Question

Two wires of the same material and the same radius have their lengths in the ratio 2:3. They are connected in parallel to a battery which supplies a current of 15 A. Find the current through the wires.

Hint:

For parallel resistors, the current divides inversely proportional to their resistances. The longer wire (higher resistance) will carry less current.

Answer 1

Provided Information:

• Identical material and radius imply uniform resistivity \(\rho\) and cross-sectional area \(A\).
• The lengths of the wires are in the ratio \(L_1 : L_2 = 2 : 3\).
• The total current supplied by the battery is \(I_{\text{total}} = 15\ \mathrm{A}\).
• The wires are connected in parallel.

Step 1: Resistance-Length Relationship

The resistance \(R\) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] As \(\rho\) and \(A\) are constant, the resistance is directly proportional to the length. Therefore, the resistances of the two wires are in the same ratio as their lengths: \[ R_1 : R_2 = L_1 : L_2 = 2 : 3. \] We can represent the resistances as \(R_1 = 2k\) and \(R_2 = 3k\) for some constant \(k\).

Step 2: Current Distribution in Parallel Connection

For components connected in parallel, the current through each component is determined by the reciprocal of its resistance relative to the total reciprocal resistance: \[ I_1 = \frac{\frac{1}{R_1}}{\frac{1}{R_1} + \frac{1}{R_2}} \times I_{\text{total}}, \quad I_2 = \frac{\frac{1}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2}} \times I_{\text{total}}. \] Substituting the values \(R_1 = 2k\) and \(R_2 = 3k\): \[ I_1 = \frac{\frac{1}{2k}}{\frac{1}{2k} + \frac{1}{3k}} \times 15 = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{3}} \times 15 = \frac{\frac{1}{2}}{\frac{3+2}{6}} \times 15 = \frac{\frac{1}{2}}{\frac{5}{6}} \times 15. \] Simplifying the expression for \(I_1\): \[ I_1 = \frac{1}{2} \cdot \frac{6}{5} \times 15 = \frac{3}{5} \times 15 = 9\ \mathrm{A}. \] The current through the second wire, \(I_2\), can be found by subtracting \(I_1\) from the total current: \[ I_2 = 15 - I_1 = 15 - 9 = 6\ \mathrm{A}. \]

Conclusion:
\(I_1 = 9\ \mathrm{A}\) flows through the wire with length \(L_1\),
\(I_2 = 6\ \mathrm{A}\) flows through the wire with length \(L_2\).

Observation: In a parallel circuit, the path with lower resistance (corresponding to the shorter wire) conducts a greater amount of current.