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You've re-stated basically what the diagram is depicting, but how did you come to the conclusion that the battery dissipates -10W, or the resistor +10W?Why did you not conclude that the battery was dissipating +10W and the resisitor -10W?
There are two possible conventions. Either works mathematically provided one applies it consistently. I chose the convention used throughout industry and academia.
They do not oppose. Both probes are in phase. In order to oppose one must be CCW and the other CW from the reference node.
Which is the reason that I object to your choice of positive power as that supplied by a source when the common convention for positive power is the quantity dissipated by loads.
If one is intent on educating folks, which is a good thing, teaching them to go against accepted conventions is a recipe for confusion and dissent.
Would you mind showing your calculations?
Mark, please quote me where I stated that a measured source power computes to a positive value.Agreed, which is why I'm continuing with this discussion.
Please show diagrammatically what you mean.
Take the test case I offered driving a 1 Ohm load with a series 1 mOhm CSR. Use the diagram of that I posted. We are assuming lossless wiring and no hidden circuit elements:Label the nodes: Source negative terminal / CSR bottom Node 0CSR / load Node 1Source positive terminal / load top Node 2Vload = V2 - V1Vcsr = V1Iloop = V2/(Rload + Rcsr)Pload = (V2^2 - V2*V1) / (Rload + Rcsr)Since Rcsr << Rload, then Rload ~= Rload + Rcsr and V2 ~= V2 - V1, then Pload ~= V2*V1/Rcsr
That falls out of a consequence of your procedure that amounts to measuring the voltage rail and the current through the load as your source power. As I have read your posts you seem quite adamant that you are measuring the source power with that method. For the positive polarity power supply, both the voltage measured and the positive convention current are positive. Therefore their product: the measured power is positive.
I have no problem with this method of measuring the load power, because it works. However, it is NOT in the purest theoretical sense the ideal way to perform it. We do it this way because it is practical for most of us.What remains is an explanation as to why or how you can conclude that the source dissipation is negative. Do your calculations work out there?
Hi poynt,just noticed this topic. Glad you're making these much needed video demo'sI've noticed MarkE scope shot has AC coupling selected on channel 2. Is this not an issue?Thanks for all your work to make this happen. I'm sure over time you'll get much appreciation for doing this and in the long run you'll save time on explaining to different individuals like me Looking forward to the show Luc