Four resistors P, Q, R, S arranged in a diamond, with a galvanometer between the midpoints. Balanced condition: P/Q = $\frac{R}{S}$ (equivalently PS = QR). When balanced, no current flows through the galvanometer, and it can be removed from the circuit. The equivalent resistance is then: two series arms (P+R and Q+S) in parallel: $R_{eq}$ = (P+R)$\frac{Q+S}{(P+Q+R+S)}$. Sensitivity depends on galvanometer resistance and the relative values of P, Q, R, S — maximum when all four are roughly equal. Unbalanced bridge: cannot be solved by simple series-parallel; use Kirchhoff's laws or star-delta transformation. JEE tip: always check if a complex network can be redrawn as a Wheatstone bridge.