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ece140: guess thevinin

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eggy 2023-02-07 11:16:26 -05:00
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@ -168,3 +168,22 @@ In linear circuits, a voltage source in series with a resistor can be replaced b
$$v_1=i_2R$$ $$v_1=i_2R$$
The arrow of the current source must point in the positive direction of the voltage source. This can also be used with dependent sources. 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.