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

Answer:

The current will be 18 A

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;

$R = \frac{\rho l}{A}$

where;

ρ is resistivity

l is length

A is area

$R = \frac{\rho l}{A} \\\\R = \frac{\rho l}{\pi r^2} = \frac{\rho l}{\pi (\frac{d}{2}) ^2} = \frac{\rho l}{\pi (\frac{d^2}{4}) }\\\\R = \frac{4*\rho l}{\pi d^2} \\\\R = (\frac{4*\rho l}{\pi } )\frac{1}{d^2} \\\\R = (k)\frac{1}{d^2} \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = \frac{R_1d_1^2}{d_2^2}$

When the diameter of the resistor is tripled

d₂ = 3d₁

$R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms$

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