Solution for Part (a)
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
The resistance of a wire is given by $R = \rho \frac{l}{A}$. Since the material ($\rho$) and radius ($A$) are the same, the resistance is directly proportional to length ($R \propto l$).
Given: $l_1 : l_2 = 2 : 3 \implies R_1 : R_2 = 2 : 3$. Let $R_1 = 2x$ and $R_2 = 3x$.
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
In a parallel connection, the potential difference ($V$) across both wires is the same. According to Ohm's Law ($V = IR$), the current is inversely proportional to resistance ($I \propto \frac{1}{R}$):
$$ \frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{2} $$Given total current $I = I_1 + I_2 = 15 \text{ A}$. Using the ratio method:
$$ I_1 = \left( \frac{3}{3+2} \right) \times 15 = \frac{3}{5} \times 15 = 9 \text{ A} $$ $$ I_2 = \left( \frac{2}{3+2} \right) \times 15 = \frac{2}{5} \times 15 = 6 \text{ A} $$Step 3 (Final Concluded Answer): [0.5 Mark]
The current through the first wire is $9 \text{ A}$ and the current through the second wire is $6 \text{ A}$.
💡 EL10 Concept Insight
Think of electricity like water flowing through pipes: if two pipes have the same width but one is longer, the longer one is harder to push water through. Because the current has to choose between the two paths, it naturally prefers the "easier" path (the shorter wire) and sends more current that way!