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                 \[C={{C}_{1}}+{{C}_{2}}=2+\frac{8}{3}=\frac{14}{3}\mu F\]