I\'m in electricity and magnetism physics. I am having trouble understanding thi
ID: 1629393 • Letter: I
Question
I'm in electricity and magnetism physics. I am having trouble understanding this problem. Can you please explain each step. And show how to check this problem. Thank you!!Variation of 17.51: Consider an aluminum wire of diameter 0.600mm and length 15.0 m. The resistivity of aluminum at 20.0 is 2.82 x 10-8 -m. a) Find the resistance of this wire at 20.0 . b) Ifa 9.00 V battery is connected across the ends of the wire, find the current in the wire. c) Given that the temperature coefficient of resistivity is 3.9 x 10"()", and the temperature is 100°C, what is the current when the 9.00 V battery is connected across the ends of the wire?
Explanation / Answer
a) resistance = resistivity *area/length
resistivity = 2.82*10-8 Ohm - m at 20oC
diameter =d= 0.6mm
Area of cross section = pi *d2 /4
= 3.14*0.62 /4 = .0.283 mm2 = 0.283*10-6 m2
length of wire = l = 15m
resistance = (2.82*10-8 * 15)/(0.283*10-6 ) = 1.49 Ohm
b) voltage of battery = 9 V
Per Ohm's Law, V=iR
i = V/R = 9/1.49 = 6.02 amperes
c)Given temperature coefficient of resistivity = 3.9*10-3oC-1
resistivity at 100oC = 2.82*10-8 *{1 + [3.9*10-3 *(100-20)]}
= 3.7*10-8 Ohm m
resistance of wire at 100oC = (3.7*10-8*15)/(0.283*10-6 )
= 1.96 Ohm
Per Ohm' law, V=ir
i =V/R = 9/1.96 = 4.59 Amperes
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