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

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SUMMARY

The discussion focuses on calculating the cross-sectional area of a wire using the conductivity constant (q=1.45*10^6 mA) and data derived from a graph of current versus the number of parallel wires. With a constant voltage of 1.00 volts and a linear equation of the form y=0.0065x+0.2197, participants analyze how the slope relates to resistance and current. The relationship between resistance, length, and cross-sectional area is established through the formula R = ρL/A, emphasizing that adding wires in parallel decreases resistance and increases effective cross-sectional area.

PREREQUISITES
  • Understanding of Ohm's Law (R=V/I)
  • Familiarity with the concept of resistance in parallel circuits
  • Knowledge of the relationship between resistivity, length, and cross-sectional area
  • Basic graph interpretation skills for linear equations
NEXT STEPS
  • Explore the implications of the slope in linear equations in physics
  • Study the effects of parallel resistors on total resistance
  • Learn about the significance of conductivity constants in electrical engineering
  • Investigate practical applications of wire cross-sectional area in circuit design
USEFUL FOR

Students in physics or electrical engineering, educators teaching circuit theory, and anyone involved in practical applications of electrical resistance and conductivity.

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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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