Question

Derive the balancing condition of a Wheatstone bridge.

Answer 1

Wheatstone bridge (balanced) - Let $i$ be the current from battery $E$ .At point A , current $i_1$ flows through resistance $P$ and current $i-i_1$ flows through $R$. In balanced state , no current flows through BD , hence point B and D are at same potential .Therefore current $i_1$ flows through resistance $Q$ also and current $i-i_1$ flows through $S$.

Applying Kirchhoff's loop rule in closed mesh $ABDA$ ,

$i_{1}P-(i-i_1)R=0$

or $i_{1}P=(i-i_1)R$ ...................eq1

In closed mesh $BCDB$ ,

$i_{1}Q-(i-i_1)S=0$

or $i_{1}Q=(i-i_1)S$ ...................eq2

Dividing eq1 from eq2 , we get

$P/Q=R/S$

This is the condition for balance in a Wheatstone's bridge .