Step 1:
\The diameter of the aluminum wire is d1= 2.5 mm.
\Let the length of the aluminum wire is l.
\Since the length are same, the length of the copper wire is l.
\Specific resistivity of the aluminum is .
Specific resistivity of the copper .
Total current in the circuit is it = 10 A.
\Current through copper wire is i2 = 2.5 A.
\The aluminum wire and copper wire are connected in parallel.
\Total Current = i1 + i2 .
\10 = i1 + 2.5
\i1 =10 - 2.5
\i1 = 7.5 A.
\Current through aluminum wire is i1 = 7.5 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 1.1902 mm.
\Step 4:
\Find the length of the aluminum conductor .
\Resistance of the aluminum conductor .
Voltage drop across the aluminum is .
An aluminum conductor has a voltage of 75V.
\Now find the resistance of aluminum conductor.
\Substitute resistance of aluminum the .
Length of aluminum conductor is 1.96 km.
\So the length of copper conductor is 1.96 km.
\Step 5:
\Find the resistance of copper conductor.
\Resistance offered by copper wire is .
Resistance of the copper conductor is 30 ohms.
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Solution:
\The diameter of the copper wire is 8.02 mm.
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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:
\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.