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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)?

Respuesta :

hamzaahmeds

Answer:

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

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