- #1

hnicholls

- 49

- 1

## Homework Statement

Hi,

One end of a metal wire (say, NiCr) of a given length and cross-sectional area is attached to an end of another NiCr wire of of a given length and cross-sectional area. If the free end of the longer wire (R1) is at a given electric potential, and the free end of the shorter wire (R2) is at a lower electric potential, what is the potential at the junction of the two wires?

## Homework Equations

V1 = E*R1/(R1+R2)

## The Attempt at a Solution

I take this to be a kind of voltage divider. The wires themselves serving as the relevant resistors (R1) and (R2). The impressed voltage (E) is the difference between the potential at the ends of the two wires. The problem I am having is that in a voltage divider the voltage is measured across the resistor (either R1 or R2), but in this problem is the junction part of R1 or R2 or neither?

I believe the answer is V2 = E*R2/(R1+R2), i.e. the voltage across the wire with the lower potential (plus the difference between the lower potential and ground), but I am not sure why.

Thanks.