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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 ' $\mathrm{P}$ '?
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Verified Answer
The correct answer is:
$300 \mathrm{~V}$
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 $\mathrm{V}_1=\frac{\mathrm{q}}{\mathrm{C}_1}=\frac{3600}{6}=600 \mathrm{~V}$
$$
\begin{aligned}
& \therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{P}}=600 \quad \therefore 900-\mathrm{V}_{\mathrm{P}}=600 \\
& \therefore \mathrm{V}_{\mathrm{P}}=300 \mathrm{~V}
\end{aligned}
$$
$\mathrm{C}_4=\mathrm{C}_2+\mathrm{C}_3=8+4=12 \mu \mathrm{F}$
$\mathrm{C}_4$ 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 $\mathrm{V}_1=\frac{\mathrm{q}}{\mathrm{C}_1}=\frac{3600}{6}=600 \mathrm{~V}$
$$
\begin{aligned}
& \therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{P}}=600 \quad \therefore 900-\mathrm{V}_{\mathrm{P}}=600 \\
& \therefore \mathrm{V}_{\mathrm{P}}=300 \mathrm{~V}
\end{aligned}
$$
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