Solution:
To solve the problem of finding the minimum capacitance we can form from three capacitors of capacitances [tex]$1600 \, \text{pF}$[/tex], [tex]$5800 \, \text{pF}$[/tex], and [tex]$0.011 \, \mu \text{F}$[/tex], we need to consider the way capacitors are connected.
### Part C:
Minimum Capacitance Calculation:
1. Convert all capacitances to the same unit.
For consistency, let's convert [tex]$0.011 \, \mu \text{F}$[/tex] to picoFarads (pF).^
[tex]\[
0.011 \, \mu \text{F} = 0.011 \times 10^6 \, \text{pF} = 11000 \, \text{pF}
\][/tex]
2. Identify the formula for capacitors in series.
For capacitors connected in series, the total or equivalent capacitance [tex]\(C_{\text{total}}\)[/tex] is given by:
[tex]\[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
\][/tex]
3. Substitute the given values into the formula.
[tex]\[
\frac{1}{C_{\text{total}}} = \frac{1}{1600} + \frac{1}{5800} + \frac{1}{11000}
\][/tex]
4. Calculate the total capacitance.
- Sum the reciprocals:
[tex]\[
\frac{1}{1600} + \frac{1}{5800} + \frac{1}{11000} = 0.000625 + 0.0001724 + 0.00009091 \approx 0.00088831
\][/tex]
- Take the reciprocal to find [tex]\(C_{\text{total}}\)[/tex]:
[tex]\[
C_{\text{total}} = \frac{1}{0.00088831} \approx 1125.72 \, \text{pF}
\][/tex]
5. Express the result to two significant figures:
[tex]\[
C_{\text{min}} = 1125.72 \, \text{pF}
\][/tex]
### Part D:
Connection of the Capacitors:
To achieve the minimum capacitance, the capacitors should be connected in series. This is because the effective capacitance in a series connection is always less than the smallest capacitance of any individual capacitor when multiple capacitors are involved.
So, connect all three capacitors (with capacitances of [tex]$1600 \, \text{pF}$[/tex], [tex]$5800 \, \text{pF}$[/tex], and [tex]$0.011 \, \mu \text{F}$[/tex]) in series to achieve the minimum capacitance of [tex]\(C_{\text{min}} = 1125.72 \, \text{pF}\)[/tex].
[tex]\[
\boxed{1125.72 \, \text{pF}}
\][/tex]
