- #1

Kittlinljd

- 9

- 0

## 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!