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Homework Statement
Please see a picture
Homework Equations
KVL
The Attempt at a Solution
V1-I1Z1-(I1-I2)Z4=0
Z2(I2-I4)+Z5(I2-I3)-(I2-I1)Z4=0
Z3I3+V2+Z5(I3-I2)=0
V3+Z2(I2-I4)=0
Any clues are appreciated..
When you write your mesh equations make sure to first indicate your assumed current directions, then make sure that you are consistent in summing either potential drops or potential rises. I note, for example, that last term in your second equation seems to be inconsistent. The first two terms look like potential drops due to assumed clockwise mesh currents, while the last term looks like a potential rise. Similarly for your last equation, the currents in the second term don't don't match the assumed mesh current directions if the V3 term is taken to be a potential drop for KVL going clockwise around the loop.M P said:Homework Statement
Please see a picture
View attachment 76921
Homework Equations
KVL
The Attempt at a Solution
V1-I1Z1-(I1-I2)Z4=0
Z2(I2-I4)+Z5(I2-I3)-(I2-I1)Z4=0
Z3I3+V2+Z5(I3-I2)=0
V3+Z2(I2-I4)=0
Any clues are appreciated..
gneill said:When you write your mesh equations make sure to first indicate your assumed current directions, then make sure that you are consistent in summing either potential drops or potential rises. I note, for example, that last term in your second equation seems to be inconsistent. The first two terms look like potential drops due to assumed clockwise mesh currents, while the last term looks like a potential rise. Similarly for your last equation, the currents in the second term don't don't match the assumed mesh current directions if the V3 term is taken to be a potential drop for KVL going clockwise around the loop.
gneill said:Mesh currents should be indicated by curved arrows inside the given loop, not in the wiring itself. Otherwise there can be confusion as to the direction on the shared boarders of loops -- which arrow pertains to which mesh?
Usually it's convenient to have all the mesh currents be either clockwise or counterclockwise (there's a good reason for this, having to do with a very simple way to write the mesh equations in matrix form by inspection which I'm sure you'll cover in your course at some point).
You seem to have chosen the following:
I1 : clockwise
I2 : clockwise
I3 : counterclockwise
I4 : counterclockwise
That's fine, but now you'll need to write out your mesh equations accordingly.
Okay, so you've decided that Z2 is irrelevant to the second loop since its potential difference is constrained by V3. That's a good observation! It reduces the number of loops that you have to deal with. However, it means that Z2 should not appear in the second term of your second equation above. Make that change and try solving the system for the mesh currents.M P said:V1 -(Z1+Z4)J1+Z4J2 =0
Z4J1-(Z4+Z2+Z5)J2+Z5J3-V3=0
Z5I2-(Z5+Z3)I3-V2=0
You don't need the fourth equation to find the desired ##I## . As I mentioned above, V3 fixed the potential drop across Z2, and I4 plays no role in determining ##I##.M P said:V1 -(Z1+Z4)J1+Z4J2 =0
Z4J1-(Z4+Z5)J2+Z5J3-V3=0
Z5I2-(Z5+Z3)I3-V2=0
V3+Z2(I4-I2)=0
Sorry I am a bit confused so I have now 4 equations as above with no Z2 in 2nd equation? does it make sense now??
Hmm. Those don't look like the values that I'm seeing. What method are you using to solve the system of equations? I know it can be a tough slog to do the complex arithmetic by hand... If you look around online you can find "calculators" that do complex math, which you can use to confirm your steps.M P said:Trying to find out if I am on the right track:
I1 = 0.613 - j26.4 A
I2 = 9.96 - j50.7 A
I3 = 26.8 - j20.3 A
I can't comment on what I cannot see. So you'll have to find a way to display your steps.M P said:So I calculated V1 and V2 and V3(in rectangular)and I put values instead od (Z) afterwards I used a calculator given to us by tutor and all the values with no number I left as zero, that was result I have achieved. Now I am thinking where did that go wrong?
Yes, those results look much better. You should now be in a position to calculate the desired ##I##.M P said:I think I know what went wrong let me try once again I1= 16.8 - j22.9
I3= 25.9 - j40.2
I2= 18.1 - j22.1
You're very welcome.M P said:I cannot thank you enough for helping me:)
gneill said:You're very welcome.
Let us know how the nodal analysis version of the problem goes.
I'm not seeing how your equations pertain to the circuit. Perhaps it would help if you were to indicate on your circuit diagram the nodes. Remember that you need to identify a reference node ("ground") and the essential nodes. Hint: V3 makes life easier once again in a super way.M P said:The Attempt at a Solution
node 1 = V10 etc.
V10-V30/Z1 + V40-V30/Z2 + 0-V30/Z4 = 0
V20-V40/Z3 + V30-V40/Z2 + 0-V40/Z5 = 0
thank you in advance
Getting closer. However...M P said:V30-V1/Z1 + V30-V40/Z2+V30/Z4+V40-V30/Z2+V40-V2/Z3+V40/Z5 =0
V40=V30+V3
gneill said:Getting closer. However...
V3 forces a fixed potential difference between V30 and V40. This creates a special situation. Note that the voltage source V3 has no impedance associated with it, so you cannot treat it like another branch and write a current for it via Ohm's Law. Yet it connects V30 and V40, and must influence KCL at those nodes.
This is a case where you consider that the fixed relationship between those nodes imposed by that fixed voltage source creates what is called a supernode. A supernode is made up of a set of nodes with potential differences that are fixed by voltage sources connected between them.
Check your text, notes, or web for the supernode concept. See if you can't identify the supernode on your circuit diagram and draw a boundary around it. When you write the node equation for a supernode you consider only the currents entering or leaving the supernode as a whole (passing through the boundary, as it were). Currents internal to the node are ignored, and only the fixed potential differences have an impact (setting the potentials at the wires protruding from the supernode).
Nope. In fact you'll find that, as was the case for mesh analysis, Z2 plays no role in the solution thanks to V3.M P said:Lets stick to web. Let me just ask a quick question. my equation need to be equal to V30-V40/Z2?
That's looking good! Solve for V30.M P said:V30-V1/Z1+V30/Z4+V40/Z5+V40-V2/Z3 = 0
V30-V40=V3
that is what I understood
Yes, it is also a valid equation for the supernode. Of course, you could confirm that yourself by seeing if it yields the same results as the mesh analysis version. I'd still like to see more parentheses in the equationM P said:one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?