- #36
M P
- 88
- 1
Back on track after break with V40 = -60.1+5.31j and V30 = -45.9+19.51j in a need of your physics assistance
Your value for V30 doesn't look right, and it's the important value here since it determines the current through Z4.M P said:Back on track after break with V40 = -60.1+5.31j and V30 = -45.9+19.51j in a need of your physics assistance
Yes! Showing your work is highly encouraged. Trying to determine where things are going awry from the results alone requires a good deal more effort :) Show your calculations, preferably with notations as to what the steps are trying to accomplish.M P said:Is it ok to show calcs? I did check it and I am not sure what is causing damage.
gneill said:Yes! Showing your work is highly encouraged. Trying to determine where things are going awry from the results alone requires a good deal more effort :) Show your calculations, preferably with notations as to what the steps are trying to accomplish.
You should be in a position to determine the node potentials from the results of your previous mesh calculations. V30, for example, is given by the potential across Z4. The mesh currents will yield that from Ohm's Law. Since your mesh calculations were successful, you have a "standard result" against which you can compare new results.M P said:I have another one to check V40 = 60.1j - 214.5 V30 = 74.3j - 200.3 ?
If by "first results" you mean the mesh analysis portion of the problem, then that's not required since I believe that you found the correct results there. If you are having difficulty achieving the same results with nodal analysis, then those are the calculations that need to be examined.M P said:Now what calcs you prefer from first results or second ?
I was trying to do complex numbers on this ...M P said:one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?
and on this ? that is what I understood to obtain V3 and V4 of this with complex numbers..M P said:one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?
before I go further -55.74 - 20.06j + 0.25 V40 - 0.05 j V40 ?gneill said:
I'm not seeing where that's coming from. Sorry.M P said:before I go further -55.74 - 20.06j + 0.25 V40 - 0.05 j V40 ?
That'll work. Or,Ebies said:I3(-Z3) - V2 - Z5(I3-I2) = 0...Eq 3 new and revised eq 3
While you can use either direction to achieve valid equations, you must be consistent in your choice. Once you've made a selection of current direction for a mesh you must not change it, otherwise where loops "touch" the current sums through the shared components will be compromised.Ebies said:when using mesh analysis to determine the voltage drop direction, do you use conventional current flow or do you use the clockwise current you chosen at the beginning when assigned your clockwise mesh currents...? I am asking so I know when to add or subtract when doing the "KVL" walk... I am a bit confused about what the sign needs to be when walking the "KVL" loop and you encounter components... spent a lot of time going through my books and watching online tutorials and its confused me even more now