Question

Assertion (A) : In a Wheatstone bridge circuit, if we interchange the position of the cell and the galvanometer, the balance condition \( \frac{P}{Q} = \frac{R}{S} \) remains unchanged.
Reason (R) : \( \frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{Q}{S} = \frac{P}{R} \) so balance condition remains same.

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Both Assertion (A) and Reason (R) are false.

Hint:

The balance condition of a Wheatstone bridge depends on the product of opposite arms being equal (\( P \times S = Q \times R \)). Because multiplication is commutative, swapping the input (battery) and output (galvanometer) terminals never breaks this fundamental equality.

Answer 1

The Correct Option is A

Concept:
• A Wheatstone bridge consists of four resistive arms (P, Q, R, S), a voltage source (cell), and a null detector (galvanometer).
• The bridge is in a "balanced" state when the potential difference across the galvanometer is zero, resulting in no current flow through it.
• For a standard configuration with the cell across one diagonal and the galvanometer across the other, the balance condition is \( \frac{P}{Q} = \frac{R}{S} \).
• The Reciprocity Theorem for electrical networks dictates that interchanging the voltage source and the detector in a linear network will not alter the balanced state of the bridge.

Step 1: Analyze the Assertion (A)
Assertion (A) states that interchanging the cell and galvanometer leaves the balance condition unchanged. Consider the standard bridge: arms AB=P, BC=Q, AD=R, CD=S. The cell is across AC, and the galvanometer is across BD. Balance requires equal potentials at B and D, yielding the condition \( \frac{P}{Q} = \frac{R}{S} \). If we interchange the components (cell across BD, galvanometer across AC), the bridge configuration is simply rotated. By the reciprocity theorem, it will remain balanced. Thus, Assertion (A) is True.

Step 2: Analyze the Reason (R) and its explanatory link
Reason (R) provides the mathematical justification: \( \frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{Q}{S} = \frac{P}{R} \).
Let's analyze the new circuit with the cell connected across nodes B and D.
Current enters at B and splits into two paths: path B-A-D (resistances P and R) and path B-C-D (resistances Q and S). For the galvanometer connected across A and C to show zero current, the voltage at A must equal the voltage at C.
This requires the ratio of the voltage drops across the first resistors in each branch to be equal: \( \frac{V_P}{V_R} = \frac{V_Q}{V_S} \), which simplifies to the resistance ratio \( \frac{P}{R} = \frac{Q}{S} \).
Does the original condition mathematically guarantee this new requirement? Yes.
Taking the original condition \( \frac{P}{Q} = \frac{R}{S} \) and cross-multiplying gives \( PS = QR \). Rearranging this equation yields exactly \( \frac{P}{R} = \frac{Q}{S} \).
Because the mathematical requirement for the interchanged setup is inherently satisfied by the algebraic equivalence of the original condition, the bridge remains balanced.
Thus, Reason (R) is True and perfectly explains the physical phenomenon stated in Assertion (A).

Step 3: Conclusion
Both statements are true, and the algebraic manipulation shown in (R) represents the exact reason why the interchanged circuit stays balanced. Therefore, option (A) is the correct choice.