Question

Derive the condition for which a Wheatstone Bridge is balanced.

Hint:

For a balanced Wheatstone bridge, \[ \frac{P}{Q}=\frac{R}{S}. \] If this condition is satisfied, no current flows through the central branch.

Answer 1

Condition for Balance of Wheatstone Bridge Concept: A Wheatstone bridge consists of four resistances arranged in the form of a bridge. A galvanometer is connected between the two intermediate junctions. The bridge is said to be balanced when no current flows through the galvanometer branch. Under this condition, the potential difference across the galvanometer becomes zero.

Step 1: Consider the Wheatstone bridge. Let the four arms of the bridge have resistances \[ P,\quad Q,\quad R,\quad S \] and let the galvanometer connect the junctions between these arms. Suppose currents \(I_1\) and \(I_2\) flow through the two branches. For a balanced bridge, \[ I_g=0. \] Hence the two junctions connected by the galvanometer are at the same potential.

Step 2: Apply the condition of equal potentials. Potential drop from one end to the galvanometer junction through resistance \(P\): \[ I_1P. \] Potential drop through resistance \(R\): \[ I_2R. \] Since the two junctions are at the same potential, \[ I_1P=I_2R. \] Similarly, considering the other side, \[ I_1Q=I_2S. \]

Step 3: Divide the two equations. \[ \frac{I_1P}{I_1Q} = \frac{I_2R}{I_2S}. \] Cancelling currents, \[ \frac{P}{Q} = \frac{R}{S}. \] Therefore the condition for balance of Wheatstone bridge is \[ \boxed{ \frac{P}{Q} = \frac{R}{S} } \] or \[ \boxed{ PS=QR. } \]