# R(eq) with two shorts

Hi all,
I am kind of puzzled with this problem:
calculate R(eq) in the following circuit:
Code:
 R(eq)
\ \___R4_______R5______
|          |           |
R1          |          R6
|__________ | _________|
|          |           |
R2          |          R7
|__R3______|___R8______|
If it were just one short wire, then the R(eq) would see just the closest loop. But with two shorts, I am not sure. The short wires do not have a connection point in the middle, they are 'one over the other'.

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Astronuc
Staff Emeritus
Look at the junction point (and potential at that point) between R1 and R2, and the junction point between R4 and R5.

The shorts mean the two points at either end are effectively at the same potential. So to solve this problem, one wants to find what is parallel and what is in series.

Potential between R1 and R2, V(R1,R2) = potential between R6 and R7, V(R6,R7).

Potential between R4 and R5, V(R4,R5) = potential between R3 and R8, V(R3, R8).

Just figure what connects to the junction pt between R4,R5 and the pt between R1 and R2.

Req = R1 + {(R2+R3),(R5+R6),(R7+R8)} + R4

and you should be able to figure the relationship of the resistances in the braces.

Thank you for the hints. But I do not quite see how (R2+R3)||(R5+R6)||(R7+R8) even though it works out for the answer. Is there a way to redraw it somehow? I redrew it with combined resistances for series, but parallel part is not obvious.

Astronuc
Staff Emeritus