The discussion revolves around a circuit problem involving Kirchhoff's Current Law, specifically focusing on the correctness of a solution to a previous problem and the interpretation of a new circuit setup. Participants express uncertainty about applying the law to the circuit and the implications of the circuit's configuration.
The discussion is active, with participants providing hints and guidance on applying Kirchhoff's Current Law. There is an exploration of different interpretations of the circuit, and some participants are beginning to clarify their understanding of the junctions and equations needed.
Participants mention constraints such as the need for clarity in communication and the challenges posed by the circuit's complexity. There is also a focus on the necessity of visual aids, like screenshots, to facilitate discussion.
What you found is the total resistance of the circuit.Marcin H said:
Oh, true... Hmm. So how else can I find the resistance? Should I find the voltage drop across the 20 ohm resistor and then subtract that from the source voltage and use that V in the equation? Or is their a different way.Doc Al said:
Oh... lol. I keep over thinking everything. 100-20-60=20ohms.Doc Al said:Well, you found the total resistance. The only unknown is R, so solve for it.
I am so used to harder circuits from E&M that now I'm making dumb mistakes over easier problems. It all works out now. Thanks!
Hello again! Sorry have one last question! I thought I had this right, but then I started thinking about it and I'm not sure anymore. My thought was since the voltage was changing and the current was constant, that the source was a voltage source. I kinda thought of it as you changing the voltage on a power supply or something keeping the current constant. Is that right? I feel like it's not.Doc Al said:Well, you found the total resistance. The only unknown is R, so solve for it.
You had it right the first time. Read this: Voltage and Current SourcesMarcin H said:
I am very confused right now. So are you saying that the graph shows a constant voltage source? My professor said that if you have a constant current, then it is a current source, but i don't think that's right. The slope for the graph on that problem would be zero which means zero resistance. The hyperphysics link agrees with that too. Here are my professors notes:Doc Al said:
No. Current is on the vertical axis and is constant.Marcin H said:
Your professor is correct.Marcin H said:
Note that is is a Current vs Voltage graph, so be careful how you interpret the slope.Marcin H said:
Not sure what you mean here. Hyperphysics agrees with your professor (and with me).Marcin H said:
How can you interpret the slope then? Slope is rise over run, so i/v. Oh, which is not resistance. REsistance is v/i. So what does the slope tell us exactly?Doc Al said:
You can think of it as 1/R going to zero.Marcin H said: