Physics - Capacitor Question with Solution | TestHub

The given circuit shows a balanced Wheatstone bridge. Now, the circuit becomesIn the branch $AB,$ both capacitor are arranged in the series combination. Hence, its cquivalent capacitance is given by $$\begin{array}{l}\frac{1}{C_{AB}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10}\\ C_{AB}=10\mu \mathrm{F}\end{array}$$ Similarly, in branch $\mathrm{CD}$, $$C_{CD}=10\mu \mathrm{F}$$ Now, $C_{AB}$ and $C_{CD}$ are connected in parallel combination. Hence, the equivalent capacitance of the circuit, $$C_{\mathrm{eq}}=10+10=20\mu \mathrm{F}$$ We know that, charge $Q=C_{\mathrm{eq}}V$ $$Q=20\times 10^{-6}\times 150\mathrm{V}$$ Amount of charge stored $Q=3\times 10^{-3}C$