Question

Three capacitors each of $3\mu \text{F}$ are connected in a circuit. The maximum and minimum equivalent capacitance that can be obtained using these capacitors are

• A. $8\mu \text{F},2\mu \text{F}$
• B. $9\mu \text{F},1\mu \text{F}$
• C. $9\mu \text{F},0\mu \text{F}$
• D. $3\mu \text{F},2\mu \text{F}$

Answer 1

The correct option is B $9\mu \text{F},1\mu \text{F}$

Maximum effective capacitance is obtained when capacitors are connected in parallel, and minimum effective capacitance is obtained when capacitors are connected in series.

In parallel, the maximum capacitance is given by

$C_{eq}=C_1+C_2+C_3=3+3+3=9\mu \text{F}$

Whereas, in series, the minimum capacitance is given by

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{3}{3}=1$

$\Rightarrow C_{eq}=1\mu \text{F}$

Hence, option $\left(a\right)$ is correct.

Key Concept : Maximum and minimum capacitance for a given set of capacitors.