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## Homework Statement

a cylindrical bar of copper has a resistance of 0.02 ohms

a) if the power of the bar is 10000 watts, what is the maximum current the bar can carry?

b) the bar is now stretched to 3 times its original length, while the total volume of the bar remains constant. what is the new resistance of the bar?

## Homework Equations

power P = IV = I^2R = V^2/R where I is current, R is resistance, V is electric potential

I = VA/L*rho = V / ((rho*L)/A) where A is area, L is length, rho is restivity

V = IR

## The Attempt at a Solution

i think i got part a correctly:

P = I^2R

10000 = I^2(0.02)

sqrt(10000/0.02) = I

I = 707.11 amperes

as for part b, do i need to use the resistivity of copper, rho_copper = 1.68*10^-8 ohm-m :

V = IR = 707.11(0.02) = 14.14 volts

I = VA/L*rho = 14.14A/3L(1.68*10^-8) ---> holding A and L constant

I = 14.14/(3*(1.68*10^-8)) = 2.81*10^8 ampere ---> that is massive!!

i think i did part a correctly, very doubtful of part be though

please tell me what needs to be fixed