Question

In the figure shown, a potential of $+12V$ is given to point $A$, and point $B$ is earthed. What is the potential at the point $P$ in $V$?

Answer 1

The equivalent capacitance is $C_{eq}=\frac{3(4+2)}{3+(4+2)}=\frac{18}{9}=2\mu F$

$\therefore Q_{eq}=C_{eq}{V}_{AB}=2(V_A-V_B)\mu C$

As the capacitors $3$ and $(2,4)$ are in series so charge on both is equal to $Q_{eq}$.

Thus,potential across $3\mu F$ is $V_{AP}=\frac{Q_{eq}}{3}\Rightarrow (V_A-V_P)=\frac{2}{3}(V_A-V_B)$

$\therefore (12-V_P)=\frac{2}{3}(12-0)$ as $V_A=12V$ and $V_B=0V$

$V_P=12-8=4V$