Step 1: Understanding the Concept:
A Wheatstone bridge is an incredibly precise electrical circuit configuration designed specifically for measuring unknown electrical resistances.
Historically developed by Samuel Hunter Christie and popularized by Sir Charles Wheatstone, the circuit is composed of four resistors arranged in a diamond-like quadrilateral loop.
The fundamental operational state of this circuit is known as the "null-point" or "balanced condition."
This condition is reached when the electrical potential at two opposite vertices of the bridge becomes exactly equal, resulting in zero potential difference across the bridge's diagonal.
When a sensitive galvanometer is placed across these vertices, it shows no deflection, indicating that no current (\(I_g = 0\)) is flowing through that branch.
The problem also highlights the "conjugate" property of the bridge.
This property states that the bridge remains balanced even if the positions of the galvanometer and the input battery are swapped.
This symmetry is a direct consequence of the network's linear topology and Kirchhoff's Laws, ensuring that the ratio-based relationship between the four resistors remains consistent regardless of which diagonal serves as the input or output.
Step 2: Key Formula or Approach:
In a state of perfect balance, the resistances in the four arms of the bridge (\(P, Q, R, S\)) obey a specific mathematical ratio.
The standard balancing equation for a bridge with adjacent arms \(P\) and \(Q\), and \(R\) and \(S\), is given by:
\[ \frac{P}{Q} = \frac{R}{S} \]
Alternatively, this can be visualized as the product of the resistances in the opposite arms being equal:
\[ P \cdot S = Q \cdot R \]
Step 3: Detailed Explanation:
The problem provides us with three known resistance values: \(P = 10\Omega\), \(Q = 20\Omega\), and \(R = 15\Omega\).
Our objective is to solve for the fourth, unknown resistance \(S\) that maintains the zero-current state in the galvanometer.
Using the ratio method, we substitute the known variables into our balancing equation:
\[ \frac{10}{20} = \frac{15}{S} \]
First, we observe that the left side of the equation simplifies to a basic fraction:
\[ \frac{1}{2} = \frac{15}{S} \]
This means that for the bridge to be balanced, the resistance in the bottom arm (\(S\)) must be exactly twice the resistance in the top arm (\(R\)), just as \(Q\) is twice \(P\).
To solve for \(S\) algebraically, we perform a cross-multiplication:
\[ S \times 1 = 15 \times 2 \]
\[ S = 30\Omega \]
The interchange of the battery and galvanometer mentioned in the question is a qualitative check on our understanding of the bridge's stability.
Even if we move the battery to the galvanometer's spot and the galvanometer to the battery's spot, the potential distribution updates in such a way that the ratio \(P/R = Q/S\) (which is effectively the same as \(P/Q = R/S\)) still holds true.
Checking our answer with the opposite-arm product rule:
\[ (10\Omega) \times (30\Omega) = 300\Omega^2 \]
\[ (20\Omega) \times (15\Omega) = 300\Omega^2 \]
Since both products are equal to \(300\), the bridge is confirmed to be balanced with \(S = 30\Omega\).
Step 4: Final Answer:
The value of the unknown resistance \(S\) required to maintain the balanced condition of the Wheatstone bridge is \(30\Omega\).