- #1
Kittlinljd
- 9
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Homework Statement
A 60.0 hollow nichrome tube of inner diameter 1.60 , outer diameter 4.20 is connected to a 4.00 battery.What is the current in the tube?
[tex]\rho[/tex] = 1.5*10^-6 ohm for nichrome
Homework Equations
I=dletaV/R
R=[tex]\rho[/tex]L/A -a should be the crossectional area
The Attempt at a Solution
need R so use R=pL/A
A=(4.2E-3)^2(pi)-(1.6E-3)^2(pi)=4.7375E-5m
plug into R
R=((1.5E-6ohm)(0.6m))/(4.7375E-5m)=0.0189972 ohms
then use the resistance to solve for current of the wire
I=4V/0.0189972ohm=210.6A
This isn't the right answer and I don't know what I did wrong! Could someone help me out here, please!