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 \[{{C}_{2}}=\frac{4\times 8}{4+8}\] \[=\frac{32}{12}=\frac{8}{3}\mu F\] Hence, the resultant capacitance in parallel is given by \[C={{C}_{1}}+{{C}_{2}}\] \[=2+\frac{8}{3}=\frac{14}{3}\mu F\]