question_answer

Four capacitors are connected as shown. Each capacitor is of \[5\,\mu F\]. The equivalent capacitance between A and B is:

A)  \[30.6\mu F\]        

B)  \[20.33\mu F\]

C)  \[6.67\mu F\]

D)  \[3.33\mu F\]

Correct Answer: C

Solution :

Key Idea: Capacitors in series have same amount of charge pacitors In the given circuit between points A and B, a capacitor \[{{C}_{1}}\]is connected and all other capacitor \[{{C}_{2}},{{C}_{3}},{{C}_{4}}\] are connected in series. Hence, equivalent capacitance in series is \[\frac{1}{C}=\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\frac{1}{{{C}_{4}}}\] \[\frac{1}{C}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\] \[\frac{1}{C}=\frac{1}{5}\] \[\Rightarrow \] \[C=\frac{5}{3}\mu F\] This is in parallel with \[{{C}_{1}}\], \[\therefore \] \[C=C+{{C}_{1}}=\frac{5}{3}+5=6.67\,\,\mu F\]