Please only respond if you know the node voltage method.
I need to write equations for this circuit. My problem is that 1 has two paths towards 3, so when I write equations do I write both of those paths or only one.
U30(1(-jXc1)+1/(jXl1+R1)+1/(jXl2)-U10((1/(-jXc1)+1/(R1+jXl1))-U20.... (the...
So my attempt was
I1*jXl1+I2*jXm=I2*(jXl2-jXc2) + I1*jXm
because they are parallel so they should have the same voltage.
I got
I1=-3*I2.
I know that J=6-j4 and that J=I1+I2 so I just plugged in what I got and I ended up getting the result for I2 and I1.
But they are not correct.
Why?
I am given...
Okay i have another problem. I have to determine complex power of the Xc1.
I have calculated that U10=5 (this Is correct)
I know that Xc1=5
So logically I=U10/-jXc1=j
Sc=1/2*-jXc1*|I|^2
Sc=1/2*-j5
Sc=1/2*-j5 - this Is not correct, whyy I am supposed to get 25*(-j)/2
okay so I do have numerical values, I did not show them because I am just interested in just knowing how to solve for (Uab)0.
As for the nodes, I don't know what to tell you. That's just how we are told to solve problems. Here is the "formula":
I might be making a mistake somewhere, but I do not see it nor do I understand it. So if you could tell me where I am wrong so I can try to see why I would be thankful because I had been at this problem for 5 hours now and I really do not see where I am making mistakes.
U10*(1/(jXl1+R1)+1/(-jXc))-U20*(1/jXl1) = -E1/(jXl1+R1)
U20(1/(R1+jXl1)+1/(-jXc2))-U30(1/-jXc2)-U10(1/(jXl1+R1)= J+ E1/(jXl1+R1) + E2/(-jXc2)
are these correct then? I removed E3 and R3
I have seen that in similar cases when calculating (Uab)0 they also add the generator. But that doesn't make really sense to me. There is no current flowing through it, and I get that te generator doesn't necessarily need it, but where does the voltage come from? Voltage is current * resistance.
So I have calculated Zab which is 3-3j (it's correct).
Now I have to calculate (Uab)0. This is where I just get completely lost.
In my opinion:
(Uab)0=E3+I3*R3+Ic1*Xc1+E4
Note: I recognize these currents are not shown in the picture, but based on their indexes and basic logic I hope you...