Question

What is the equivalent capacitance between $A$ and $B$ in given figure ?

• A. $\frac{1}{31}F$
• B. $\frac{48}{13}F$
• C. $\frac{10}{13}F$
• D. $\frac{240}{71}F$

Answer 1

Answer: (D)

Since capacitors between $C$ and $D$ are connected in series, therefore their equivalent capacitance
$\frac{1}{C_{CD}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$ or $C_{CD}=\frac{8}{3}F$
Now, $C_{CD}$ is in a parallel combination with $C_{EF}$. Therefore equivalent capacitance of the combination of $C_{CD}$ and
$C_{EF}=\frac{8}{3}+4=\frac{20}{3}F$
Now, this equivalent capacitance is in series with the capacitors of $12F$ and $16F$,
Therefore equivalent capacitance between $A$ and $B$
$=\frac{1}{C_{AB}}=\frac{1}{12}+\frac{3}{20}+\frac{1}{16}=\frac{71}{240}$ or
$C_{AB}=\frac{240}{71}F$