# Homework circuit need help understanding where to start

#### Bastian O

Joined May 20, 2021
3

First of all sorry the pictures are in spanish as as I'm taking the course in this language I could translate if necesary. Second sorry if
my grammar isn't perfect english aint my mother language.

So I've got this circuit and I'm told that "cola de alimentador" has V = 13.8[kV] and I need the tension (voltage) of points 1, 2, 3, 4, 5 and 6. And to find the power factor of "cabecera del alimentador"

So I've tried to go from the end to the front calculating the electric current in every node and then tryng to get the voltages with that and the equivalent impedances that i had previously calculated but so far i get that the voltage in "cabecera de alimentador" is 94.06 with -40.40 degrees and i dont think thats right considering Voltage at the end (cola de alimentador) is 14.8 kV.

I would love if anyone could point me in a direction to solve this problem as it's the first time i work in a circuit with multiple grounds and I'm not sure how to approach it. Maybe a redrowing of the circuit where it would be easier to see whats parallel to what or anything tbh. Thanks in advance.

#### neonstrobe

Joined May 15, 2009
153
I take it that 1&2, 3&4, 5&6 are connected? Therefore have the same potential (but different currents)?
Are you able (and allowed) to use numerical methods or are you supposed to be using algebra?

#### Bastian O

Joined May 20, 2021
3
It's not stablished but i would say it's safe to assume that 1-2, 3-4 and 5-6 are connected.
I can use MatLab and such at a later point of the assignment but I'm supposed to use algebra for this.

#### MrAl

Joined Jun 17, 2014
8,262
View attachment 239168View attachment 239169
First of all sorry the pictures are in spanish as as I'm taking the course in this language I could translate if necesary. Second sorry if
my grammar isn't perfect english aint my mother language.

So I've got this circuit and I'm told that "cola de alimentador" has V = 13.8[kV] and I need the tension (voltage) of points 1, 2, 3, 4, 5 and 6. And to find the power factor of "cabecera del alimentador"

So I've tried to go from the end to the front calculating the electric current in every node and then tryng to get the voltages with that and the equivalent impedances that i had previously calculated but so far i get that the voltage in "cabecera de alimentador" is 94.06 with -40.40 degrees and i dont think thats right considering Voltage at the end (cola de alimentador) is 14.8 kV.

I would love if anyone could point me in a direction to solve this problem as it's the first time i work in a circuit with multiple grounds and I'm not sure how to approach it. Maybe a redrowing of the circuit where it would be easier to see whats parallel to what or anything tbh. Thanks in advance.
This looks like a transmission line problem although it may be somewhat simplified.

The way i think i would go about this is to first convert everything into component values at a given frequency i would think is either 50 or 60Hz but that could make a difference too.
Because of the way it is drawn it looks like they are allowing the use of lumped circuit elements but i cant be sure about that part. You'd have to know from previous exercises.

#### Bastian O

Joined May 20, 2021
3
This looks like a transmission line problem although it may be somewhat simplified.

The way i think i would go about this is to first convert everything into component values at a given frequency i would think is either 50 or 60Hz but that could make a difference too.
Because of the way it is drawn it looks like they are allowing the use of lumped circuit elements but i cant be sure about that part. You'd have to know from previous exercises.
Yes, I can use lumped circuits elements I tried that and got the results I posted V=94.06<-40.40º which I dont think is right as it's 3 orden of magnitudes bellow what i previously had. The problem I have is that I'm not sure what's in parallel or in series. The grounds are really messing with my mind lol.

#### MrAl

Joined Jun 17, 2014
8,262
Yes, I can use lumped circuits elements I tried that and got the results I posted V=94.06<-40.40º which I dont think is right as it's 3 orden of magnitudes bellow what i previously had. The problem I have is that I'm not sure what's in parallel or in series. The grounds are really messing with my mind lol.
Well it looks like the problem is trying to mimic a power distribution system. In that case we would see several loads in parallel at different points along the line. Each point could be a house or business with different loads connected to the line at those different points. That would mean anything connected to a vertical line is in parallel, and anything between is in series with those.
That puts Z1 and Z2 in parallel, Z3 and Z4 in parallel, and Z5 and Z6 in parallel.
That also means that Z1 and Z2 are connected to Z3 and Z4 with a series combination of R_Line and Z_Line and the distance between those loads is given, and same goes for Z3 and Z4 in series with Z5 and Z6 and the distance for that part of the line given too.
The distance values allow for the calculation of R_Line and X_Line for the given part of the line.

Would you happen to know what the frequency is i dont see that anywhere.
Maybe you could show a little work too (neatly written that is).

#### BobaMosfet

Joined Jul 1, 2009
1,780
View attachment 239168View attachment 239169
First of all sorry the pictures are in spanish as as I'm taking the course in this language I could translate if necesary. Second sorry if
my grammar isn't perfect english aint my mother language.

So I've got this circuit and I'm told that "cola de alimentador" has V = 13.8[kV] and I need the tension (voltage) of points 1, 2, 3, 4, 5 and 6. And to find the power factor of "cabecera del alimentador"

So I've tried to go from the end to the front calculating the electric current in every node and then tryng to get the voltages with that and the equivalent impedances that i had previously calculated but so far i get that the voltage in "cabecera de alimentador" is 94.06 with -40.40 degrees and i dont think thats right considering Voltage at the end (cola de alimentador) is 14.8 kV.

I would love if anyone could point me in a direction to solve this problem as it's the first time i work in a circuit with multiple grounds and I'm not sure how to approach it. Maybe a redrowing of the circuit where it would be easier to see whats parallel to what or anything tbh. Thanks in advance.
Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3