<Dd> R x = R 3 ⋅ R 2 R 1 (\ displaystyle R_ (x) = ((R_ (3) \ cdot R_ (2)) \ over (R_ (1)))) </Dd> <P> If all four resistor values and the supply voltage (V) are known, and the resistance of the galvanometer is high enough that I is negligible, the voltage across the bridge (V) can be found by working out the voltage from each potential divider and subtracting one from the other . The equation for this is: </P> <Dl> <Dd> V G = (R 2 R 1 + R 2 − R x R x + R 3) V s (\ displaystyle V_ (G) = \ left ((R_ (2) \ over (R_ (1) + R_ (2))) - (R_ (x) \ over (R_ (x) + R_ (3))) \ right) V_ (s)) </Dd> </Dl> <Dd> V G = (R 2 R 1 + R 2 − R x R x + R 3) V s (\ displaystyle V_ (G) = \ left ((R_ (2) \ over (R_ (1) + R_ (2))) - (R_ (x) \ over (R_ (x) + R_ (3))) \ right) V_ (s)) </Dd>

What is the need of resistance bridge for a resistive sensor