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 tripled, the current will be

Answer 1

The current will be 18 A

Step-by-step explanation:

Given;

potential difference, V = 10 V

current between the resistor, I = 2 A

Apply ohm's law;

V = IR

R = V / I

R = 10 / 2

R = 5Ω

Resistance is given as;

`R = (\rho l)/(A)`

where;

ρ is resistivity

l is length

A is area

`R = (\rho l)/(A) \\\\R = (\rho l)/(\pi r^2) = (\rho l)/(\pi ((d)/(2)) ^2) = (\rho l)/(\pi ((d^2)/(4)) )\\\\R = (4*\rho l)/(\pi d^2) \\\\R = ((4*\rho l)/(\pi ) )(1)/(d^2) \\\\R = (k)(1)/(d^2) \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = (R_1d_1^2)/(d_2^2)`

When the diameter of the resistor is tripled

d₂ = 3d₁

`R_2 = (5*d_1^2)/((3d_1)^2) \\\\R_2 = (5d_1^2)/(9d_1^2) \\\\R_2 = 0.556 \ ohms`

The current is now calculated as;

Apply ohms law;

V = IR

I = V / R

I = 10 / 0.556

I = 17.99 A

I = 18 A

Therefore, the current will be 18 A