The equivalent capacitance between points $A$ and $B$ of the circuit shown will be
A$\frac{2}{3}\,\mu F$
B$\frac{5}{3}\,\mu F$
C$\frac{8}{3}\,\mu F$
D$\frac{7}{3}\,\mu F$
Medium
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C$\frac{8}{3}\,\mu F$
c
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