For Q1, I disagree with your conclusion that the specification of the polarity of Vo actually tells you anything about directions and whether the current due to one source is greater than that from another. We can arbitrarily define current direction arrow and voltage polarities before we solve the problem. If the answer for a given variable comes out negative, then we know that our starting assumption was backward in a certain sense. But, this is no problem. It is only after we solve the problem numerically that we really know the directions.Hi
Please have a look on the attachment; it also has my queries there. Please help me with them. Thank you.
Yes, you can do that no problem. You can either change the direction of the current source, or you can redefine the voltage source to be positive up and negative down by adding 180 degrees to the phase.I wasn't able to transform the second voltage source into a current source because it had its positive terminal connected to the ground, 0 V, and it had an impedance connected in series with its negative terminal. Can we do the transformation to a current even when a voltage source has a series resistance connected to its negative terminal? Please let me know.
I'm not sure if your thinking is correct. You are getting the correct answer, so it is possible you are correct. But, honestly I can't follow the logic of what you are trying to do. Whatever you are doing, it seems to be a roundabout way, so I don't recommend it, even if it is correct.Hi
Please have a look on the attachment. Please let me know if my thinking is correct. Thank you.
With best regards
Ah, OK, I understand what you did now. I'm pretty sure that what you did is correct. I can't find any flaw in your logic and it gives the correct answer.Thanks a lot, Steve.
Please have a look on the attachment. I have tried to explain it what I was doing. This is important for me to know if my thinking was correct so that in future I can be careful. Thank you for the help and always being so kind to me.
Q1: Almost yes, but it's really -90 deg not +90 deg.Hi
Please have a look on the attachment and it would be very kind of if you could help me with the queries there. My knowledge about basic circuit analysis techniques is quite rusty right now so please excuse me if any of the queries is absurdly silly. Thanks.
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by Steve Arar
by Jake Hertz