Question

The minimum number of capacitors each of $3\mu F$ required to make a circuit with an equivalent capacitance $2.25\mu F$ is :

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (B)

$4$

There are four capacitors in which 3 capacitors are parallel and in series with 1 capacitor.

Net Capacitance for 3 parallel capacitors = 9 $\mu F$

Equivalent Capacitance for 3 parallel capacitors and in series with 1 capacitor = $\frac{9\times 3}{9+3}=2.25$ $\mu F$