Question

Homework 7 Adaptive Follow-Up Problem 25.68 Item 4 Constants A cylindrical copper cable 3.30 km long is connected across a 220.0-V potential difference. Part A What should be its diameter so that it produces heat at a rate of 50.0 W ? Express your answer with the appropriate units. TemplatesSymbols undoredoresetkeyboard shortcutshelp D = nothing nothing Request Answer Part B What is the electric field inside the cable under these conditions? Express your answer with the appropriate units. TemplatesSymbols undoredoresetkeyboard shortcutshelp E = nothing nothing Request Answer Provide Feedback

Answer 1

Answer:

 0.446 mm  

0.066 V/m

Explanation:

Given  

We are given the length of the copper cable L = 3.30 km and the potential difference is V = 220 V  

Solution  

(a) We want to find the diameter d of the cable when the dissipated power is P = 50W. The power consumed by the cable depends on its resistance R and it is given by equation in the form

P= V^2/R                                                (1)

Where V is the voltage in the cable. Now let us solve equation (1) for R and plug our values for V and P into equation (1) to get R  

R = V^2/P = (220)^2/(50) = 968Ω

Now we can determine the diameter of the copper wire. The resistance R of the wires depends on the area of the wire, resistivity and the length of the cable. Where equation gives us the relationship between these variables in the form  

R = pL/π*r^2                             (solve for r)

r = √pL/πR                             (2)

Now we can plug our values for Rep and L into equation (2) to get the radius of the cable where p for copper equals 1.72 x 10-8 Ω m

r =√pL/πR                    

 = √1.72 x 10-8 *3300m/968

 = 0.234 mm

Therefore, the diameter is d= 2r = 2(0.234  mm) =  0.446 mm  

(b) To determine the electric field we can use the values for the potential difference across the cable and the length of the cable, where the electric field is inversely proportional to the length of the cable as next  

E =V/L

  =220/3300m

  = 0.066 V/m