Resistance and Resistivity problem

AI Thread Summary
The discussion revolves around calculating the length of Nichrome wire needed for a heater that dissipates 575 W at 120 V, given its resistivity. The user initially calculated the current as 4.792 Amperes and the cross-sectional area as 0.000232 m². However, they encountered an error in their calculation for the length of the wire, initially arriving at 0.1079 m. Another participant corrected the approach, indicating the need to rearrange the resistance formula to properly solve for length. The problem was ultimately resolved with the correct formula application.
Snowman2526
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Homework Statement



A piece of Nichrome wire has a radius of 8.6 x 10^-3 m. It is used in a laboratory to make a heater that dissipates 575 W of power when connected to a voltage source of 120 V. Ignoring the effect of temperature on resistance, estimate the necessary length of wire.

Nichrome wire had resistivity of 100x10^-8

Homework Equations



p=resistivity
P=power
L=length of wire
A=cross sectional area of wire
I=Current
V=Voltage

R=pL/A
R=V/I
P=IV

The Attempt at a Solution




I first solved for current, since i'd need to plug resistance into the first equation to make L the only unknown. So P/V=I gives I=4.792 Amperes. Then I needed to take the radius of the wire and make it into cross sectional area, so A=(pi)r^2...A=.000232 m^2.

Now to solve the first equation for L...L=pV/IA.

L=(100x10^-8 ohm-m)(120 V)/(4.792 A)(.000232 m^2)


now, i know I'm getting the wrong answer of .1079 m. can anyone tell me what I'm doing wrong?
 
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Hi Snowman2526! :wink:

(have a rho: ρ :wink:)
Snowman2526 said:
Now to solve the first equation for L...L=pV/IA.

Noooo … ρ = (A/L)(V/I), so L = … ? :smile:
 
thank you sir :) problem solved
 
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