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 \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{1}{3}+\frac{1}{1}=\frac{4}{3}\mu F\] \[\Rightarrow \] \[C=\frac{3}{4}\mu F\]You need to login to perform this action.
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