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;
[tex]R = \frac{\rho l}{A}[/tex]
where;
ρ is resistivity
l is length
A is area
[tex]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}[/tex]
When the diameter of the resistor is tripled
d₂ = 3d₁
[tex]R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms[/tex]
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
