Circuit Theorems

A circuit is linear if it contains only linear elements (resistors, ideal sources, linear dependent sources, ideal capacitors and inductors). Linear means two things:

• Homogeneity: if you scale every independent source by a factor $k$, every voltage and current in the circuit scales by the same $k$.
• Additivity: the response to several sources acting together equals the sum of the responses to each source acting alone.

The superposition theorem is the additivity property made into a calculation method. For a linear circuit with multiple independent sources, the voltage across (or current through) any element is the algebraic sum of contributions from each independent source considered separately, with all other independent sources deactivated.

Deactivating a source:

• Independent voltage source → replace with a short circuit (0 V).
• Independent current source → replace with an open circuit (0 A).
• Dependent sources are *never* deactivated — they stay as-is throughout.

When to use it: when several independent sources of different types or frequencies drive a linear network, especially when one source is much easier to analyse alone.

Any practical voltage source (an ideal source $V_s$ in series with $R_s$) is equivalent — at its external terminals — to a practical current source (ideal source $I_s$ in parallel with the same $R_s$). The transformation is:

$$I_s=\frac{V_s}{R_s},R_s\text{ unchanged}$$

and the reverse:

$$V_s=I_s\cdot R_s$$

Polarity matters: the current source's arrow points toward the + terminal of the voltage source it replaces.

Why it's useful: by repeatedly transforming sources, you can often collapse a complex network of resistors and sources into a single source plus a single resistance — the same endpoint as Thévenin / Norton, but reached step by step.

Limitations:

• Doesn't apply to an ideal source (no internal resistance). You need a finite $R_s$.
• The transformed sources are equivalent only at the external terminals — internal currents and powers are different. So the power dissipated in $R_s$ is different in the two forms, even though everything outside is identical.

Thévenin's theorem states that any two-terminal linear network — no matter how many sources and resistors — can be replaced at those two terminals by:

• A single voltage source $V_{Th}$ (the Thévenin voltage)
• in series with a single resistance $R_{Th}$ (the Thévenin resistance).

Finding $V_{Th}$: remove the load and measure (or compute) the open-circuit voltage at the terminals: $V_{Th}=V_{oc}$.

Finding $R_{Th}$ (three methods, pick whichever fits):

1. No dependent sources — deactivate all independent sources (voltage → short, current → open) and compute the equivalent resistance at the terminals using series/parallel reduction.
2. Has dependent sources — apply a test source $V_T$ at the terminals (with all independent sources still deactivated), compute the resulting test current $I_T$, then $R_{Th}=V_T/I_T$.
3. Open-short method — $R_{Th}=V_{oc}/I_{sc}$ where $I_{sc}$ is the short-circuit current at the terminals. Works whether or not dependent sources are present.

Why it's useful: once you have $(V_{Th},R_{Th})$, you can hook up any load $R_L$ and instantly find load current and voltage:

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

No need to re-solve the whole network for every new load value.

Norton's theorem is the dual of Thévenin. Any two-terminal linear network is equivalent to:

• A single current source $I_N$ (the Norton current)
• in parallel with a single resistance $R_N$.

Finding $I_N$: short the terminals and measure the short-circuit current: $I_N=I_{sc}$.

Finding $R_N$: same as $R_{Th}$ — by Thévenin / Norton equivalence, $R_N=R_{Th}$ always.

Conversion between Thévenin and Norton (it's just a source transformation):

$$V_{Th}=I_N\cdot R_N,I_N=\frac{V_{Th}}{R_{Th}},R_{Th}=R_N$$

Use Thévenin when the rest of the circuit attached to the terminals is dominated by voltage-source thinking (series loops). Use Norton when current-source thinking (parallel branches) is more natural. They contain identical information; pick whichever simplifies the next step.

Given a fixed Thévenin source $(V_{Th},R_{Th})$ driving a variable load $R_L$, how do you choose $R_L$ to extract the maximum power?

Power delivered to $R_L$:

$$P_L=i^2{R}_L={\left(\frac{V_{Th}}{R_{Th}+R_L}\right)}^2{R}_L$$

Differentiate with respect to $R_L$, set $d{P}_L/d{R}_L=0$, and you get the Maximum Power Transfer Theorem:

$$\boxed{R_L=R_{Th}}$$

At this load:

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

Efficiency at maximum power transfer is only 50% — half the source power goes into $R_L$, the other half is dissipated in $R_{Th}$. This is fine for signal-transfer applications (audio, RF), where what matters is the signal *amplitude* delivered. It's not what you want for power-delivery applications (electric grid), where efficiency matters more than raw power — there you want $R_L\gg R_{Th}$, giving low loss in the source but also low absolute power.

Procedure to apply it:

1. Find $V_{Th}$ and $R_{Th}$ at the load's terminals.
2. Set $R_L=R_{Th}$.
3. $P_{\text{max}}=V_{Th}^2/(4R_{Th})$.

Source modelling. Real batteries, generators, and amplifier outputs are not ideal sources. A 9 V battery doesn't deliver 9 V at every load — drop a serious current and the terminal voltage sags. Modelling a real source as a Thévenin (or Norton) equivalent lets you predict that sag:

$$V_{\text{terminal}}=V_{Th}-I\cdot R_{Th}$$

A typical AA battery: $V_{Th}\approx 1.5$ V, $R_{Th}\approx 0.15$ Ω. At 10 A short-circuit current... actually it can't deliver 10 A; the terminal voltage would collapse first.

Resistance measurement by Wheatstone bridge. Four resistors in a diamond — when the bridge is balanced, the galvanometer (between the two midpoints) reads zero current, and $R_1/R_2=R_3/R_4$. Unknown resistance can be solved if the other three are known. The circuit is a classic textbook example of Thévenin analysis: to find the galvanometer current when the bridge is *not* balanced, find the Thévenin equivalent at the two midpoints and connect the galvanometer as a load.

Multimeter ohmmeter mode: passes a small known current through the unknown resistor, measures the voltage across it, computes $R=V/I$. Internally a current-source + voltmeter implementation of Ohm's law.