What is the formula for potential difference of a wheatstone bridge

Let the resistances be $R_1, R_2, R_3, R_4$ and the galvanometer resistance be $R_g$. Let the supply voltage be $V$.

$$V_{AB} = V \frac{R_2}{R_1 + R_2}$$ $$V_{AD} = V \frac{R_4}{R_3 + R_4}$$

The potential difference across the galvanometer ($V_g$) is the difference between these two potentials:

$$V_g = V_{AB} - V_{AD} = V \left( \frac{R_2}{R_1 + R_2} - \frac{R_4}{R_3 + R_4} \right)$$

This can be simplified to:

$$V_g = V \frac{R_2 R_3 - R_1 R_4}{(R_1 + R_2)(R_3 + R_4)}$$

The bridge is balanced when $V_g = 0$, which implies $R_2 R_3 = R_1 R_4$.