Step 1:

The diameter of the aluminum wire is d₁=10 mm.

The length of the aluminum wire is l₁ = 1 km.

The length of the copper wire is .

Specific resistivity of the aluminum .

Specific resistivity of the copper .

Total current in the circuit is i_t = 220 A.

Current through copper wire is i₂ = 120 A.

The aluminum wire and copper wire are connected in parallel.

Total Current = i₁ + i₂ .

220 = i₁ + 120

i₁ = 220-120

i₁ = 100 A.

Current through aluminum wire is i₁ = 100 A.

Step 2:

Law of Resistivity:

Resistance offered by a conductor is given by .

Resistance offered by aluminum wire is .

Area of the aluminum wire is .

Resistance offered by aluminum wire is .

Resistance offered by copper wire is .

Area of the copper wire is .

Resistance offered by copper wire is .

Ratio of the resistance is

Step 3:

In a parallel combination, the voltage across the element are same.

Substitute in equation (1).

Substitute the corresponding values in the above formula.

The diameter of the copper wire is 8.02 mm.

Solution:

The diameter of the copper wire is 8.02 mm.

Contd..

Step 2:

(b)

Specific resistivity of the aluminum .

Specific resistivity of the copper .

Voltage drop across the aluminum is .

Resistance offered by aluminum wire is .

Resistance offered by aluminum is .

Voltage drop across the aluminum:

z

Voltage drop across the aluminum is 3.57 v.

In a parallel combination, the voltage across both the elements are same.

Voltage drop across the copper is 3.57 v.

Solution:

(a) The diameter of the copper wire is 8.02 mm.

(b) Voltage drop across the conductors is 3.57 v.