Wheatstone Bridge

A Wheatstone bridge reaches balance when no current flows through the galvanometer, which happens precisely when the ratios of resistances in its two arms are equal, expressed as (\frac{P}{Q} = \frac{R}{S}). At balance the bridge points across the galvanometer are at the same potential, so the detector reads zero independent of the supply. Taking (P = 4;\Omega), (Q = 8;\Omega), (R = 6;\Omega), and solving for the fourth arm (S): (\frac{4}{8} = \frac{6}{S}), hence (S = \frac{6 \times 8}{4} = 12;\Omega). The value 3 (\Omega) inverts the ratio and divides instead of multiplying. The value 10 (\Omega) wrongly adds resistances rather than using the proportion. The value 18 (\Omega) misplaces which arm multiplies. This is the standard NCERT balance condition of the Wheatstone bridge. A plausibility check supports the answer: since the first ratio (4:8) is one-half, the second ratio (6:S) must also be one-half, requiring (S = 12;\Omega), exactly as found. A key practical advantage of this null method is that the result depends only on the four resistance ratios and not on the supply voltage or the galvanometer's sensitivity, so small fluctuations in the source or a poorly calibrated detector do not affect the measured value of the unknown arm at balance.