Four resistances 40 $$\Omega$$, 60 $$\Omega$$, 90 $$\Omega$$, 110 $$\Omega$$ and make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD in V is __________

$$\text{Let the potential at node } C \text{ be } V_C = 0\text{ V} \implies \text{Potential at node } A \text{ is } V_A = 40\text{ V}$$

$$\text{Using the voltage divider rule for the upper branch } ABC: V_B - V_C = V_A \cdot \frac{R_{BC}}{R_{AB} + R_{BC}}$$

$$V_B - 0 = 40 \cdot \frac{60}{40 + 60} = 40 \cdot \frac{60}{100} = 24\text{ V}$$

$$\text{Using the voltage divider rule for the lower branch } ADC: V_D - V_C = V_A \cdot \frac{R_{CD}}{R_{AD} + R_{CD}}$$

$$V_D - 0 = 40 \cdot \frac{110}{90 + 110} = 40 \cdot \frac{110}{200} = 22\text{ V}$$

$$\text{Potential difference across } BD: V_{BD} = V_B - V_D = 24 - 22 = 2\text{ V}$$