A) \[\text{3}\mu \text{F}\]
B) \[\text{1}.\text{5 }\mu \text{f}\]
C) \[\text{2}\mu \text{F}\]
D) \[1\mu \text{F}\]
Correct Answer: D
Solution :
In the given circuit, two \[1.5\,\,\mu F\] capacitors are connected in parallel, hence equivalent capacitance is \[C={{C}_{1}}+{{C}_{2}}=1.5\times 2=3\mu F\] Now, the equivalent capacitance of 3 capacitors in series is \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}=\frac{3}{3}=\frac{1}{1}\] \[\Rightarrow \] \[C=1\,\,\mu F\]You need to login to perform this action.
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