Question Video: Calculating Equivalent Capacitances of Capacitor Combinations

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 𝐶 sub one and the second capacitor value 8.00 microfarads as 𝐶 sub two. We want to know the total capacitance created by combining these two capacitors first in series, which we’ll call 𝐶 sub 𝑠, and then combining them in parallel, what we’ll call 𝐶 sub 𝑝.

As we start, let’s make a quick sketch of what these two arrangements look like. Arranged in series, 𝐶 sub one and 𝐶 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 𝐶 sub one and 𝐶 sub two, we want to solve for the capacitance of each arrangement.

We can begin solving for 𝐶 sub 𝑠 by recalling the rule for adding capacitors in series. Arranged in series, one over the total capacitance 𝐶 sub 𝑡 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 𝐶 sub 𝑛.

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

Since we’re given 𝐶 sub one and 𝐶 sub two in the problem statement, we can plug in and solve for 𝐶 sub 𝑠. When we do, we’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, 𝐶 sub 𝑠 is 3.08 microfarads. That’s the total capacitance when these two capacitors are combined in series.

Now we move on to solving for 𝐶 sub 𝑝, the capacitance when the capacitors are arranged in parallel. To do that, we’ll recall the rule for adding capacitors in parallel. Arranged this way, the total capacitance 𝐶 sub 𝑡 is simply the linear sum of each of the individual capacitances, 𝐶 sub one, 𝐶 sub two, all the way up to the last capacitor in the circuit, 𝐶 sub 𝑛.

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