# Current density in wires connected in series

## Homework Statement

Wires A and B, having equal lengths of 40.0 m and equal diameters of 2.60 mm, are connected in series. A potential difference of 60.0 V is applied between the ends of the composite wire. The resistances of the wires are 0.127 and 0.729 , respectively.

Determine:
(a) the current density in each wire [A/m2]

(b) the potential difference across each wire. [V]

V = iR

## The Attempt at a Solution

I found the volume of 1 wire, SA times Length, 2.6 mm diameter is a .0013 m radius, squared times pi yields 5.31 E-6 for the surface area, times the length of 40m is volume of 2.124E-4.
I tried V=iR, but for the first wire I had 60=.127R, which would make the current 472.44, which I'm not sure how to relate to the volume.
I don't know how to find the potential difference either.

Redbelly98
Staff Emeritus
Homework Helper

## The Attempt at a Solution

I found the volume of 1 wire, SA times Length, 2.6 mm diameter is a .0013 m radius, squared times pi yields 5.31 E-6 for the surface area,
That's actually the cross-sectional area, not the surface area.

...times the length of 40m is volume of 2.124E-4.
The volume is not needed here.
I tried V=iR, but for the first wire I had 60=.127R, which would make the current 472.44, which I'm not sure how to relate to the volume.
V is not 60V for either wire, it is 60V across the series combination of the two wires.

What is the total resistance of the two wires connected in series? You can use that to find the current.
I don't know how to find the potential difference either.
We'll worry about that later, let's get part (a) first.