Question

Effective capacitance between A and B in the figure shown is (all capacitances are in $\mu F$).

• A. $\frac{3}{14}\mu F$
• B. $\frac{14}{3}\mu F$
• C. $21\mu F$
• D. $23\mu F$

Answer 1

Answer: (B)

$\frac{14}{3}\mu F$

The points C and D will be at same potentials since, $\frac{3}{6}=\frac{4}{8}$. Therefore, capacitance of $2\mu F$ will be unaffected. So, the equivalent circuit can be shown as the effective capacitance in upper in series, is given by
$C_1=\frac{3\times 6}{3+6}=\frac{18}{9}=2\mu F$
The effective capacitance in lower arm in series is given by
$C_2=\frac{4\times 8}{4+8}=\frac{32}{12}=\frac{8}{3}\mu F$
Hence, the resultant capacitance in parallel given by
$C=C_1+C_2=2+\frac{8}{3}=\frac{14}{3}\mu F$.