Question

When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now 1/3 the original diameter, the current will be?

Answer 1

Current would decrease 9 times.

Step-by-step explanation:

Assume all others (potential difference U, resistivity ρ, and length L) are the same, only change in the diameter, we would have the following ratio

`(I_1)/(I_2) = (U/R_1)/(U/R_2) = (U)/(U)(R_2)/(R_1) = (\rho L/A_2)/(\rho L / A_1) = (\rho L)/(\rho L)(A_1)/(A_2) = (\pi d_1^2/4)/(\pi d_2^2/4) = (4\pi)/(4\pi)\left((d_1)/(d_2)\right)^2 = 3^2 = 9`

`I_2 = I_1/9`

So the current would decrease 9 times.