Question

Derive the condition for a balanced Wheatstone Bridge.

Hint:

In a balanced Wheatstone bridge: - No current flows through the galvanometer. - Ratio of resistances in opposite arms are equal. - Used in meter bridge and strain gauge measurements.

Answer 1

Concept: A Wheatstone bridge is used to determine an unknown resistance by comparing it with known resistances. It consists of four resistors arranged in a diamond shape with a galvanometer connecting the middle junctions. Let:

• \( P, Q \) be resistances in one branch
• \( R, S \) be resistances in the other branch
• Galvanometer connects junctions \( B \) and \( D \)

The bridge is said to be balanced when no current flows through the galvanometer.
Step 1: Balanced Condition If no current flows through the galvanometer: \[ I_g = 0 \] This means the potentials at points \( B \) and \( D \) are equal: \[ V_B = V_D \]
Step 2: Apply Kirchhoff’s Law Let current \( I_1 \) flow through branch \( P \to Q \), and current \( I_2 \) flow through branch \( R \to S \). Potential drop from \( A \) to \( B \): \[ V_{AB} = I_1 P \] Potential drop from \( A \) to \( D \): \[ V_{AD} = I_2 R \] Since \( V_B = V_D \): \[ I_1 P = I_2 R \quad \cdots (1) \]
Step 3: Consider lower arms Potential drop from \( B \) to \( C \): \[ V_{BC} = I_1 Q \] Potential drop from \( D \) to \( C \): \[ V_{DC} = I_2 S \] Again using equal potentials: \[ I_1 Q = I_2 S \quad \cdots (2) \]
Step 4: Divide equations (1) and (2) \[ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \] Cancel currents: \[ \frac{P}{Q} = \frac{R}{S} \] \[ \boxed{\frac{P}{Q} = \frac{R}{S}} \] This is the condition for a balanced Wheatstone bridge.
Physical Meaning:

• No current flows through the galvanometer.
• Potential difference between its terminals is zero.
• The bridge behaves like two independent voltage dividers.

Application: If one resistance is unknown (say \( S \)): \[ S = \frac{RQ}{P} \] Thus Wheatstone bridge is used for precise resistance measurement.