question_answer

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)  \[23\,\mu F\]

Correct Answer: B

Solution :

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 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                 \[{{C}_{2}}=\frac{4\times 8}{4+8}\]                          \[=\frac{32}{18}=\frac{8}{3}\mu F\]  Hence, the resultant capacitance in parallel is given by                 \[C={{C}_{1}}+{{C}_{2}}\]                 \[=2+\frac{8}{3}\]                 \[=\frac{14}{3}\mu \,F\]