If three capacitors each of capacity $1\,\mu F$ are connected in such a way that the resultant capacity is $1.5\,\mu F$, then
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(d) The circuit can be drawn as follows
Equivalent capacitance ${C_{eq}} = \frac{1}{2} + 1 = \frac{3}{2}\,\mu F$
Equivalent capacitance ${C_{eq}} = \frac{1}{2} + 1 = \frac{3}{2}\,\mu F$
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