
PART B: keep 5 mA, SHORT the 10 V source
(voltage source off = wire)
I_B through 4 kΩ ... solve -> +1.40 mA (downward)
TOTAL: I = I_A + I_B = 1.00 + 1.40 = 2.40 mA downward
(numbers illustrative — note we SUM the partial currents)Maximum power transfer: the impedance handshake
Now that any source collapses to V_th behind R_th, a beautiful design question becomes answerable. You attach a load resistor R_L to the terminals. How big should R_L be to draw the most power from the box? Make R_L tiny and lots of current flows, but almost all the voltage is wasted inside R_th. Make R_L huge and the full voltage appears across it, but barely any current flows. There must be a sweet spot in between — and there is.
The maximum power transfer theorem gives the elegant answer: power delivered to the load is greatest when R_L = R_th — the load matches the source's internal resistance. At that match, exactly half the total power is delivered to the load and half is burned inside R_th, so the efficiency is only 50%. That tells you *when* to use this rule: in signal and communications paths — antennas, audio lines, RF stages — where extracting the maximum *signal* power matters more than wasting some heat. In a power grid you do the opposite, keeping R_L ≫ R_th to stay efficient.
WHY R_L = R_th — the curve has a single peak
P_L(R_L) = V_th^2 * R_L / (R_th + R_L)^2
Set dP_L/dR_L = 0 -> R_L = R_th (the maximum)
P_L ^
(mW) | .--*--. peak at R_L = R_th
| .-' | '-. P_max = V_th^2 / (4 R_th)
| .' | '-.
| .' | '--.___
| .' |
+-------------+------------------> R_L
0 R_th
At the peak: P_max = V_th^2 / (4 R_th), efficiency = 50%Why it matters: shortcuts that scale
These theorems aren't classroom curiosities — they are how real circuits get designed without losing your mind. A multi-stage amplifier is analyzed by Thévenin-izing each stage so the next sees a clean source-plus-resistance. A sensor's datasheet quotes its output impedance — that's just R_th of the chip's internals, telling you how much it will sag when you load it. Battery engineers model a cell as an ideal EMF in series with an internal resistance: pure Thévenin, and the reason your phone dims under heavy load is the voltage drop across that R_th.
Notice the chain of dependencies you've now built: nodal and mesh analysis from rung 3 are the *engines* that extract V_th and R_th; the voltage divider is just the simplest Thévenin you already knew; and Ohm's law ties V_th, R_th and I_N together. Master these four moves — open-circuit voltage, source-free resistance, superposition, conjugate match — and most of analog electronics becomes a matter of swapping in the right equivalent and turning a small crank.