Question

Derive the condition for which a Wheatstone Bridge is balanced.

Hint:

Wheatstone bridge balance condition: \[ \frac{P}{Q} = \frac{R}{S} \] At balance → galvanometer shows zero deflection.

Answer 1

Concept: A Wheatstone bridge is used to compare resistances. It consists of four resistors arranged in a diamond shape with a galvanometer between the middle junctions. Let the four resistances be: \[ P, \; Q, \; R, \; S \]
Step 1: Bridge configuration.

\( P \) and \( Q \) in one branch
\( R \) and \( S \) in the other branch
Galvanometer connected between junctions B and D
Bridge is balanced when no current flows through galvanometer.
Step 2: Condition for balance. If no current flows through galvanometer:

Potential at junction B = potential at junction D
So: \[ V_B = V_D \]
Step 3: Using potential division. Current through branch \( P, Q \) = \( I_1 \) Current through branch \( R, S \) = \( I_2 \) Potential at B (between \( P \) and \( Q \)): \[ V_B = I_1 P \] Potential at D (between \( R \) and \( S \)): \[ V_D = I_2 R \] For balance: \[ I_1 P = I_2 R \quad \cdots (1) \]
Step 4: Total potential drop in branches. Since both branches are across the same battery: \[ I_1 (P + Q) = I_2 (R + S) \quad \cdots (2) \]
Step 5: Divide (1) by (2). \[ \frac{I_1 P}{I_1 (P+Q)} = \frac{I_2 R}{I_2 (R+S)} \] \[ \frac{P}{P+Q} = \frac{R}{R+S} \]
Step 6: Simplify. Cross multiplying: \[ P(R+S) = R(P+Q) \] \[ PR + PS = PR + RQ \] \[ PS = RQ \] Final Condition: \[ \boxed{\frac{P}{Q} = \frac{R}{S}} \] This is the condition for a balanced Wheatstone bridge. Conclusion: When the ratio of resistances in one branch equals the ratio in the other branch, no current flows through the galvanometer and the bridge is balanced.