[[The Bare Essentials of Electrical Engineering.pdf#page=127&offset=99,702|The Bare Essentials of Electrical Engineering, Powerful Circuit Theorems]] # Introduction - Thevenin & Norton equivalent circuits: finding the Thevenin or Norton Equivalent circuits for a given network is a great way to analyze a circuit when we are only interested in its in/output ports. Circuits are usually very complicated and we often only interact w/ one or two of its ports, in/output. Thevenin and Norton theorems provide us with a way to find a simple representation of an electrical network w/ respect to the desired ports - Procedure is not reversible, meaning if you have the Thevenin or Norton circuit, you cannot recover the original circuit # Linearity - Linearity: mathematical principle implies additivity and homogeneity - saves us from resolving a circuit when we have its input(s) - Linearity has 2 important properties - Homogeneity or scalability: $f(\alpha x)=\alpha f(x)$ for any scalar $\alpha$ - Additivity: $f(x+y)=f(x)+f(y)$ - See [[The Bare Essentials of Electrical Engineering.pdf#page=128|Example 4.2.1]] - Reminder that voltage division is the distribution of a source voltage among components that carry the same current (i.e. elements in series). Defined by $V_{k}=V_{s} \frac{R_{k}}{\sum R}$ where $V_{k}$ is voltage across each branch of parallel network, $\sum R$ is sum of resistors in series, $V_{s}$ is source voltage. Voltage division is only valid if resistors carry the same current - Above example demonstrates that if an input to a linear system is scaled by a factor $\alpha$, output will be scaled by the same factor # Superposition - Superposition: allows us to examine the impact of each input by itself. Adding the resulting voltage and current (but not power) derived by considering each input independently allows us to find the total result due to all inputs - Useful in circuits w/ multiple independent inputs - allows us to assess the effects of each input separately then add the together to find the total output - Provides the advantage of understanding the impact of each source on the output separately - See [[The Bare Essentials of Electrical Engineering.pdf#page=129|Example 4.3.1]] - Reminder: current division applies to resistors in parallel and uses formula $I_{1}=I_{T} \frac{R_{2}}{R_{1}+R_{2}}$ or in conductance form: $I_{k}=I_{t} \frac{G_{k}}{\sum G}$ - Essential idea is to take out every independent source except the one you're currently focusing on, find what you're looking for in each source, then add them all together - For concrete process, see: 1. Turn off all ind. sources except one - don't modify dependent sources 2. Find desired variable by solving the circuit for it 3. Repeat process for all ind. sources (voltage or current) 4. Final result will be addition of all partial results - While you can add partial voltages and currents, you cannot do so with power because it is a nonlinear quality -> can be found by multiplying total voltage with total current $P_{tot}=V_{tot}I_{tot}=(V_{partial,1}+V_{partial,2})\cdot(I_{partial,1}+I_{partial,2})\neq V_{partial,1}\cdot I_{partial,1}+V_{partial,2}\cdot I_{partial,2}$ # Source transformation - Source transformation can be used to change an originally given circuit to an equivalent, simpler one. Particularly useful when trying to find the Thevenin or Norton equivalent circuits - Allow us to manipulate source-resistor pairings to create additional possibilities for creating parallel and series combinations of sources and resistors - See [[The Bare Essentials of Electrical Engineering.pdf#page=133|Figure 4.4]] analysis - Combination of $V_{s}$ with internal resistor $R$ used to differentiate it from an ideal voltage source as a realistic voltage source - Original voltage source $V_{s}$ with series resistor $R$ is equivalent to current source $I_{s}=\frac{V_{s}}{R}$ in parallel with the same resistor because $I=\frac{V_{s}-V_{ab}}{R}=\frac{V_{s}}{R}-\frac{V_{ab}}{R}$, exactly like a KCL between two parallel elements - Also works with a current source in parallel with a resistor, useful model for realistic current sources - Key idea: voltage source $V$ in series with $R\Longleftrightarrow$ current source $I$ in parallel with $R$, and the reverse $V=ir$ - See [[The Bare Essentials of Electrical Engineering.pdf#page=134|Example 4.5.1]] 1. $30V+30\Omega\to1A\parallel30\Omega$ 2. $30\Omega \parallel60\Omega, R_{eq}=20\Omega \parallel1A$ 3. $1A\parallel20\Omega\to20V+20\Omega$ 4. $20\Omega+20\Omega,R_{eq}=40\Omega+20V$ 5. $40\Omega+20V\to0.5A\parallel 40\Omega$ 6. Combine current sources, $-0.5A\parallel40\Omega$ 7. $-0.5A\parallel40\Omega\to-20V+40\Omega$ 8. Solve circuit for $V_{x}$ via voltage division: $V_{net}=40-20=20V,I=\frac{V_{net}}{\sum R}=0.25A\to V_{x}=IR=.25(40)=10V$ - Use of source transform is essentially cycling between transforming voltage sources in series with current sources in parallel # Thevenin and Norton Theorems - [[The Bare Essentials of Electrical Engineering.pdf#page=135|Understanding Thevenin and Norton by the Lab Approach]] - Thevenin and Norton Theorem: any linear circuit, no matter how complicated inside, behaves at its terminal exactly like *one* voltage source and *one* resistor **or equivalently** it behaves like *one* current source and *one* resistor - Specifically, a voltage source in series w/ resistor or current source in parallel with resistor - See [[The Bare Essentials of Electrical Engineering.pdf#page=140|Figure 4.19]]: consider $R_{L}$ varies from $1\Omega$ to $100\Omega$ with $1\Omega$ increments and we need to measure $I_{L}$ every time the resistor is changed. Nodal analysis would be a low-cost approach but would involve solving a 3x3 matrix a hundred times - Instead, find its Thevenin/Norton equivalent circuit (see [[The Bare Essentials of Electrical Engineering.pdf#page=141|Figure 4.20]]) at the ports attaching to $R_{L}$ -> would require us to find $R_{eq}$ seen at port a-b and either open-circuit voltage $V_{oc}=V_{Th}$ (Thevenin equivalent) or short-circuit current $I_{c}=I_{N}$ (Norton equivalent) - First find $R_{eq}$ by turning off all independent sources - Draw open-circuit voltage & short-circuit current schematics focused around the Norton and Thevenin-equivalent circuits and determine whether to find $V_{Th}$ or $I_{N}$ based on the number of unknowns (see [[The Bare Essentials of Electrical Engineering.pdf#page=134|Figure 2.22]]) - To find the Thevenin or Norton equivalent circuits for any linear circuit we need to find 2 of the 3 following qtys. - Equivalent resistance at the desired port - Turn off all indepdent sources and calculate resulting resistance at desired port. May have to apply the i-v test if resistors cannot be combined through series and parallel connections, or if the circuit includes dependent sources. Also $R_{eq}=R_{Th}=R_{N}$ - Open-circuit voltage $V_{Th}$ across the desired port - Leave desired port open-circuited and find the voltage across it - Short-circuit current $I_{N}$ through the desired port - Short circuit the desired port and find the current through it - See [[The Bare Essentials of Electrical Engineering.pdf#page=145|Example 4.6.1]], [[The Bare Essentials of Electrical Engineering.pdf#page=140|Example 4.6.3]] for equivalent resistance & short-circuit methods -