Question Video: Calculating Equivalent Capacitances of Capacitor Combinations

What total capacitance can you make by connecting a 5.00-๐œ‡F capacitor and an 8.00-๐œ‡F capacitor in series? What total capacitance can you make by connecting a 5.00-๐œ‡F capacitor and an 8.00-๐œ‡F capacitor in parallel?

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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 ๐ถ sub one and the second capacitor value 8.00 microfarads as ๐ถ sub two. We want to know the total capacitance created by combining these two capacitors first in series, which weโ€™ll call ๐ถ sub ๐‘ , and then combining them in parallel, what weโ€™ll call ๐ถ sub ๐‘.

As we start, letโ€™s make a quick sketch of what these two arrangements look like. Arranged in series, ๐ถ sub one and ๐ถ 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 ๐ถ sub one and ๐ถ sub two, we want to solve for the capacitance of each arrangement.

We can begin solving for ๐ถ sub ๐‘  by recalling the rule for adding capacitors in series. Arranged in series, one over the total capacitance ๐ถ sub ๐‘ก 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 ๐ถ sub ๐‘›.

Since our circuit has two capacitors in it, we can write one over ๐ถ sub ๐‘  equals one over ๐ถ one plus one over ๐ถ two. Algebraically rearranging this equation to solve for ๐ถ sub ๐‘ , we find itโ€™s one over one over ๐ถ sub one plus one over ๐ถ sub two.

Since weโ€™re given ๐ถ sub one and ๐ถ sub two in the problem statement, we can plug in and solve for ๐ถ sub ๐‘ . When we do, weโ€™re 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, ๐ถ sub ๐‘  is 3.08 microfarads. Thatโ€™s the total capacitance when these two capacitors are combined in series.

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

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

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