Three capacitors each of capacitance $1\,\mu F$ are connected in parallel. To this combination, a fourth capacitor of capacitance $1\,\mu F$ is connected in series. The resultant capacitance of the system is.......$\mu F$
Medium
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(d) The circuit can be drawn as follows
$ \Rightarrow \,{C_{AB}} = \frac{{3 \times 1}}{{3 + 1}} = \frac{3}{4}\,\mu F$
$ \Rightarrow \,{C_{AB}} = \frac{{3 \times 1}}{{3 + 1}} = \frac{3}{4}\,\mu F$
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