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)

The points $C$ and $D$ will be at same potenitals 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 arm 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 is given by
$C=C_1+C_2=2+\frac{8}{3}=\frac{14}{3}\mu F$