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(D) The given circuit is a balanced Wheatstone bridge because the ratio of capacitances in the arms is equal $(2 \mu F / 2 \mu F = 2 \mu F / 2 \mu F)$

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(D) The given circuit is a balanced Wheatstone bridge because the ratio of capacitances in the arms is equal $(2 \mu F / 2 \mu F = 2 \mu F / 2 \mu F)$.In a balanced Wheatstone bridge,the central capacitor does not store any charge,so it can be removed from the circuit.After removing the central capacitor,the circuit consists of two parallel branches,each containing two $2 \mu F$ capacitors in series.For the upper branch,the equivalent capacitance $C_1$ is given by:$1/C_1 = 1/2 + 1/2 = 1 \implies C_1 = 1 \mu F$.Similarly,for the lower branch,the equivalent capacitance $C_2$ is:$1/C_2 = 1/2 + 1/2 = 1 \implies C_2 = 1 \mu F$.Since these two branches are in parallel,the total equivalent capacitance $C_{AB}$ is:$C_{AB} = C_1 + C_2 = 1 \mu F + 1 \mu F = 2 \mu F$.

In the following circuit,the equivalent capacitance between terminal $A$ and terminal $B$ is: (in $\mu F$)

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