Question

1. A 0.1m long cylindrical resistor is connected across a 12V battery. The resistor has a diameter of 0.1cm. The resistor is made of copper (resistivity = 68 x 10^-8 W m).

1. Find the current in the resistor.
2. If the resistor is replaced with a resistor of diameter 0.05cm, but otherwise identical to the first resistor, what is the new current in the resistor

Answer 1

Note: plz review the value of resistivity of copper.

a)

resistance

R = rho * L / (pi /4 d^2)

current

i = V/r = V ( pi/4 d^2) / ( rho * L)

i = 12* (3.14/4* (0.1* 10^-2)^2 / ( 68* 10^-8* 0.1)

i = 138.529 A

=======

b)

i = 12* 3.14/4* ( 0.05* 10^-2)^2 / ( 68* 10^-8* 0.1)

i = 34.632 A

=====

Comment before rate in case any doubt, will reply for sure.. goodluck