Question

When a voltage difference is applied to a piece of metal wire, a 8-mA 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

Answer 1

Answer:

Explanation:

The resistance of a wire can be given by the following expression

$R=\frac{\rho\times L}{A }$

where R is resistance , ρ is specific resistance , L is length of wire and A is cross sectional area

specific resistance of metals are almost the same . So in the present case ρ and l are same . Hence the formula becomes

R = k / A where k is a constant .

The diameter of wire becomes two times hence area of cross section becomes 4 times or 4A .

Resistance becomes 1/4 times . Hence if resistance of metal wire is R , resistance of silver wire will be R / 4 .

current = voltage / resistance

In case of metal wire

8 x 10⁻³ = V / R

In case of silver wire

I = V / (R / 4 ) , I is current , V is potential difference .

I = 4 x V/R

= 4 x 8 x 10⁻³ A

= 32 mA.