In the given figure potential at point ' \(\mathrm{A}\) ' is 900 volt and point ' \(\mathrm{B}\) ' is earthed. What will be the potential at point ' \(\mathrm{P}\) '?

Capacitors \(\mathrm{C}_2\) an \(\mathrm{C}_3\) are in parallel. Hence their equivalent capacitance
\(\mathrm{C}_4=\mathrm{C}_2+\mathrm{C}_3=8+4=12 \mu \mathrm{F}\)
\(\mathrm{C}_4\) and \(\mathrm{C}_1\) are in series.
Their equivalent capacitance \(\mathrm{C}=\frac{12 \times 6}{12+6}=\frac{72}{18}=4 \mu \mathrm{F}\)
Charge stored by the combination \(\mathrm{q}=\mathrm{CV}=4 \times 900=3600 \mu \mathrm{C}\)
In the series combination, charge in same on each capacitor.
\(\therefore\) Change on \(\mathrm{C}_1=3600 \mu \mathrm{C}\)
P.D. across \(\mathrm{C}_1\) is \(\mathrm{V}_1=\frac{\mathrm{q}}{\mathrm{C}_1}=\frac{3600}{6}=600 \mathrm{~V}\)
\(
\begin{aligned}
& \therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{P}}=600 \quad \therefore 900-\mathrm{V}_{\mathrm{P}}=600 \\
& \therefore \mathrm{V}_{\mathrm{P}}=300 \mathrm{~V}
\end{aligned}
\)