A) 4\[\mu \]F
B) 3\[\mu \]F
C) 2\[\mu \]F
D) 1\[\mu \]F
Correct Answer: C
Solution :
Equivalent capacitance of \[2\mu F\] and \[2\mu F\] connected in series is given by \[=\frac{2\times 2}{2+2}=1\mu F\] Now, equivalent capacitance ot \[1\mu F\] and \[1\mu F\] connected in parallel is \[=1+1=2\mu F\] equivalent circuit Now \[2\mu F\] and \[2\mu F\] are in connected in series \[=\frac{2\times 2}{2+2}=1\mu F\] Equivalent capacitance between A and B is given by, \[C=1+1=2\mu F\]You need to login to perform this action.
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