Question

When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire? the lengths of both wires are the same, and the voltage difference remains unchanged. (the resistivity of the original metal is 1.68 × 10-8 ω ∙ m, and the resistivity of silver is 1.59 × 10-8 ω ∙ m.)?

Answer 1

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$.

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ .

Comparing two wires,

$R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}$

Dividing the above 2 equations,

$\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{A_2 }{A_1} \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{d_2^2 }{d_1^2}$

Since $d_2=2d_1$

The above ratio is

$\frac{R_1}{R_2}=\frac{1.68(10^{-8}) }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264$

We also have,

$\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1$

The current through the Silver wire will be 4.23 times the current through the original wire.