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 \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{1}{3}+\frac{1}{1}=\frac{4}{3}\mu F\] \[\Rightarrow \] \[C'=\frac{3}{4}\mu F\]