A plastic tube 25.0 m long and 4.00 cm in diameter is dipped into a silver solution, depositing a layer of silver 0.100 mm thick uniformly over the outer surface of the tube. If this coated tube is then connected across a 12.0-V battery, what will be the current?
R = (p*L)/A
V = I*R, where I = V/R
The Attempt at a Solution
I’m a bit uncertain about this problem since there is both a semiconductor (Ag) and plastic involved.
A = 2*pi*(r1 + r2)*L – 2*pi*(r1)*L = 2*pi*r2* L = 2*pi*(0.100* 10^-3 m)*25 m = 0.0157 m^2???
R = (p_Ag*L)/A = [(1.47*10^-8 ohm*m)*(25 m)]/(0.0157 m^2) = 2.34*10^-5 ohm???
I = 12 V/(2.34*10^-5 m^2 ohm) = 5.129*10^5 A ???
If the above is incorrect, please direct me. Any help is appreciated.