The two ends of a long copper wire are connected to the terminals of a battery o
ID: 2172234 • Letter: T
Question
The two ends of a long copper wire are connected to the terminals of a battery of EMF 0.800volts forming a closed circuit; the copper wire has a length of 2.30m, and a diameter of 0.620mm, with copper having a resistivity of p=1.70x10^-8 ohm-meter at 20oC, and the temperature coefficient of resistivity a = 3.90 x10^-3 (oC)-1. After being connected to the battery for a while, the equilibrium, temperature of the wire is 80.0oC.
a. At this equilibrium temperature, find the copper resistivity.
b. At this equilibrium temperature, find the copper wire resistance and the current of the wire.
c. If a resistor R
Explanation / Answer
at t=80 degree resistivity=1.7*10^(-8)*(1+3.9*10^(-3)*(80-20))=2.0978*10^(-8) ohm-meter (b.)area of the copper wire=pi*(0.620*0.001)^2/4=3.019*10^(-7) m^2 resistance= resistivity*length/area=2.0978*10^(-8)*2.3/(3.019*10^(-7))=0.1598 ohms. current=potential difference/resistance=0.8/0.1598=5 A. (c.)total reistance after series resistance is added=0.6598 ohms so current=0.8/0.6598=1.2124 A power=1.2124^2*0.5=0.735 W
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