From 7b99e1ccf5768105608fd60c7919b93c52e82216 Mon Sep 17 00:00:00 2001 From: eggy Date: Tue, 7 Feb 2023 11:16:26 -0500 Subject: [PATCH] ece140: guess thevinin --- docs/1b/ece140.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/docs/1b/ece140.md b/docs/1b/ece140.md index 20e056a..4faac55 100644 --- a/docs/1b/ece140.md +++ b/docs/1b/ece140.md @@ -168,3 +168,22 @@ In linear circuits, a voltage source in series with a resistor can be replaced b $$v_1=i_2R$$ The arrow of the current source must point in the positive direction of the voltage source. This can also be used with dependent sources. + +### Thevenin's theorem + +Any part of a circuit including an independent source can be replaced with exactly one voltage source and a resistor in series. Two circuits are **Thevenin equivalent** if their $\lambda$ are equal in $V=\lambda I$. + +1. Cut off the load. +2. Group the rest of the circuit together, removing all independent sources (short / open). +3. The Thevenin resistance of the new resistor is the same as the load $R_{Th}=R_L$. + +If dependent sources exist, the load should be replaced with an independent source of arbitrary value (e.g., 1 V) and the other variable determined to find $R_{Th}=V_{Th}/I$, where $V_{Th}$ is the voltage drop across the load. + +Across the load: + +$$ +I_L=\frac{V_{Th}}{R_{Th}+R_L} \\ +V_L=R_LI_L = \frac{R_L}{R_{Th}+R_L}V_{Th} + +!!! warning + A negative resistance $R_{L}$ indicates that the load supplies power.