question_answer

Four capacitors each of 1\[\mu F\] are connected as shown. The equivalent capacitance between P and Q is:

A)  4\[\mu F\]                                        

B) \[\frac{1}{4}\mu F\]

C) \[\frac{3}{4}\mu F\]                                       

D) \[\frac{4}{3}\mu F\]

Correct Answer: C

Solution :

                When capacitors are connected in parallel, equivalent capacitance is equal to sum of individual  capacitances  and   in  series equivalent capacitance is \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+.....\] In the given circuit three capacitors of\[1\mu F\]each are connected in parallel, hence, equivalent capacitance is                 \[C={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]                 \[=1\mu F+1\mu F+1\mu F=3\mu F\] Also,\[3\mu F\]and\[1\mu F\]are in series, hence, equivalent capacitance is                 \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{1}{3}+\frac{1}{1}=\frac{4}{3}\mu F\] \[\Rightarrow \]               \[C=\frac{3}{4}\mu F\]