5. Network Theorems: Powerful Shortcuts

For analyzing circuits by hand, mesh and nodal methods are the brute force tools. But four classic theorems give you serious shortcuts.

5.1 Thevenin's theorem

Any linear two-terminal network can be replaced by an equivalent voltage source $V_{Th}$ in series with a resistance $R_{Th}$, viewed from those two terminals.

   complicated network ─── A
                          ●     ⇔   ───[V_Th]──[R_Th]── A
                          ●                              ●
   complicated network ─── B                              B

The two terminals A and B see the same V-I behavior either way. Replace the whole tangled mess with a simple battery + resistor.

Why does Thevenin work?

The intuition is worth dwelling on, because it is one of the most useful results in all of electronics. A linear two-terminal network's behavior is entirely captured by the relationship between the voltage across its terminals and the current flowing out of them. Apply some external test current $I$ and measure the resulting voltage $V$, or apply some test voltage $V$ and measure the current $I$. Because the network is linear (resistors, capacitors, inductors, and sources whose values do not depend on $V$ or $I$), the relationship is a straight line in $V$-$I$ space:

$V = a - b I$

for some constants $a$ and $b$. There are only two numbers describing the entire external behavior of the network, no matter how complicated its insides are. We just need to identify what those two numbers mean. Set $I = 0$ (open circuit): $V = a$. So $a$ is the open-circuit voltage. Set $V = 0$ (short circuit): $I = a/b$. So $b$ has units of resistance and equals $V_{oc}/I_{sc}$.

A simple voltage-source-plus-resistor circuit produces exactly this $V$-$I$ relationship: open-circuit voltage equals the source value; short-circuit current equals source divided by resistor. So any linear network is indistinguishable (from the outside) from a voltage source $V_{Th} = a$ in series with a resistance $R_{Th} = b$. That is Thevenin's theorem.

The key word is "linear." If the network contains a diode or any nonlinear element, the $V$-$I$ curve is not a straight line and Thevenin does not apply. Linearity is the assumption that makes this magic possible.

Procedure to find $V_{Th}$ and $R_{Th}$

1. Open-circuit voltage $V_{Th}$: with terminals A and B open (nothing connected externally), measure (or compute) the voltage between them.
2. Thevenin resistance $R_{Th}$: deactivate all independent sources (replace voltage sources with short circuits, current sources with open circuits) and compute the equivalent resistance looking back into the terminals.

Alternative for $R_{Th}$ when there are dependent sources: short-circuit the terminals, measure (or compute) the short-circuit current $I_{sc}$. Then $R_{Th} = V_{Th}/I_{sc}$.

Yet another method, especially handy when dependent sources prevent the "deactivate all sources" technique from giving a clean answer (because the dependent sources do not get deactivated): apply a test source. With all independent sources deactivated, drive the terminals with a test voltage $V_{test}$ and measure the resulting current $I_{test}$ (or vice versa). Then $R_{Th} = V_{test}/I_{test}$. The test-source method is the universal fallback.

Process diagram

Worked example with independent sources only

Source $V_s = 12$ V in series with $R_1 = 1$ kΩ, then a parallel combination of $R_2 = 3$ kΩ to ground and the output terminals. Find Thevenin equivalent at the output.

• $V_{Th}$ = voltage at output with nothing connected = $V_s \cdot R_2/(R_1+R_2) = 12 \cdot 3/4 = 9$ V.
• $R_{Th}$: deactivate source ($V_s$ becomes a short), look back. From output terminals, you see $R_2$ in parallel with $R_1$ (because $R_1$ now connects to GND through the shorted source). $R_{Th} = R_1 R_2/(R_1+R_2) = 750$ Ω.

So that whole network looks, from the output, like a 9 V source in series with 750 Ω. If you connect a 750 Ω load, current flows = 9/(750+750) = 6 mA, output voltage = 4.5 V. No need to redo the whole network analysis.

Worked example with a dependent source

Consider a 12 V source feeding a 3 kΩ resistor to node A, with a current-controlled current source pulling $2 I_x$ from node A to ground, where $I_x$ is the current flowing toward node A through the 3 kΩ. Output terminals are node A and ground.

For $V_{Th}$: open terminals, no external current. Let $I_x = (12 - V_A)/3000$ be the current toward node A. KCL at A says $I_x = 2 I_x$, which forces $I_x = 0$, so $V_A = 12$ V. Hence $V_{Th} = 12$ V.

For $R_{Th}$: deactivate the 12 V source (short it), keep the dependent source active, apply test voltage $V_{test}$ at the terminals. With the source shorted, the 3 kΩ goes node-A-to-ground; current out of A through it is $V_{test}/3000$, so $I_x = -V_{test}/3000$ (defined positive toward A). The dependent source pulls $2I_x = -2V_{test}/3000$ from A, i.e., pushes $2V_{test}/3000$ into A.

KCL at A: $I_{test} + 2V_{test}/3000 = V_{test}/3000$, giving $I_{test} = -V_{test}/3000$, so $R_{Th} = V_{test}/I_{test} = -3000$ Ω. A negative Thevenin resistance, which is strange but possible with dependent sources. Negative resistance means the network delivers energy in proportion to applied voltage, which is exactly what amplifiers do. The dependent source models the active amplification. This is the kind of insight Thevenin analysis is uniquely good at extracting.

Why Thevenin matters. Imagine you have a complex bias network feeding a single transistor's base, and you want to know how the network behaves when the transistor draws current. Find the Thevenin equivalent of everything up to the transistor's base node, one number for $V_{Th}$, one for $R_{Th}$. Now your "circuit" is just $V_{Th}$, $R_{Th}$, and the transistor. Trivially analyzable. Without Thevenin, you would write KCL at every node every time you wanted to know one thing.

Hardware-security angle: power-trace analysis

When an attacker probes a chip's power supply pin to extract a cryptographic key, what they measure is the chip's instantaneous current draw, modulated by the impedance of the package, the bondwire, the PCB traces, and the surrounding decoupling network. From the attacker's perspective, the chip-plus-supply-network is a Thevenin source: the chip's transient current draw is the "open-circuit signal" they care about, and the network's impedance is the "Thevenin resistance" that distorts what they measure. Defenders can place stiff decoupling caps very close to the chip, dropping the network's effective impedance and making the source-side signal harder to recover. The duel between attacker and defender is, in a real sense, a Thevenin source-impedance game.

5.2 Norton's theorem

The dual of Thevenin: any linear two-terminal network can be replaced by a current source $I_N$ in parallel with a resistance $R_N$.

The relations: $I_N = I_{sc}$ (the short-circuit current), and $R_N = R_{Th}$ (same value as Thevenin resistance). Also $V_{Th} = I_N R_N$.

Use whichever form is more convenient for the problem at hand. Sometimes Thevenin is cleaner, sometimes Norton. Norton is especially natural for "what current can this network deliver?" types of questions, while Thevenin is natural for "what voltage will appear across this load?"

5.3 Superposition

In a linear circuit with multiple sources, the response (current or voltage) is the sum of the responses to each source acting alone, with the others zeroed (voltage sources shorted, current sources opened).

Wave-superposition analogy. Drop two stones in a still pond; where the ripples meet, the surface is the sum of what each set would have done alone.

Caveat: superposition does not apply to power. Power is quadratic in $V$ or $I$, hence not linear. Compute total $V$ or $I$ first, then square.

5.4 Maximum power transfer

A specific and useful result: a load resistance $R_L$ extracts the maximum power from a source with Thevenin resistance $R_{Th}$ when $R_L = R_{Th}$.

Derivation

Treat the source as its Thevenin equivalent: $V_{Th}$ in series with $R_{Th}$, driving a load $R_L$. The current through the load is

$I_L = \frac{V_{Th}}{R_{Th} + R_L}$

The power delivered to $R_L$:

$P_L = I_L^2 R_L = \frac{V_{Th}^2 R_L}{(R_{Th} + R_L)^2}$

This is a function of $R_L$ for fixed source. Differentiate with respect to $R_L$ and set the derivative to zero. Using the quotient rule:

$\frac{dP_L}{dR_L} = \frac{V_{Th}^2 (R_{Th} + R_L)^2 - V_{Th}^2 R_L \cdot 2(R_{Th} + R_L)}{(R_{Th} + R_L)^4}$

$= \frac{V_{Th}^2 \left[ (R_{Th} + R_L) - 2 R_L \right]}{(R_{Th} + R_L)^3} = \frac{V_{Th}^2 (R_{Th} - R_L)}{(R_{Th} + R_L)^3}$

Set equal to zero: $R_{Th} - R_L = 0$, so $R_L = R_{Th}$. The load matches the source resistance, and that is the optimum.

Plug $R_L = R_{Th}$ back in to find the maximum power:

$P_{max} = \frac{V_{Th}^2 R_{Th}}{(2 R_{Th})^2} = \frac{V_{Th}^2}{4 R_{Th}}$

Why efficiency is exactly 50% at maximum power transfer

At $R_L = R_{Th}$, the load and the source resistance carry the same current and have the same value. So they dissipate the same power. The total power delivered by the source is split exactly in half: half in $R_{Th}$ (wasted as heat inside the source), half in $R_L$ (the "useful" load). Efficiency = useful / total = 50%.

This is fine for very small loads where you want to extract every bit of signal energy and don't care about dissipation in the source. A microphone preamp wants to extract every bit of signal from the microphone, and the microphone's small Thevenin resistance is matched (or near-matched) by the preamp's input resistance.

But 50% efficiency is useless for power transmission. You would not run a power grid at 50% efficiency. The power grid wants $R_L \gg R_{Th}$ so almost all the energy goes to the load and very little wastes in the wires. For a generator with 0.1 Ω source resistance feeding 1 Ω of load through the grid, the efficiency is $1/(0.1 + 1) = 91$%, much better than 50.

Different domains have different priorities

So in different contexts:

• Power systems want $R_L \gg R_{Th}$. Most of the energy goes to the load. Efficiency matters more than absolute power transferred.
• Audio amplifier output stages want $R_L \gg R_{Th}$. Voltage gain matters; you want the speaker to "see" the amplifier as a stiff voltage source. Speakers are 4-8 Ω; amplifiers have much lower output impedance, often <0.1 Ω. The speaker is the load that gets nearly all the power.
• RF amplifiers and antennas want $R_L = R_{Th}$. At high frequencies, any mismatch causes reflections back along the cable, which can ruin a signal or fry an output stage. Impedance matching is mandatory. This is why antennas, transmission lines, and RF amplifiers are all 50 Ω (or 75 Ω for video).
• Microphone preamps want $R_L = R_{Th}$ for maximum signal transfer. A condenser mic with 200 Ω source resistance pairs with a 200 Ω input impedance for max signal power. Efficiency is irrelevant; the absolute signal level is what matters.

Audio and RF differ not because they violate Maximum Power Transfer but because they are optimizing different things. Audio amplifiers optimize voltage across the speaker (because speakers convert voltage to motion); RF amplifiers optimize power delivered to the load (because they care about radiation efficiency at the antenna).

5.5 Other theorems worth knowing

• Reciprocity: in a linear network, swapping source and meter gives the same reading. Useful in reverse-engineering.
• Millman's theorem: the voltage at a common node from many series-resistor sources is $V = \sum (V_i/R_i) / \sum (1/R_i)$. A specialized nodal result.
• Tellegen's theorem: sum of $v_b i_b$ over all branches is zero. Energy conservation in disguise.
• Substitution: known branch voltage (or current) lets you replace the branch with a source of that value; the rest of the circuit is unchanged.
• Compensation: a small change $\Delta R$ in a resistor is equivalent to inserting a voltage source $\Delta V = I \Delta R$ in series with the original. Used for sensitivity analysis.