Specific Resistance of Conductors
Basic Electricity
Examine the following specific resistance table for various metals:
Metal type ρ in Ω · cmil / ft @ 32^{o}F ρ in Ω · cmil / ft @ 75^{o}F
Zinc (very pure) 34.595 37.957
Tin (pure) 78.489 86.748
Copper (pure annealed) 9.390 10.351
Copper (harddrawn) 9.810 10.745
Copper (annealed) 9.590 10.505
Platinum (pure) 65.670 71.418
Silver (pure annealed) 8.831 9.674
Nickel 74.128 85.138
Steel (wire) 81.179 90.150
Iron (approx. pure) 54.529 62.643
Gold (99.9 % pure) 13.216 14.404
Aluminum (99.5 % pure) 15.219 16.758
Of the metals shown, which is the best conductor of electricity? Which is the worst? What do you notice about the resistivity of these metals as temperature is increased from 32^{o}F to 75 ^{o}F?
A spool holds an unknown length of aluminum wire. The size of the wire is 4 AWG. Fortunately, both ends of the wire are available for contact with an ohmmeter, to measure the resistance of the entire spool. When measured, the wire’s total resistance is 0.135 Ω. How much wire is on the spool (assuming the spool is at room temperature)?
Suppose a power system were delivering AC power to a resistive load drawing 150 amps:

Calculate the load voltage, load power dissipation, the power dissipated by the wire resistance (R_{wire}), and the overall power efficiency, indicated by the Greek letter “eta” (η = [(P_{load})/(P_{source})]).
 E_{load} =
 P_{load} =
 P_{lines} =
 η =
Now, suppose we were to redesign both the generator and the load to operate at 2400 volts instead of 240 volts. This tenfold increase in voltage allows just onetenth the current to convey the same amount of power. Rather than replace all the wire with different wire, we decide to use the exact same wire as before, having the exact same resistance (0.1 Ω per length) as before. Recalculate load voltage, load power, wasted power, and overall efficiency of this (higher voltage) system:

 E_{load} =
 P_{load} =
 P_{lines} =
 η =
The efficiency (η) of a simple power system with losses occurring over the wires is a function of circuit current, wire resistance, and total source power:

A simple formula for calculating efficiency is given here:

Where,
P_{source} = the power output by the voltage source, in watts (W)
I = the circuit current, in amperes (A)
R = the total wire resistance (R_{wire1} + R_{wire2}), in ohms (Ω)
Algebraically manipulate this equation to solve for wire resistance (R) in terms of all the other variables, and then calculate the maximum amount of allowable wire resistance for a power system where a source outputting 200 kW operates at a circuit current of 48 amps, at a minimum efficiency of 90%.
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