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