A) \[6\mu F\]
B) \[3/2\,\mu F\]
C) \[2/3\,\mu F\]
D) \[5\,\mu F\]
Correct Answer: C
Solution :
The resultant capacitance of three capacitors in series is \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\] Given, \[{{C}_{1}}={{C}_{2}}={{C}_{3}}=2\mu F\] \[\therefore \] \[\frac{1}{C}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\] \[\Rightarrow \] \[C=\frac{2}{3}\mu F\]You need to login to perform this action.
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