AC Circuit analysis problem

Friends,
I try to analyse the attach analysis circuit but the result of all nodes come to 0, can you please let me know where i am wrong.
Here are three equations of nodal analysis, which always come with 0 value. The complex number of AC source (60√90) is 0+60i
V1(1-60i/(5-2i) +1/3+1/5i) - V2 (1/5i) = 0
V2(1/5i+1/2i+1/1-2i) - V1 (1/5i) - V3(1/2-2i) = 0
V3( 1/(2-2i)+1/3i) - V2 (i/2-2i) = 0

Well yes, why don't you 1st as in FIRST send the schematic and let someone else check your set of equations. It would be like your instructor gave you an assignment and didn't give the specific question or page number.

The schematic is attached already, can you verify please

Well 1st of all there is no V2 or V3, where did that come from. It looks as if yu want to use KVL ( voltage drops in the loops), is that correct.
Last what are you tying to solve for? What does the question specifically ask you to soleve for and provide as an answer.

Clearly though the V2 & V3, you will never get any correct answer. Also to check your answer, there is simple fre software to solve the matrix and provide you the currents & voltages, but you have to get the equations correct 1st because that's what you put in the software.

Send me the info I requested so I can point you in the correct direction. I will not though give you the answer.

Ok so here you go. I went through the loops and wrote out the equations. Hopefully I didn't make an algebraic mistake. Please look at it and make sure it makes sense to you. Do you know how to do the complex algebra? Sometimes rectangular works better than Polar. This one is mixed as it has the mag & phase yet it use ,"i" which should be like 5 (angle 90) if it's a 5Ohm inductor, but you need to sort that out.

Sorry made a mistake on my equation, the second one which as a "-3 3i," whicjh is incorrect, should be ,-3,". Anyway I have included the marked up version.

Made a mistake on equation #2. Updated equations included.

Reduced form, again make sure there are no algebra mistakes, like incorrect signs. This is the hard part because this is usually where mistakes are made, in the mundane algebra.

Really thanks for your effort.

I actually need V1, V2 and V3 with nodal analysis. The updated schematic attached with mentioned V1,V2 and V3.

OK here are the KCL equations. Again check the algebra.

You can take these same KCL equations and solve, turn them into KVL, either one will enable you to determine V1 (say node 1 for instance) and V2 (node 2).

Yes, the same equations i got earlier if you see my first message. But when i solve V1,V2,V3 then all came to 0. Do you think this is possible??? whether it could be because AC source is 60√90.

It is possible to get zero volts if the node V1 is zero Ohms. You can easily check that by looking at the voltage between the 5 Ohm resistor and 2 Ohm cap on the left side and if you see a voltage greater than 60V then that is good. You should have already solved for that branch current. There is another alternative method that will take time but it eliminates all nodes to just one loop.

Try looking at the R/C in indicated 1st. We know at zero Hz, those nodes are zero volts and at infinate frequency, V1 is about 22.5V, V2 & V3 are zero volts.

I went through it and did the math combining all of the impedances series/parallel. The current came out to 10A (106 deg), Now the impedance of the 5 Ohm resistor and the -2 Ohm cap come out to 5.38 Ohms ( -21.80 deg), so take the current mentioned 7 multiply the R/C impedance, subtract from V1, and the voltage at V1 came out to 6.20V & slightly less than 90 deg.

Ok that's close enough to zero to recheck the numbers, but on the other hand you may have a math error on your part.

Do you know how to do both Polar & Rectangular, then go back & fourth to convert? If your answer at V1 is zero then the current from 60 (90 deg) has o be 11.15A.

Albin, Thanks for your kind support. I have attached the PDF file with complete analysis, if you please check and help me where i am wrong.

Well you have made a mistake. You can verify the branch currents in the fashion I indicated the other day, that is if you can' t figure out by examination. Node V1 wold have to be a short and you can see it isn't. Start with the three branch currents at Node V1, I already gave you one branc current @ that node. Node V1 is where you have at least one problem & if node V1 is zero, all the rest will be as we'll.

Good luck.

Earl, why you don't add 3i (inductor in the bottom) in your second KCL equation as it should be (v2/2-2i+3i). Can you make reference node 0 even one inductor (3i) is there??

Because when you add the cap and the inductor it comes out 1i.

Good luck. You have a mistake I told you how to sort it it. I told you where the mistake is. What more do you want?

Appreciated.

Ok I actually took soe time t your matrix. You have it set up incorrect, The Y column is zero so you are solving for the trivial solution, zero. You have one voltage in the problem, j60, that belongs in the Y column so you need to separate that out and bring it to the left side:

j60 = "......" , whereby this may be Equation 1, that will give your Y matrix at least one non zero entry, which is required for a non trivial solution, i.e., not zero. If you don't want to use the matrix then do row reduction till you have j60 = A*V1 (A = some resulting impedance) and sub that in to solve for V2, etc.

You can't have a zero Y matrix!

I have included a general form of matrix, but you probably know it?