the resistivity of copper is 1.7 x 10^-8
i think you're supposed to use the equation P=IV or P=(I^2)(R) or P=(V^2)/R but i'm not sure how to use them
2007-03-15 13:51:13 · 2 answers · asked by paulinatran10 1 in Science & Mathematics ➔ Physics
If 2 Watts are dissipated when 300 A flows through 1 meter of the wire, you can find the resistance of one meter of it from P=(I^2)(R).
Knowing the resistance of 1 meter of wire you can use the equation
R = rho*l/A
where rho is the resistivity (rho=1.7 x 10^-8ohm-m), l is the length (l=1 m), and A is the area. Solve for A and then compute the radius.
2007-03-15 15:25:53 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
A meter of cable is meant to lose ability of two W even as the present I is 450 A. So P = I^2 * R = 2. Plug in I = 450 and remedy for the resistance R of this meter of cable. in view that R is on the placement of ? with assistance from R = ?L/A and also you be conscious of ?, R and L (a million meter), this facilitates you to remedy for the flow sectional section A of the cable. And upon getting A for the round flow area, you've the radius.
2016-11-25 22:52:53 · answer #2 · answered by Anonymous · 0⤊ 0⤋