014 points | Previous Answers Tipler6 25.P.044. My Not While remodeling your gar

Question

014 points | Previous Answers Tipler6 25.P.044. My Not While remodeling your garage, you need to temporarily splice, end to end, an 80 m long copper wire that is 0.980 mm in diameter with a 49 m long aluminum wire that has the same diarneter. The maximun current in the wires is 2.00 A. (a) Find the potential drop across each wire of this system when the current is 1.69 A v (copper wire) V (aluminum wire) (b) Find the electric field in each wire when the current is 1.69 A mV/m (copper wire) mV/m (aluminum wire) eBook

Explanation / Answer

resistivity values:

copper=1.68*10^(-8) ohm.m

aluminium=2.65*10^(-8) ohm.m

resistance of each cable=resistivity*length/area=resistivity*length/(pi*radius^2)

resistance of copper cable:

length=80 m

radius=diameter/2=0.98 mm/2=0.49 mm

so resistance=R1=1.68*10^(-8)*80/(pi*(0.49*0.001)^2)=1.7818 ohms

resistance of aluminium cable:

length=49 m

radius=0.98 mm/2=0.49 mm

resistance=R2=2.65*10^(-8)*49/(pi*(0.49*0.001)^2)=1.7215 ohms

part A:

potential drop across copper wire=current*resistance

=1.69*1.7818

=3.0112 volts

potential drop across aluminium wire=current*resistance

=1.69*1.7215

=2.9093 volts

part b:

electric field in copper wire=potential difference/distance between cables

=3.0112/80

=37.64 mV/m

electric field in aluminium wire=potential difference across the aluminium wire/distance between the wire=2.9093/49

=59.373 mV/m

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