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1.4 — Thevenin, Norton, Superposition and Maximum Power
You have a circuit board with forty components on it, and two terminals coming out. You want to know what happens when you plug different loads into those two terminals. Do you solve the forty-component circuit again for every load?
No. Whatever is inside that box, however complicated, behaves at those two terminals exactly like one voltage source in series with one resistor. Always, provided the box contains only linear elements. Measure two numbers once and you are done forever.
That is Thevenin's theorem, and it is the single most useful result in circuit theory.
This chapter rests on one property of the circuits in Chapter 1.3: they are linear, meaning doubling every source doubles every response. That is what makes it legal to consider sources one at a time and add the results, and every technique below is a consequence of it.
1. Superposition, which everything else rests on
Start with the property that makes the rest possible. A linear circuit is one whose elements obey relations where doubling the cause doubles the effect: v = iR qualifies, i = C\,dv/dt qualifies, and v = Li' qualifies. A diode's exponential curve does not, which is why none of this chapter applies to Part 2's devices without extra care.
The superposition theorem: in a linear circuit with several independent sources, the current or voltage anywhere is the sum of the contributions each source would produce acting alone, with all other sources switched off.
"Switched off" needs a precise meaning:
- A voltage source is turned off by replacing it with a short circuit (a plain wire). A source set to 0 V is a thing with zero volts across it no matter the current, which is exactly what a wire is.
- A current source is turned off by replacing it with an open circuit (a break). A source set to 0 A is a thing that carries no current no matter the voltage, which is exactly what a gap is.
Getting these backwards is the standard mistake. Say them out loud once: zero volts is a wire, zero amps is a gap.
Worked example
A node is fed by a 12 V source through 4\ \Omega and by a 6 V source through 6\ \Omega, with a 12\ \Omega resistor from the node to ground. Find the node voltage by superposition.
Contribution of the 12 V source alone. Kill the 6 V source by shorting it, which puts its 6\ \Omega from the node straight to ground. So the node sees 6\ \Omega \parallel 12\ \Omega to ground, fed through 4\ \Omega:
6\parallel12 = \frac{6\times12}{18} = 4\ \Omega
Voltage divider:
V' = 12\cdot\frac{4}{4+4} = 6\ \text{V}
Contribution of the 6 V source alone. Kill the 12 V source by shorting it, putting its 4\ \Omega from the node to ground. The node now sees 4\parallel12 = 3\ \Omega to ground, fed through 6\ \Omega:
V'' = 6\cdot\frac{3}{6+3} = 2\ \text{V}
Add them.
V = 6 + 2 = 8\ \text{V}
Check directly by nodal analysis:
\frac{V-12}{4} + \frac{V-6}{6} + \frac{V}{12} = 0
Multiply by 12: 3V - 36 + 2V - 12 + V = 0 \Rightarrow 6V = 48 \Rightarrow V = 8 ✓.
Where superposition does not help
Two warnings.
Power does not superpose. Power depends on the square of current or voltage, and squares do not add. If source A alone would put 4 W into a resistor and source B alone would put 9 W, the two together do not put 13 W in. You must find the total current first, then square it. This trips up almost everyone once.
It is slower than nodal analysis for most problems. With n sources you solve the circuit n times. Superposition earns its place when you want to understand a contribution — for example, how much of an amplifier's output is signal and how much is the DC bias — rather than when you want a single number fast. Chapter 2.3's small-signal analysis is exactly this: split the transistor circuit into a DC problem and an AC problem, solve them separately, and add.
2. Thevenin's theorem
Any network of linear elements and sources, seen from two terminals, is equivalent to a single voltage source V_{Th} in series with a single resistance R_{Th}.
Equivalent means something precise: no measurement you can make at those two terminals can tell the difference. Connect any load whatever, and the voltage and current will match what the original network would have produced.
Finding the two numbers
V_{Th} is the open-circuit voltage. Disconnect the load entirely and measure (or calculate) the voltage across the terminals. That is V_{Th}, because with no load there is no current, so no voltage is lost across R_{Th}, so the terminal voltage equals the internal source.
R_{Th} is the resistance looking back into the terminals with all independent sources killed — voltage sources shorted, current sources opened, using the same rules as superposition. Compute it by series-parallel reduction from the terminals inward.
There is a third way that works even when the network contains dependent sources, where you cannot simply kill things:
R_{Th} = \frac{V_{oc}}{I_{sc}}
Open-circuit voltage divided by short-circuit current. Short the terminals, find the current that flows, and divide. This always works, and it is how you would measure R_{Th} on a real board if it were safe to short it.
Worked example — a loaded divider
A 24 V source feeds R_1 = 8\ \Omega down to a node; from that node R_2 = 4\ \Omega goes to ground; the load is connected across R_2. Find the Thevenin equivalent, then find the current for loads of 1, 4 and 100 Ω.
Step 1 — open-circuit voltage. With the load removed, this is a plain divider:
V_{Th} = 24\cdot\frac{4}{8+4} = 8\ \text{V}
Step 2 — Thevenin resistance. Kill the 24 V source by shorting it. Now R_1 runs from the node to ground (through the short) and R_2 runs from the node to ground. They are in parallel:
R_{Th} = 8\parallel4 = \frac{32}{12} = 2.667\ \Omega
Step 3 — the equivalent is 8 V in series with 2.667 Ω, and now every load is a one-line calculation:
| Load R_L | I = \dfrac{8}{2.667+R_L} | V_L = IR_L |
|---|---|---|
| 1 Ω | 2.18 A | 2.18 V |
| 4 Ω | 1.20 A | 4.80 V |
| 100 Ω | 0.078 A | 7.79 V |
Check the 4 Ω case against the original circuit. The load 4 Ω is in parallel with R_2's 4 Ω, giving 2 Ω, in series with 8 Ω, so total current from the source is 24/10 = 2.4 A, and the voltage at the node is 2.4\times2 = 4.8 V ✓. Matches.
That check is worth doing once with your own hand, because it is the moment Thevenin stops being a formula and starts being obviously true.
3. Norton's theorem
The same statement, told the other way round. Any linear network at two terminals is equivalent to a current source I_N in parallel with a resistance R_N.
- I_N is the short-circuit current — connect the terminals with a wire and measure what flows.
- R_N = R_{Th}, exactly the same number, found the same way.
The two forms convert into each other by Ohm's law:
V_{Th} = I_N R_N, \qquad I_N = \frac{V_{Th}}{R_{Th}}, \qquad R_N = R_{Th}
In the worked example above, I_N = 8/2.667 = 3 A. Check directly: shorting the terminals puts R_2 out of action (it is shorted out) and the current from the source is 24/8 = 3 A ✓.
Which form to use is only ever a matter of convenience. Thevenin suits circuits you are analysing with KVL and voltage dividers; Norton suits KCL, nodal analysis and parallel combinations. Transistor models in Part 2 are usually drawn Norton-style because a transistor genuinely behaves like a current source.
Edward Lawry Norton worked at Bell Labs and published this in an internal technical memorandum in 1926, never in a journal; Hans Ferdinand Mayer published the same result independently in Germany the same year. Thevenin's own paper was 1883, and even that was a rediscovery — Hermann von Helmholtz had the result in 1853, buried in a paper about muscle physiology.
4. Source transformation, the everyday shortcut
Because the two forms are equivalent, you can swap between them mid-circuit to make things collapse. A voltage source V in series with R becomes a current source V/R in parallel with R, and back again.
This is genuinely useful: converting all voltage sources to current sources often turns a chain of series elements into a bundle of parallel ones that you can combine in one step. Worked pattern:
A 10 V source in series with 5\ \Omega, feeding a node that also has a 2 A source injected and a 20\ \Omega to ground.
Transform the voltage source: it becomes a 2 A current source in parallel with 5\ \Omega.
Now the node has two current sources injecting 2 + 2 = 4 A total, into 5\parallel20 = 4\ \Omega:
V_{node} = 4\times4 = 16\ \text{V}
Three lines, no simultaneous equations.
The one thing you may not do: the resistor being transformed must genuinely be in series with the voltage source, and after transformation it is in parallel — so the voltage across it changes, and any quantity you wanted inside the transformed pair is lost. Transform only the parts of the circuit you do not need internal details of.
5. Maximum power transfer
Now the question that Thevenin was built to answer. You have a source with internal resistance R_{Th}. What load resistance draws the most power from it?
The load current is
I = \frac{V_{Th}}{R_{Th}+R_L}
and the load power is
P_L = I^2R_L = \frac{V_{Th}^2\,R_L}{(R_{Th}+R_L)^2}
Think about the two extremes before doing calculus. With R_L = 0 (a short), the current is maximum but the voltage across the load is zero, so the power is zero. With R_L = \infty (open), the voltage is maximum but no current flows, so the power is again zero. Zero at both ends means there is a maximum somewhere in between, and the only question is where.
Differentiate with respect to R_L and set to zero. Using the quotient rule:
\frac{dP_L}{dR_L} = V_{Th}^2\cdot\frac{(R_{Th}+R_L)^2 - R_L\cdot2(R_{Th}+R_L)}{(R_{Th}+R_L)^4}
Cancel one factor of (R_{Th}+R_L) from top and bottom:
= V_{Th}^2\cdot\frac{(R_{Th}+R_L) - 2R_L}{(R_{Th}+R_L)^3} = V_{Th}^2\cdot\frac{R_{Th}-R_L}{(R_{Th}+R_L)^3}
This is zero when the numerator is zero:
\boxed{R_L = R_{Th}}
Maximum power is delivered to the load when the load resistance equals the source resistance. This is called impedance matching, and the AC version of it in Chapter 7.7 is why coaxial cable is sold as "50 ohm" or "75 ohm".
The power at that point is
P_{max} = \frac{V_{Th}^2 R_{Th}}{(2R_{Th})^2} = \frac{V_{Th}^2}{4R_{Th}}
The efficiency catch nobody mentions in the exam
At the matched point, the source's internal resistance is dissipating exactly as much power as the load, because equal resistances carrying equal current dissipate equally. So the efficiency is
\eta = \frac{P_{load}}{P_{total}} = \frac{R_L}{R_{Th}+R_L} = 50\%
Half the energy is wasted inside the source. That is fine for a radio receiver, where the signal is measured in picowatts and you want every scrap of it. It is a catastrophe for a power station, where half of everything generated would heat the generator.
So the design rule splits by domain:
| Domain | Goal | Rule |
|---|---|---|
| Signal, antenna, audio input | Maximum power | Match: R_L = R_{Th} |
| Power delivery, mains, batteries | Maximum efficiency | Make R_{Th} tiny: R_L \gg R_{Th} |
A mains supply has an internal resistance measured in milliohms and drives loads measured in ohms, so it operates deliberately and enormously mismatched, at efficiency near 100%. Matching is for signals; mismatching is for power. That sentence resolves the apparent contradiction between this chapter and every power engineering chapter that follows.
6. Reciprocity, briefly
One more theorem, because it explains a surprising experimental fact. In a linear network with a single source, swapping the source's position with the position where you measure the response leaves the response unchanged.
Put a 1 V source at terminal pair A and measure 3 mA at pair B; move the source to B and you will measure exactly 3 mA at A. This is why an antenna has the same radiation pattern transmitting as receiving (Chapter 7.6), which is not obvious and saves an enormous amount of measurement work.
Reciprocity fails the moment a dependent source is present, which is why an amplifier is emphatically not reciprocal — you cannot drive a loudspeaker output and get a microphone signal out of the input.
Every circuit so far has settled to its final answer instantly. Real circuits do not: connect a battery and something rises, taking microseconds or seconds to get there. That behaviour comes from two components that store energy instead of burning it, and it is Chapter 1.5.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Thevenin, Norton and maximum power
V_{Th} = V_{oc} \qquad R_{Th} = \frac{V_{oc}}{I_{sc}} \qquad I_N = I_{sc} \qquad R_N = R_{Th}
Kill sources to find R_{Th} directly: voltage sources become short circuits (zero volts across them at any current is what a wire does), current sources become open circuits (zero amps at any voltage is what a gap does). This shortcut fails if dependent sources are present; then use V_{oc}/I_{sc}.
V_{Th} = I_NR_N \qquad I_N = \frac{V_{Th}}{R_{Th}}
Maximum power transfer
P_L = \frac{V_{Th}^2R_L}{(R_{Th}+R_L)^2}
Differentiate with respect to R_L:
\frac{dP_L}{dR_L} = V_{Th}^2\frac{(R_{Th}+R_L)^2 - 2R_L(R_{Th}+R_L)}{(R_{Th}+R_L)^4} = V_{Th}^2\frac{R_{Th}-R_L}{(R_{Th}+R_L)^3}
Zero when R_L = R_{Th}, giving
P_{max} = \frac{V_{Th}^2}{4R_{Th}}, \qquad \eta = 50\%
Half the power is lost inside the source at the matched point, which is why matching is for signals and never for power delivery.
What the next chapter fixes
Every component so far has obeyed Ohm's law: put a voltage across it and a current appears instantly, in proportion. Capacitors and inductors do not. They respond to how fast things are changing, which turns algebra into differential equations and makes circuits able to remember, delay and oscillate. Chapter 1.5 introduces both, derives their transient behaviour, and explains where the time constant comes from.