I just learned this and it's still a bit confusing. Can't I just V=IR and, using the current and resistor, make the 2 sections into voltage sources? Or do I have to add those resistors in series in my thevenin circuit?NascentOxygen said:
So would I just add a 100ohm resistor in series with my picture on the bottom left and then change the 400ohms to 500ohms on the picture on the right?NascentOxygen said:
No. You should not convert 400 into 500.Marcin H said:
Oh, they would be in parallel. So would I find my thevenin voltage by doing 80V(400ohms/500ohms)=64V?cnh1995 said:
Right!Marcin H said:
Can you explain the voltage division part? I never really understood where that equation comes from. The Vout=V1(R1/(R1+R2)) (check link). So that equation gives us the voltage across the terminals right? But why? I'm not following the math there.cnh1995 said:
The current? And then you multiply by the resistance R2 to find the voltage there and that gives you the voltage Vout?cnh1995 said:
Right! It is derived from Ohm's law,V=IR.Marcin H said:
Then why can I use that here in this circuit if they are in parallel?cnh1995 said:
For the original voltage source in your diagram, 100Ω and 400Ω are in series. For calculating Thevenin voltage, you make use of that original voltage source. To find the Thevenin resistance, you short the original voltage source and then proceed. 100Ω and 400Ω are in parallel when "viewed from port a-b".Marcin H said:
Ok so if they are parallel for the thevenin voltage source, then why can I use that same equation for finding the thevenin voltage? if they are in parallel.cnh1995 said:
Thevenin voltage will be dependent on original voltage source and resistanceMarcin H said:
oh, ok. that makes more sense now. So how would I find the norton equivalent? Would I use the do In= Vth/Rth? Would Rth be found by adding the 100ohm and 400ohm in parallel?cnh1995 said:Thevenin voltage will be dependent on original voltage source and resistance
combinations "seen" by the original voltage source. You convert this circuit into Thevenin equivalent. It doesn't affect the load.
Yes. And it will be in parallel with the Norton current source.Marcin H said:
Ok. I got .8A. The original problems says to find the Norton equivalent at terminals a-b for the circuit shown. Do I just have to find the value or do I have to redraw the circuit with a current source and norton(equivalent) resistance?cnh1995 said:
I guess they are expecting you to redraw the circuit.Marcin H said:
Norton's Theorem is a circuit analysis technique used to simplify a circuit with multiple resistors and current sources into an equivalent circuit with a single current source and a single resistor. This simplification is useful for calculating the current and voltage in a particular part of the circuit.
To calculate the Norton equivalent current (IN), you need to find the short-circuit current (ISC) in the original circuit. This can be done by replacing all the resistors in the original circuit with their respective Norton equivalent resistors and then applying Ohm's law (ISC = VTH/RTH). The resulting IN will be the same as ISC.
The main difference between Norton and Thevenin equivalent circuits is that Norton uses a current source and Thevenin uses a voltage source. In Norton's Theorem, the current source is in parallel with the Norton equivalent resistor, while in Thevenin's Theorem, the voltage source is in series with the Thevenin equivalent resistor.
Yes, Norton's Theorem can be applied to both DC and AC circuits. However, the calculations for the Norton equivalent current and resistor may differ depending on the type of circuit. For AC circuits, impedance (Z) is used instead of resistance (R) in the calculations.
To use Norton's Theorem to solve a circuit problem, you first need to identify the part of the circuit you want to analyze. Then, you can replace the circuit with its Norton equivalent circuit, calculate the Norton equivalent current and resistor, and use them to determine the voltage and current in the desired part of the circuit. Finally, you can convert the solution back to the original circuit if needed.