Engineering Find the Thevenin equivalent circuit:

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img809/9307/homeworkprobsg210omgthi.jpg


Find the Thevenin equivalent circuit with respect to the terminals.

Homework Equations

V = IR,

KVL, KCL

voltage division, current division

Thevenin formula / procedure

The Attempt at a Solution

I think the Thevenin resistance is from A to B so with the 72V source surpressed, then 5 and 20 are in parallel which is 4, and then that would be in series with 8, which is then 12. Then 12 and the original 12 are in parallel, so that would mean

R(Th, A to B) = 6Ω

But I'm having trouble finding V(Th) from a to B (if that's what I'm supposed to find)


I forgot this, but from the diagram, are the 5 ohm and 20 ohm in parallel resistors in parallel with respect to the 72V source? I think the potential from A to B would be the voltage "through" the 8 and 20 ohm resistors, but not sure.

 

[gneill]

Homework Statement

http://imageshack.us/a/img809/9307/homeworkprobsg210omgthi.jpg


Find the Thevenin equivalent circuit with respect to the terminals.

Homework Equations

V = IR,

KVL, KCL

voltage division, current division

Thevenin formula / procedure

The Attempt at a Solution

I think the Thevenin resistance is from A to B so with the 72V source surpressed, then 5 and 20 are in parallel which is 4, and then that would be in series with 8, which is then 12. Then 12 and the original 12 are in parallel, so that would mean

R(Th, A to B) = 6Ω

Yes, that looks good for R_th.

But I'm having trouble finding V(Th) from a to B (if that's what I'm supposed to find)


I forgot this, but from the diagram, are the 5 ohm and 20 ohm in parallel resistors in parallel with respect to the 72V source? I think the potential from A to B would be the voltage "through" the 8 and 20 ohm resistors, but not sure.

No, they're not in parallel. And voltage doesn't go "through". Current goes "through"; voltage is always "across" or between two points.

For this problem I would suggest applying a method such as nodal analysis to find the potential at a with respect to b.

 

[Color_of_Cyan]

Okay, so using nodal analysis with respect to b do you mean using b (the node by it) as a reference node? I mean, since the a and b points were part of the load and constituted the load, aren't they cut off while doing this? So I would find the voltages at the points in the circuit where they would be connected to then instead, right (ie. just at the "node" that would connect point a)?Because if a and b are cut off from the rest of the circuit then the current doesn't go anywhere and then I would be able to put the 12 and 8 resistors in series no matter what.

 

[gneill]

Okay, so using nodal analysis with respect to b do you mean using b (the node by it) as a reference node?

Sure. Since you are looking for the potential at a with respect to b, it would make sense to choose b as the reference node.

I mean, since the a and b points were part of the load and constituted the load, aren't they cut off while doing this? So I would find the voltages at the points in the circuit where they would be connected to then instead right (ie just for point a)?

Terminals a and b are where a load would be connected. You are looking for the Thevenin equivalent of the network from the point of view of terminals a and b. In other words, imagine that whatever load was previously connected between terminals a and b has been removed for analysis, and the diagram as given is the result.

 

[Color_of_Cyan]

So potential at point (node) A is to start from the battery across the resistors and to the point then, and then find out how MUCH more there is to still cross before getting back to the battery then.It's seems there's 2 points though... would V(Th) be different if points A and B were switched? Or does it just matter where the potential from the battery goes across to the first point?
Thanks yet again, by the way.

 

[gneill]

So potential at point (node) A is to start from the battery across the resistors and to the point then, and then find out how MUCH more there is to cross before getting back to the battery then, right?

If that's a way of saying you want to find the potential at a with respect to b, then yes. Otherwise, I guess I don't understand your paragraph

Thanks yet again, by the way.

You're welcome