- #1
Numbskull
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This relates to a homework question which I have spent considerable time on and although I understand the concepts, the process of getting to the answer is difficult because of several different 'versions' of the right answer I see.
The relevant threads are https://www.physicsforums.com/threads/thevenins-theorem.775385/ and https://www.physicsforums.com/threads/superposition-theorem.789030/#post-4955422
My textbooks (for Thevenin) demonstrate that we would find the open terminal voltage (where the load connects), and then calculate the current through the load. That's clear.
In the second part of the question, we are asked to use superposition to prove that the algebraic sum of the current through the load is equal to that supplied by the individual voltage sources.
Looking at the diagram in the above thread, when we remove either V1 or V2 then we create a parallel combination with the load and one of the inductors. However, I don't believe that we should be calculating the Thevenin voltage with the load attached.
So where we might use the voltage divider formula to calculate the voltage at the junction of the two inductors, the value of the load has also been included in the calculations that I have seen.
Can someone help me out please?
The relevant threads are https://www.physicsforums.com/threads/thevenins-theorem.775385/ and https://www.physicsforums.com/threads/superposition-theorem.789030/#post-4955422
My textbooks (for Thevenin) demonstrate that we would find the open terminal voltage (where the load connects), and then calculate the current through the load. That's clear.
In the second part of the question, we are asked to use superposition to prove that the algebraic sum of the current through the load is equal to that supplied by the individual voltage sources.
Looking at the diagram in the above thread, when we remove either V1 or V2 then we create a parallel combination with the load and one of the inductors. However, I don't believe that we should be calculating the Thevenin voltage with the load attached.
So where we might use the voltage divider formula to calculate the voltage at the junction of the two inductors, the value of the load has also been included in the calculations that I have seen.
Can someone help me out please?