### Video Transcript

What total capacitance can you make by connecting a 5.00-microfarad capacitor and an 8.00-microfarad capacitor in series? What total capacitance can you make by connecting a 5.00-microfarad capacitor and an 8.00-microfarad capacitor in parallel?

We can refer to the first capacitor value 5.00 microfarads as ๐ถ sub one and the second capacitor value 8.00 microfarads as ๐ถ sub two. We want to know the total capacitance created by combining these two capacitors first in series, which weโll call ๐ถ sub ๐ , and then combining them in parallel, what weโll call ๐ถ sub ๐.

As we start, letโs make a quick sketch of what these two arrangements look like. Arranged in series, ๐ถ sub one and ๐ถ sub two would be on the same wire as shown. In parallel, they would be arranged on different branches of the circuit. Considering these two ways of combining ๐ถ sub one and ๐ถ sub two, we want to solve for the capacitance of each arrangement.

We can begin solving for ๐ถ sub ๐ by recalling the rule for adding capacitors in series. Arranged in series, one over the total capacitance ๐ถ sub ๐ก is equal to one over the first capacitor value plus one over the second capacitor value, and so on up to the one over the last capacitor value ๐ถ sub ๐.

Since our circuit has two capacitors in it, we can write one over ๐ถ sub ๐ equals one over ๐ถ one plus one over ๐ถ two. Algebraically rearranging this equation to solve for ๐ถ sub ๐ , we find itโs one over one over ๐ถ sub one plus one over ๐ถ sub two.

Since weโre given ๐ถ sub one and ๐ถ sub two in the problem statement, we can plug in and solve for ๐ถ sub ๐ . When we do, weโre careful to write our capacitor values in units of farads. When we enter these values on a calculator, we find that, to three significant figures, ๐ถ sub ๐ is 3.08 microfarads. Thatโs the total capacitance when these two capacitors are combined in series.

Now we move on to solving for ๐ถ sub ๐, the capacitance when the capacitors are arranged in parallel. To do that, weโll recall the rule for adding capacitors in parallel. Arranged this way, the total capacitance ๐ถ sub ๐ก is simply the linear sum of each of the individual capacitances, ๐ถ sub one, ๐ถ sub two, all the way up to the last capacitor in the circuit, ๐ถ sub ๐.

In our case, since we have two capacitors in our circuit, we can write ๐ถ sub ๐ equals ๐ถ sub one plus ๐ถ sub two. Knowing ๐ถ sub one and ๐ถ sub two, we can now plug in those values. When we do and add them together, we find that ๐ถ sub ๐, to three significant figures, is 13.0 microfarads. Thatโs the total capacitance when these two capacitors are arranged in parallel.