# Current vs Number of wires (finding cross section of a wire)

## Homework Statement

Using conductivity constant (q=1.45*10^6mA) deduce the cross-Sectional area of the wire from the slope of the Data found in the graph.

Voltage was kept constant 1.00 volts

The wires are each 1m long
After graphing the data the
Current(mA) # of parralel lines
121.4 1
268.8 2
442.2 3
578 4

The equation of the graph is y=.0065x+.2197

## Homework Equations

I=qVA/(L)--------- I is current, q is conductivity constant=1/resistivity(p), V is voltage, A is cross sectional area, L is length of the wire.

R=pL/A

R=V/I

## The Attempt at a Solution

I was thinking that since the number of wires are increasing and the voltage is constant the resistance is dropping.
I'm not sure what the slope is representing and how i can put it to use in the formulas above.

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Spinnor
Gold Member
You have the equation you need to understand this problem. The resistance of a wire

R = rho*L/A

This tell you that the resistance of a wire of a given material is equal to a constant times the length divided by the cross-sectional area of the wire. Double the length of the wire, double the resistance. Double the cross-sectional area of the wire and halve the wires resistance. When you add the wires in parallel the resistance goes down, you are effectively increasing the area of the wire.

You have noticed that the current is nearly proportional to the number of wires?