Thank you for reporting, we will resolve it shortly
Q.
What is the equivalent capacitance between $A$ and $B$ in given figure ?
Electrostatic Potential and Capacitance
Solution:
Since capacitors between $C$ and $D$ are connected in series, therefore their equivalent capacitance
$\frac{1}{C_{C D}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8} $ or $C_{ CD }=\frac{8}{3} F$
Now, $C_{C D}$ is in a parallel combination with $C_{E F}$. Therefore equivalent capacitance of the combination of $C_{C D}$ and
$C_{E F}=\frac{8}{3}+4=\frac{20}{3} F$
Now, this equivalent capacitance is in series with the capacitors of $12 \,F$ and $16\, F$,
Therefore equivalent capacitance between $A$ and $B$
$=\frac{1}{C_{A B}}=\frac{1}{12}+\frac{3}{20}+\frac{1}{16}=\frac{71}{240}$ or
$C_{A B}=\frac{240}{71} F$
