Effective capacitance between $A$ and $B$ in the figure shown is (all capacitance are in $\mu F$)
A$21\,\mu F$
B$23\,\mu F$
C$\frac{3}{{14}}\mu F$
D$\frac{{14}}{3}\,\mu F$
Medium
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D$\frac{{14}}{3}\,\mu F$
d (d) Given circuit is balanced Whetstone bridge. So capacitor of $2\ \mu F$ can be dropped from the circuit
${C_{AB}} = 2 + \frac{8}{3} = \frac{{14}}{3}\,\mu F$
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