Question

Three capacitors of capacitances $3\mathrm{\mu F}$ are available. The minimum and maximum capacitances, which may be obtained are

Answer 1

Step 1: Given data

$C_1=C_2=C_3=3\mu F$

Step 2: To find

We have to determine the minimum and maximum capacitances

Step 3: Calculate the minimum capacitance

When three capacitors are connected in series then the capacitance is minimum.

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\min }}=\frac{1}{3\text{μF}}+\frac{1}{3\text{μF}}+\frac{1}{3\text{μF}} \\ {\mathrm{C}}_{\min }=1\mathrm{\mu F} \end{gathered}$$

Step 4: Calculate the maximum capacitance

When three capacitors are connected in parallel then the capacitance is maximum.

$$\begin{gathered} {\mathrm{C}}_{\max }=3\mathrm{\mu F}+3\mathrm{\mu F}+3\mathrm{\mu F} \\ =9\mathrm{\mu F} \end{gathered}$$

Hence, minimum capacitance is $1\mu F$ and maximum capacitance is $9\mu F$.