Step 1: Understanding the Concept:
When no current passes through the galvanometer (Wheatstone bridge balanced): \(\dfrac{P}{Q} = \dfrac{R}{S}\). Also apply KCL at junctions.
Step 2: Detailed Explanation:
From the circuit: Current through 10 \(\Omega\) = 3 A; through \(X\) = 1 A.
Using Kirchhoff's laws and bridge balance condition with given resistances (24 \(\Omega\), 84 \(\Omega\), 30 \(\Omega\), 10 \(\Omega\), \(X\), \(Y\)):
By voltage division and Wheatstone balance: \(X = 6\ \Omega\), \(Y = 12\ \Omega\).
Step 3: Final Answer:
\(X = \mathbf{6\ \Omega}\) and \(Y = \mathbf{12\ \Omega}\).