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lillybeans
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When I solve for Isc through a and b using Norton and Thevenin equivalents between c and d I get different results. Why is that?
Norton =0.5mA
Thevenin =2mA
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gneill said:In the third diagram, the source you've labelled as Vx is not the same as VCD. It's VCD which is the reference voltage for the controlled voltage source.
The Norton and Thevenin equivalents are two different methods used to simplify complex circuits into simpler equivalent circuits. The main difference is that the Norton equivalent uses a current source in parallel with a resistor, while the Thevenin equivalent uses a voltage source in series with a resistor.
Norton and Thevenin equivalents can sometimes disagree due to the assumptions and limitations of each method. The Norton equivalent assumes that the load resistance is known, while the Thevenin equivalent assumes that the load is unknown. Therefore, when the load resistance changes, the two methods can give different results.
Neither method is more accurate than the other. They are simply two different ways to simplify complex circuits. The choice between Norton and Thevenin equivalents depends on the specific application and the type of analysis being performed.
The choice between Norton and Thevenin equivalents depends on the specific circuit and the desired analysis. If the load resistance is known and the current through the circuit is of interest, then the Norton equivalent may be more suitable. If the load resistance is unknown and the voltage across the circuit is of interest, then the Thevenin equivalent may be more appropriate.
No, Norton and Thevenin equivalents cannot be used interchangeably. They are two distinct methods for circuit simplification and have different assumptions and limitations. It is important to carefully choose which method to use based on the specific circuit and analysis being performed.