Lesson Explainer: Capacitors in Series and in Parallel Physics

In this explainer, we will learn how to calculate the total capacitance of multiple capacitors connected in series and in parallel combinations.

To begin, let us recall Kirchhoff’s laws, which will help us understand the effects of combining capacitors in different ways:

  1. The current into a junction equals the current out of the junction.
  2. The sum of voltage differences around a closed loop equals zero.

We will begin our discussion by looking at capacitors combined in parallel, as shown in the diagram below.

Note that both capacitors are on their own separate branch of this circuit, and recall that each branch in a parallel circuit receives the same voltage, or potential difference. Kirchhoff’s second law confirms this. Thus, the potential difference across capacitor 1, which we can call 𝑉, is equal to the potential difference of the second capacitor, 𝑉, and the potential difference supplied by the battery, 𝑉total. This general relationship is stated mathematically below.

In the diagram above, we have two capacitors, but the dotted continuation of the equation below (and others in this explainer) implies that the relationship continues for however many capacitors we might want to consider: 𝑉=𝑉=𝑉=⋯.total

Kirchhoff’s first law states that the current into and that out of a branch in a circuit are equal. Also, remember that the amount of charge that flows through a circuit branch is the product of the current in the branch and the time for which the charge flows. Thus, when the circuit is closed and the capacitors are able to charge up for some time, the charges on all capacitors (𝑄 and 𝑄 here) add up to the total charge in the entire circuit, 𝑄total, as follows: 𝑄=𝑄+𝑄+⋯.total

We already know that we can relate potential difference and charge to capacitance using the equation 𝐶=𝑄𝑉, which can be rewritten as 𝑄=𝐶𝑉.

Let us apply this to the equation for charge given above by substituting “𝑄” for “𝐶𝑉,” as follows: 𝑄=𝐶𝑉=𝐶𝑉+𝐶𝑉+⋯.totaltotaltotal

Remember that the potential difference values across all elements in a parallel combination are equivalent, so we can divide this entire equation by potential difference. This results in the equation we use to relate the value of total capacitance to that of each capacitor in a parallel combination.

Definition: Total Capacitance for Parallel Combination

The total capacitance for a parallel combination of capacitors is given by 𝐶=𝐶+𝐶+⋯.total

We will practice combining capacitors in parallel in the following examples.

Example 1: Combining Capacitors in Parallel

The circuit in the diagram contains two capacitors connected in parallel. What is the total capacitance of the circuit?

Answer

Let us begin by recalling the equation for capacitors combined in parallel: 𝐶=𝐶+𝐶+⋯.total

Since we have two capacitors in parallel here and we know their values, we are ready to add them to find the total capacitance of the circuit: 𝐶=35+65=100.totalµFµFµF

Thus, we have found that the total capacitance of this circuit is 100 µF.

Example 2: Combining Capacitors in Parallel

The circuit in the diagram contains two capacitors connected in parallel. The total capacitance of the circuit is 240 µF. What is the capacitance 𝐶?

Answer

Here, we need to determine the unknown capacitance value 𝐶, and we can begin by looking at the equation for the total capacitance of a parallel combination: 𝐶=𝐶+𝐶+⋯.total

Thus, the individual capacitance values simply add to a total capacitance value. Plugging in the values we have been given, the equation becomes 240=𝐶+135.µFµF

We can solve for 𝐶 by subtracting 135 µF from both sides of the equation: 𝐶=240−135=105.µFµFµF

Thus, we have found that the capacitance 𝐶 is equal to 105 µF.

Let us now focus on combining capacitors in series, as shown in the diagram below.

Recall that current is equal at all points in a series circuit, which is confirmed by Kirchhoff’s first law. This means that capacitors in series store equal charges. Thus, for a series combination, 𝑄=𝑄=𝑄=⋯.total

Because of Kirchhoff’s second law, we know that the sum of the potential differences across elements in a closed loop equals zero. A series combination is one big closed loop, so the potential differences across the capacitors must sum to the potential difference across the battery. So, 𝑉=𝑉+𝑉+⋯.total

Once again, recall that the capacitance, potential difference, and charge for any capacitor are given by 𝐶=𝑄𝑉, which can be rearranged as 𝑉=𝑄𝐶.

We can substitute this into the potential difference equation above so that the relationship is written as 𝑉=𝑄𝐶=𝑄𝐶+𝑄𝐶+⋯.totaltotaltotal

We have already determined that the charges on all series circuit elements are equivalent, so we can divide the whole equation by charge. And we have a relationship to describe the capacitance values in a series combination.

Definition: Total Capacitance for a Series Combination

The total capacitance for a series combination of capacitors is given by 1𝐶=1𝐶+1𝐶+⋯.total

Notice the inverse nature of this equation, which means that when more capacitors are added in series, the total capacitance decreases. We will explore this concept in the next couple of examples.

Example 3: Combining Capacitors in Series

Two capacitors, 𝐶 and 𝐶, are connected in series, where 𝐶>𝐶. Which of the following statements correctly relates the total capacitance, 𝐶total, to 𝐶 and 𝐶?

  1. 𝐶=𝐶+𝐶total
  2. 𝐶=(𝐶+𝐶)total
  3. 𝐶=𝐶𝐶total
  4. 𝐶<𝐶<𝐶total
  5. 𝐶<𝐶<𝐶total

Answer

Choice A may look familiar, but this equation would be applicable if the two capacitors were added in parallel, not in series. Therefore, A is incorrect. The equation for the total capacitance for two capacitors in series is 1𝐶=1𝐶+1𝐶.total

This equation does not rearrange or simplify to either equation in B or C, so those two options are incorrect. Although the equation above is not directly stated in any answer option, we can use it to compare 𝐶total, 𝐶, and 𝐶 in size and determine whether D or E is correct.

Because of the inverse properties in the equation above, we can tell that as we add more capacitors in series, the equivalent, or total, capacitance decreases. Thus, for a series combination of capacitors, the total capacitance is less than the capacitance of any one capacitor in the circuit. This means that 𝐶<𝐶total and that 𝐶<𝐶total.

Therefore, choice E is correct.

Example 4: Combining Capacitors in Series

The circuit in the diagram contains two capacitors connected in series. What is the total capacitance of the circuit? Answer to the nearest microfarad.

Answer

We can begin by recalling the equation to find equivalent capacitance for capacitors in series: 1𝐶=1𝐶+1𝐶+⋯.total

Now, let us plug in the values for the two capacitors shown above: 1𝐶=1𝐶+1𝐶=1150+1250.totalµFµF

To add the fractions on the right side of the equation, we will use 750 µF as the least common denominator: 1𝐶=5750+3750=8750.totalµFµFµF

We can now take the inverse of, or flip, both sides of the equation to solve for the final value of 𝐶total: 𝐶=7508=93.75.totalµFµF

Rounding to the nearest microfarad, we find that the total capacitance of the circuit is 94 µF.

Since we now have seen how to add capacitors in series and in parallel, let us exercise both skills in the following examples.

Example 5: Combining Capacitors in Series and in Parallel

A 135 µF capacitor and a 264 µF capacitor can be combined either in series or in parallel. Find the ratio of the total capacitance in parallel to the total capacitance in series. Give your answer to two decimal places.

Answer

Here, we will explore the effects that combining two capacitors in different ways has on their total capacitance. We can begin by recalling the equation for combining capacitors in parallel: 𝐶=𝐶+𝐶+⋯.total

Thus, we can tell, before plugging in any values for 𝐶 or 𝐶, that the total capacitance will be greater than any single capacitance value of either capacitor. Now, plugging in the two values that we have been given, we can find the total capacitance for a parallel combination, which we can call 𝐶parallel: 𝐶=135+264=399.parallelµFµFµF

Now, we can look at the equation for combining capacitors in series: 1𝐶=1𝐶+1𝐶+⋯.total

We can see that the total capacitance for the series combination, which we can call 𝐶series, will be smaller than the individual capacitance values of 𝐶 or 𝐶. Thus, we can expect that 𝐶parallel will be greater than 𝐶series and that the ratio of their values will be greater than one.

Let us now find a value for the total capacitance in series: 1𝐶=1135+1264.seriesµFµF

Rewriting the equation so that we can add the fractions using the least common denominator, we have 1𝐶=8811880+4511880=13311880.seriesµFµFµF

Let us take the inverse of, or flip, the entire equation to solve for 𝐶series by moving it from the denominator to the numerator: 𝐶=11880133=89.32.seriesµFµF

Now that we know 𝐶parallel and 𝐶series, we can find the ratio of their values: 𝐶𝐶=39989.32=4.4671.parallelseriesµFµF

Rounding to two decimal places, we find that the ratio of the total capacitance in parallel to the total capacitance in series is 4.47.

Example 6: Combining Capacitors in Series and in Parallel

The circuit in the diagram contains capacitors connected in series and in parallel. What is the total capacitance of the circuit? Give your answer to the nearest microfarad.

Answer

Here, we have a circuit that contains capacitors connected in series and in parallel. We will work though this circuit in parts and refer to the three capacitors as A, B, and C, as shown in the diagram below.

Capacitors A and B are combined in series, so let us find their equivalent capacitance first. This equivalent capacitance, which we can call 𝐶AB, will represent the total capacitance for the middle wire of this parallel circuit. We can begin with the equation for finding the total capacitance of a series combination and plug in our values for capacitors A and B: 1𝐶=1𝐶+1𝐶+⋯1𝐶=1𝐶+1𝐶=175+155.totalABABµFµF

The least common denominator of these fractions is 825 µF: 1𝐶=11825+15825=26825.ABµFµFµF

We will now take the inverse of, or flip, the whole equation and calculate a value for the equivalent capacitance of A and B: 𝐶=82526=31.73.ABµFµF

Now we can imagine that the circuit is just made up of two capacitors connected in parallel, as shown in the diagram below.

We now effectively have two capacitors combined in parallel, so we can plug their values in and solve for the total capacitance of the entire circuit: 𝐶=𝐶+𝐶+⋯total𝐶=𝐶+𝐶=31.73+35=66.73.totalABCµFµFµF

Rounding to the nearest microfarad, we find that the total capacitance of this circuit is 67 µF.

Let us finish by summarizing some important concepts.

Key Points

  • When combining capacitors in parallel, use 𝐶=𝐶+𝐶+⋯total.
  • When combining capacitors in series, use 1𝐶=1𝐶+1𝐶+⋯total.
  • Capacitors combined in parallel have equal potential differences.
  • Capacitors combined in series store equal charges.

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