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?
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