Explain what a capacitor is and calculate the maximum, minimum, and a resultant capacitance of 3μF using three identical 2μF capacitors.

To obtain the maximum capacitance, the capacitors must be connected in parallel. The equivalent capacitance for parallel connection is the sum of individual capacitances.

$$C_{max} = C_1 + C_2 + C_3$$ $$C_{max} = 2\mu F + 2\mu F + 2\mu F$$ $$C_{max} = 6\mu F$$

To obtain the minimum capacitance, the capacitors must be connected in series. The equivalent capacitance for series connection is given by the reciprocal sum.

$$\frac{1}{C_{min}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$ $$\frac{1}{C_{min}} = \frac{1}{2\mu F} + \frac{1}{2\mu F} + \frac{1}{2\mu F}$$ $$\frac{1}{C_{min}} = \frac{3}{2\mu F}$$ $$C_{min} = \frac{2}{3}\mu F$$

We need to combine the three $2\mu F$ capacitors to get $3\mu F$.
Consider connecting two capacitors in series first:

$$C_{series} = \frac{1}{\frac{1}{2\mu F} + \frac{1}{2\mu F}} = \frac{1}{\frac{2}{2\mu F}} = 1\mu F$$

Now, connect this $1\mu F$ equivalent capacitance in parallel with the third $2\mu F$ capacitor:

$$C_{resultant} = C_{series} + C_3$$ $$C_{resultant} = 1\mu F + 2\mu F$$ $$C_{resultant} = 3\mu F$$

This arrangement yields the desired capacitance.

The diagram would show two capacitors connected in series, and this series combination is then connected in parallel with the third capacitor.

• Draw two capacitors connected end-to-end (series).
• Draw the third capacitor connected across the two terminals of the series combination (parallel).

The maximum capacitance is $\boxed{\text{6}\mu F}$.
The minimum capacitance is $\boxed{\frac{2}{3}\mu F}$.
The arrangement for $3\mu F$ is $\boxed{\text{two capacitors in series, connected in parallel with the third capacitor}}$.