One of my tools stopped working, and I found that in order to fix it, I need to replace a broken wire. The broken wire has a diameter of 5mm and a length of 15 cm. When it was connected to the battery, a current of 12.5mA flowed through the wire. The only wire that I have at home is the same material, but has a diameter of 2mm.

a) If I replace the broken piece of wire with a 15cm length of my 2mm diameter wire, what will happen to the current (increase or decrease) ?

What is the value of the current through the new wire (assume that it's connected to the same battery as the original wire) ? Hint: Think about the resistances of the old and new wires.

b) How long should the new wire (2mm diameter) be if I want to ensure that the current passing through it is the same as the original wire (12.5mA) ?

a) I₂ = 2 mA (The current has decreased)

b) L₂ = 2.4 cm

Explanation:

Consider the old wire as wire 1 and the new wire as wire 2. The data that we have from the question is:

Current through wire 1 = I₁ = 12.5 mA

Diameter of wire 1 = d₁ = 5 mm

Length of wire 1 = L₁ = 15 cm

Cross-Sectional Area of wire 1 = A₁ = πd₁²/4 = π (5 mm) ²/4 = 19.63 mm²

Diameter of wire 2 = d₂ = 2 mm

Cross-Sectional Area of wire 2 = A₂ = πd₂²/4 = π (2 mm) ²/4 = 3.14 mm²

a)

Length of wire 2 = L₂ = 15 cm

Since, the battery is same. Therefore, the voltage will be same for both wires.

V₁ = V₂

using Ohm's Law (V = IR)

I₁R₁ = I₂R₂

Since resistance of wire is given by formula: R = ρL/A

Therefore,

I₁ρ₁L₁/A₁ = I₂ρ₂L₂/A₂

where,

ρ = resistivity, and it depends upon material of wire and the material of both wires is same and the length of wires is also same.

Hence, ρ₁ = ρ₂

and L₁ = L₂

and the equation becomes:

I₁/A₁ = I₂/A₂

I₂ = I₁A₂/A₁

I₂ = (12.5 mA) (3.14 mm²) / (19.63 mm²)

I₂ = 2 mA

Thus, the current has decreased.

b)

In order to have same current the resistance of both wires must be same:

R₁ = R₂

ρ₁L₁/A₁ = ρ₂L₂/A₂

Since, ρ₁ = ρ₂

Therefore,

L₁/A₁ = L₂/A₂

L₂ = L₁A₂/A₁

L₂ = (15 cm) (3.14 mm²) / (19.63 mm²)

L₂ = 2.4 cm