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

Given:

• Same material and radius \(\Rightarrow\) resistivity \(\rho\) and cross-sectional area \(A\) are same.
• Length ratio: \(L_1 : L_2 = 2 : 3\).
• Total current from battery: \(I_{\text{total}} = 15\ \mathrm{A}\).
• Connection: Parallel.

Step 1: Relation between resistance and length

For a wire: \[ R = \rho \frac{L}{A} \] Since \(\rho\) and \(A\) are the same, the resistances are in the same ratio as lengths: \[ R_1 : R_2 = L_1 : L_2 = 2 : 3. \] Let \(R_1 = 2k\) and \(R_2 = 3k\).

Step 2: Current division in parallel

In parallel: \[ I_1 = \frac{\frac{1}{R_1}}{\frac{1}{R_1} + \frac{1}{R_2}} \times I_{\text{total}}, \quad I_2 = \frac{\frac{1}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2}} \times I_{\text{total}}. \] Substituting \(R_1 = 2k, R_2 = 3k\): \[ I_1 = \frac{\frac{1}{2k}}{\frac{1}{2k} + \frac{1}{3k}} \times 15 = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{3}} \times 15 = \frac{\frac{1}{2}}{\frac{3+2}{6}} \times 15 = \frac{\frac{1}{2}}{\frac{5}{6}} \times 15. \] Simplify: \[ I_1 = \frac{1}{2} \cdot \frac{6}{5} \times 15 = \frac{3}{5} \times 15 = 9\ \mathrm{A}. \] Similarly: \[ I_2 = 15 - I_1 = 15 - 9 = 6\ \mathrm{A}. \]

Final Answer:
\(I_1 = 9\ \mathrm{A}\) through the shorter wire (\(L_1\)),
\(I_2 = 6\ \mathrm{A}\) through the longer wire (\(L_2\)).

Note: In parallel, the wire with smaller resistance (shorter length) carries more current.

Answer 2

Step 1: The setup. The resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] Where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire. Since the two wires are of the same material and same radius, they have the same resistivity and cross-sectional area. Thus, the resistance is proportional to the length. Let the resistance of the first wire be \( R_1 \) and the second wire be \( R_2 \). Since their lengths are in the ratio 2:3, the resistances will also be in the same ratio: \[ \frac{R_1}{R_2} = \frac{L_1}{L_2} = \frac{2}{3} \] Thus, \( R_1 = \frac{2}{3} R_2 \).

Step 2: Using the formula for parallel resistances. The total resistance \( R_{\text{total}} \) for two resistors in parallel is: \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \] Substituting \( R_1 = \frac{2}{3} R_2 \) into the formula: \[ \frac{1}{R_{\text{total}}} = \frac{1}{\frac{2}{3} R_2} + \frac{1}{R_2} = \frac{3}{2R_2} + \frac{1}{R_2} = \frac{5}{2R_2} \] Thus, the total resistance is: \[ R_{\text{total}} = \frac{2R_2}{5} \] 

Step 3: Using Ohm's Law. The total current supplied by the battery is \( I = 15 \, \text{A} \). Using Ohm's law: \[ I = \frac{V}{R_{\text{total}}} \] Solving for \( V \): \[ V = I \times R_{\text{total}} = 15 \times \frac{2R_2}{5} = 6R_2 \] Now, the current through each wire can be found using Ohm’s law for each wire. For wire 1: \[ I_1 = \frac{V}{R_1} = \frac{6R_2}{\frac{2}{3} R_2} = 9 \, \text{A} \] For wire 2: \[ I_2 = \frac{V}{R_2} = \frac{6R_2}{R_2} = 6 \, \text{A} \] Thus, the current through the first wire is 9 A and through the second wire is 6 A.