Much like resistors are a pain to add in parallel, capacitors get funky when placed in series. The total capacitance of N capacitors in series is the inverse of the sum of all inverse capacitances.
If you only have two capacitors in series, you can use the "product-over-sum" method to calculate the total capacitance:. Taking that equation even further, if you have two equal-valued capacitors in series , the total capacitance is half of their value.
For example two 10F supercapacitors in series will produce a total capacitance of 5F it'll also have the benefit of doubling the voltage rating of the total capacitor, from 2.
Need Help? But how do we figure out what that amount of charge is going to be? Well, there's a trick we can use when dealing with situations like this. We can imagine replacing our three capacitors with just a single equivalent capacitor. If we choose the right value for this single capacitor, then it will store the same amount of charge as each of the three capacitors in series will.
The reason this is useful is because we know how to deal with a single capacitor. We call this imaginary single capacitor that's replacing multiple capacitors the "equivalent capacitor. And it turns out that there's a handy formula that lets you determine the equivalent capacitance.
The formula to find the equivalent capacitance of capacitors hooked up in series looks like this. And if you had more capacitors that were in that same series, you would just continue on this way until you've included all of the contributions from all of the capacitors.
We'll prove where this formula comes from in a minute, but for now, let's just get used to using it and see what we can figure out. Using the values from our example, we get that 1 over the equivalent capacitance is going to be 1 over 4 farads plus 1 over 12 farads plus 1 over 6 farads, which equals 0.
But be careful. You're not done yet. We want the equivalent capacitance, not 1 over the equivalent capacitance. So we have to take 1 over this value of 0. And if we do that, we get that the equivalent capacitance for this series of capacitors is 2 farads. Now that we've reduced our complicated multiple capacitor problem into a single capacitor problem, we can solve for the charge stored on this equivalent capacitor.
We can use the formula capacitance equals charge per voltage and plug in the value of the equivalent capacitance. And we can plug in the voltage of the battery now because the voltage across a single charged-up capacitor is going to be the same as the voltage of the battery that charged it up. Solving for the charge, we get that the charge stored on this equivalent capacitor is 18 coulombs. But we weren't trying to find the charge on the equivalent capacitor. We were trying to find the charge on the leftmost capacitor.
But that's easy now because the charge on each of the individual capacitors in series is going to be the same as the charge on the equivalent capacitor. So since the charge on the equivalent capacitor was 18 coulombs, the charge on each of the individual capacitors in series is going to be 18 coulombs.
Series connections produce a total capacitance that is less than that of any of the individual capacitors. We can find an expression for the total capacitance by considering the voltage across the individual capacitors shown in Figure 1. Entering the expressions for V 1 , V 2 , and V 3 , we get.
Canceling the Q s, we obtain the equation for the total capacitance in series C S to be. An expression of this form always results in a total capacitance C S that is less than any of the individual capacitances C 1 , C 2 , …, as Example 1 illustrates. Find the total capacitance for three capacitors connected in series, given their individual capacitances are 1. With the given information, the total capacitance can be found using the equation for capacitance in series.
The total series capacitance C s is less than the smallest individual capacitance, as promised. In series connections of capacitors, the sum is less than the parts. In fact, it is less than any individual. Note that it is sometimes possible, and more convenient, to solve an equation like the above by finding the least common denominator, which in this case showing only whole-number calculations is Figure 2a shows a parallel connection of three capacitors with a voltage applied.
Here the total capacitance is easier to find than in the series case. To find the equivalent total capacitance C p , we first note that the voltage across each capacitor is V , the same as that of the source, since they are connected directly to it through a conductor.
Conductors are equipotentials, and so the voltage across the capacitors is the same as that across the voltage source. Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source.
Figure 2. Each is connected directly to the voltage source just as if it were all alone, and so the total capacitance in parallel is just the sum of the individual capacitances. The upshot of this is that we add series capacitor values the same way we add parallel resistor values.
Both the product-over-sum and reciprocal methods are valid for adding capacitors in series. But it should be pointed out that one thing we did get is twice as much voltage or voltage ratings. Just like batteries, when we put capacitors together in series the voltages add up. Adding capacitors in parallel is like adding resistors in series: the values just add up, no tricks.
Why is this? Putting them in parallel effectively increases the size of the plates without increasing the distance between them. More area equals more capacitance.
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