Are these two resistors in Parallel?

[jwllorens]

Say I have a circuit that looks like this:

..._______
B___|_|_|_|___ A

The red and green lines are resistors.
The black lines are wires.

Assume a connection between any component line that is at a right angle to any other line, and a current source applied to points A and B.My question is this: Are the two green resistors in parallel, and thus, can I reduce them to a single resistor using (Ra*Rb)/(Ra+Rb)?

If not, any suggestions on how to calculate the equivalent resistance of this entire circuit, between points A and B?

 

[berkeman]

Say I have a circuit that looks like this:

..._______
B___|_|_|_|___ A

The red and green lines are resistors.
The black lines are wires.

Assume a connection between any component line that is at a right angle to any other line, and a current source applied to points A and B.


My question is this: Are the two green resistors in parallel, and thus, can I reduce them to a single resistor using (Ra*Rb)/(Ra+Rb)?

If not, any suggestions on how to calculate the equivalent resistance of this entire circuit, between points A and B?

Welcome to the PF.

It's a little hard to read the diagram, but if there are just wires connecting the tops and bottoms of the two green resistors, then yes, they are in parallel.

 

[phinds]

Welcome to the PF.

It's a little hard to read the diagram, but if there are just wires connecting the tops and bottoms of the two green resistors, then yes, they are in parallel.

I had to magnify the page to really see the diagram but they ARE black (lines) so your explanation is correct.

 

[jwllorens]

great, thank you. So the whole thing can be reduced to a diamond shaped circuit with a resistor in the middle, which becomes an irreducible circuit, at which point kirkoffs equations can be applied?

 

[berkeman]

great, thank you. So the whole thing can be reduced to a diamond shaped circuit with a resistor in the middle, which becomes an irreducible circuit, at which point kirkoffs equations can be applied?

There may be other simplifications, depending on the values of the resistors (like if there are symmetries). But in general yes, you would use KCL or some other technique to work on the circuit at that point.

 

[phinds]

great, thank you. So the whole thing can be reduced to a diamond shaped circuit with a resistor in the middle, which becomes an irreducible circuit, at which point kirkoffs equations can be applied?

Irreducible is a bit strong, since simple circuits like that have been done and have equations you can look up. Google "Delta-Y conversion" to see what I mean. You have a delta and want a Y and then you can go from there.

These conversion were done USING kirchhoff analysis, so of course you CAN just do that.