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

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.