Two capacitors of $100 \mu \mathrm{~F}$ and $50 \mu \mathrm{~F}$ are connected in parallel. If the potential difference across $100 \mu \mathrm{~F}$ is 20 V and across $50 \mu \mathrm{~F}$ is 40 V , then the common potential of the parallel combination will be (same polarities of the capacitor connected together)

Parallel Combination Common Potential

The initial charges are $Q_1 = C_1 V_1 = (100 \times 10^{-6}) \cdot 20 = 2 \times 10^{-3}$ C and $Q_2 = C_2 V_2 = (50 \times 10^{-6}) \cdot 40 = 2 \times 10^{-3}$ C. Using conservation of charge, the total charge becomes $Q_{\text{total}} = Q_1 + Q_2 = 4 \times 10^{-3}$ C. The equivalent capacitance is $C_{\text{eq}} = C_1 + C_2 = 150 \times 10^{-6}$ F. The common potential is calculated using $V_{\text{common}} = \frac{Q_{\text{total}}}{C_{\text{eq}}}$. $V_{\text{common}} = \frac{4 \times 10^{-3}}{150 \times 10^{-6}} = \frac{80}{3}$ V. The common potential is $\frac{80}{3}$ V.