Find the net capacitance of the combination of series and parallel capacitors shown.
As we look at this figure, we identify it as a complex arrangement of capacitors in a segment of a circuit. We can imagine current flowing through this circuit from top to bottom, moving through each capacitor in turn.
Now, it doesn’t matter whether current flows top to bottom or bottom to top. We know that in either case it will pass through all the different branches and therefore all the different capacitors shown here. And we want to solve for the net or total capacitance of this arrangement. The statement tells us that these capacitors are arranged both in series and in parallel. So we want to recall the two rules for adding capacitors in those ways.
Using those rules, here is our goal: to take all the capacitors in this complex circuit and simplify them down to one capacitance value. We’ll do that by combining the capacitors we see here through these addition rules bit by bit. We’ll simplify this circuit segment as we do ultimately arriving at an equivalent capacitance value for the whole group. That then is our mission. And like we said, we’ll use the series and parallel capacitor addition rules to carry it out.
Before going further then, let’s recall what those rules are as equations. When we add together capacitors in parallel, finding a total capacitance we’ve called here 𝐶 sub 𝑝, that involves taking the linear sum of the individual capacitance values up to the last of the set.
Notice how similar this rule is to the rule for adding resistors in series. On the other hand, when we add capacitors in series, we find that one over the total equivalent capacitance is equal to one over the individual capacitors all summed together. These are the two rules we’ll make primary use of in solving for the net capacitance of the arrangement we’ve seen.
There’s just one more relationship I want to add which is a simplification of the series capacitance rule. In cases where we have exactly two capacitors arranged in series, we can solve for their total equivalent capacitance according to this rule: we take the product of their values and divide it by their sum.
This is how the general equation for solving for capacitances in series simplifies when there are only two of them. So here then are the tools that we’ll use to solve for the total capacitance of this arrangement. Looking at this arrangement, we’ll want to start somewhere. And we may as well start with the branch to the far right by combining these two capacitors arranged in parallel.
One has a value of 0.75 microfarads and the other has a value of 15 microfarads. Since they’re arranged in parallel, we’ll go and refer to our parallel capacitance rule. And from this rule, we see that the total equivalent capacitance of these two parallel capacitors — we’ll simply call it 𝐶 — is equal to their values added together, so 0.75 microfarads plus 15 microfarads. That gives us 15.75 microfarads.
And here’s what we can do now. We can simplify our drawing of this circuit because we’ve solved for an equivalent capacitance of these two capacitors. Instead of this far-right branch looking like this then, now we can draw it like this, where our two capacitors have been replaced with a single one of value 15.75 microfarads. So far so good, let’s keep going.
Once again, we have the choice of the next step we’ll take: which two capacitors we’ll combine in this arrangement? Since we’re already working on this right branch, why don’t we choose that one? We’ll combine the 1.5-microfarad capacitor with the 15.75-microfarad equivalent capacitance. And since these capacitors are arranged in series and there are two of them, we’ll refer to our special series capacitance addition rule.
This rule tells us that when we add together two capacitors in series, their total capacitance which we can call once again simply 𝐶 is equal to the product of their capacitances divided by their sum. When we calculate this fraction, we find it’s about 2.06 microfarads. This then is the equivalent value of adding together our two remaining capacitances on our rightmost branch.
So once again, we can simplify this diagram. We can combine the two capacitors on the right branch into one with the capacitance of what we’ve just solved for, 2.06 microfarads. Excellent! We’re making good progress. Now, how about we do the same thing on the left branch by combining the 5.0- and 3.5-microfarad capacitors once again arranged in series?
And since they are arranged that way, we’ll once more use our special rule for adding capacitors in series when there are two of them. Dividing the product of their capacitances by their sum, we find an approximate value of 1.37 microfarads.
We can now rewrite the left branch of this segment to include just one equivalent capacitor. As we can see, we now have a much simplified circuit segment, consisting just of three capacitors which are in parallel with one another. We’ve nearly found the net or equivalent capacitance of all these capacitors together. And to solve for it, we’ll once more apply our parallel capacitor addition rule.
This time since our result will be the total equivalent capacitance for the whole circuit, we’ll call it 𝐶 sub 𝑇 and it’s equal to the sum of the three remaining capacitors we have. When we add these three together, to two significant figures, the result is 11 microfarads.
This means we can once more redraw and simplify our circuit segment, where now the only remaining capacitor has an equivalent capacitance of 11 microfarads. That’s the net capacitance of this combination of series and parallel capacitors.