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

AI Thread Summary
The discussion focuses on deducing the cross-sectional area of a wire using a conductivity constant and data from a graph. With a constant voltage of 1.00 volts and varying numbers of parallel wires, the current increases, indicating a decrease in resistance as more wires are added. The slope of the graph, represented by the equation y=0.0065x+0.2197, relates to the relationship between current and the number of wires, suggesting that current is nearly proportional to the number of wires. The resistance formula indicates that as the number of wires increases, the effective cross-sectional area increases, thus reducing resistance. Understanding the slope's significance in this context is crucial for calculating the wire's cross-sectional area.
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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.

If someone could please explain. Please let me know if you need any more information.
 
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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?
 
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