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Capacitors in series _ Circuits _ Physics _ Khan Academy

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Language: en

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You have to work with a single capacitor,
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connected to a battery is not that complicated,
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but when you have multiple capacitors,
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people are usually much, much more confused.
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There are many different ways
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to connect multiple capacitors.
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But if the capacitors are connected one after the other,
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we call them series-connected capacitors.
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Let's say we do the test and the test is asked of you
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find the charge of the leftmost capacitor.
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Some people may try to do this.
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Since the capacity is the charge divided by the voltage,
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they can enter the capacity of the leftmost capacitor, which is 4,
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to introduce a battery voltage that is 9 volts.
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And by looking for the charge, they will get that leftmost capacitor
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stores 36 pendants, which is a completely wrong answer.
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To try to find the why and find the right one
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to work with this kind of scenario,
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let's look at what happens in this example.
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When the battery is connected, a negative charge
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will start flowing on the right side of capacitor 3,
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which causes the negative charge to be inserted
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on the left side of capacitor 1.
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This causes the negative charge to flow
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on the right side of the capacitor 1
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to the left side of the capacitor 2.
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And this causes the negative charge to flow
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on the right side of capacitor 2
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to the left side of the capacitor 3.
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The charges will continue to do so.
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And it's important to note something here.
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Because of the way the charging process works,
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all the capacitors here should have the same amount of charge,
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stored in them.
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It has to be that way.
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Watching these capacitors charge,
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there is simply nowhere else to go to charge,
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except for the next capacitor in the series.
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It's nice.
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This means that for serial capacitors
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the charge stored in each capacitor,
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will be the same.
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That is, if you find the charge of one of the capacitors,
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you find the charge of all the capacitors.
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But how do we find out what that amount of charge is?
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There is one trick we can use,
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when dealing with situations like this.
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We can imagine that we are replacing our three capacitors
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with one equivalent capacitor.
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If we choose the right value for this one capacitor,
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it will store the same amount of charge
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like each of the three serial capacitors.
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The reason this is useful is
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because we know how to handle a capacitor.
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Let's call this imaginary single capacitor,
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which replaces multiple capacitors,
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"equivalent capacitor".
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It's called the equivalent capacitor,
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because its effect on the circuit
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is equivalent to the total effect,
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that individual capacitors have on this circuit.
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And it turns out that there is a useful formula that lets you
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determine the equivalent capacity.
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The formula for finding equivalent capacity
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of connected capacitors is this one.
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1 on the equivalent capacity
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will be equal to 1 on the first capacity
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plus 1 on the second capacity
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plus 1 on the third capacity.
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And if there were more capacitors in this series,
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you're just going to continue the same way,
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until you turn on all the capacitors.
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We will prove shortly where this formula comes from,
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but for now let's get used to it
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and see what we can find.
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Using the values in our example,
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we get that 1 on the equivalent capacity
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will be 1 on 4 farads plus 1 on 12 farads
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plus 1 on 6 fronts, which is equal to 0.5.
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But be careful.
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We're not ready yet.
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We want equivalent capacity,
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and not 1 on equivalent capacity.
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That is, we need to find out what 1 is at that value of 0.5 we found.
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And if we do that, we get that equivalent capacity
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for this series of capacitors is 2 faroe.
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Now that we have reduced our complex problem with many capacitors
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to a task with a single capacitor,
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we can find how much is stored in this one
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equivalent capacitor.
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We can use the formula the capacity is equal to
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charge on the voltage and enter the value
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of equivalent capacity.
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And we can enter the battery voltage,
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because the voltage across a single charged capacitor
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will be the same as the battery voltage that charges it.
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When we decide to find the charge, we get that the charge,
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stored in this single capacitor is 18 pcs.
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But we did not try to find the charge of the equivalent capacitor.
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We were trying to find the charge
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of the leftmost capacitor.
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But this is easy now because the charge
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of each of the individual capacitors in the series
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will be the same as the charge of the equivalent capacitor.
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Since the charge of this equivalent capacitor
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it was 18 pillars,
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the charge of each of the individual series capacitors
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will be 18 pence.
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This process can be confusing,
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so let's try another example.
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This time, let's say you have 4 connected capacitors in series
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to a 24 volt battery.
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The arrangement of these capacitors
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looks a little different than the last example,
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but all of these capacitors are still sequential,
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because they are connected one after another.
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In other words, the charge has no other choice,
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except to flow directly from a capacitor
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straight to the next capacitor.
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These capacitors are still considered sequential.
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Let's try to find out what the charge is,
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to be stored in the 16-Far Capacitor.
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We will use the same process as before.
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First, imagine replacing the four capacitors
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with a single equivalent capacitor.
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We will use the formula to find
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equivalent capacitance of sequential capacitors.
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As we enter the values, we find
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that 1 on the equivalent capacity would be 0.125.
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Watch out.
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We still have to find how much is 1 on this value,
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to get that equivalent capacity for that circuit
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there will be 8 farads.
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Now that we know the equivalent capacity,
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we can use the formula that tells us
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that the capacity is equal to the charge on the voltage.
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We can enter the value of the equivalent capacity, 8 farads.
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And since we now have only one capacity,
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the voltage on this capacity
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will be the same as the battery voltage,
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which is 24 volts.
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We find our imaginary equivalent capacitor
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will store a charge of 192 pence.
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That means everyone's charge
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of the individual capacitors, it will also be 192 pence.
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And that gives us our answer,
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that the charge of the 16-Farad capacitor would be 192 pence.
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In fact, we can take this further.
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Now that we know the charge of each capacitor,
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we can find the voltage,
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which will exist on each of the individual capacitors.
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Again we will use the fact that capacity
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is the charge on the voltage.
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If we enter the values for capacitor 1,
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we introduce a capacity of 32 farads.
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The charge stored by Capacitor 1 is 192 pence.
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And we can find the voltage across capacitor 1
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and we get 6 volts.
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If we do the same calculation for each of the other 3 capacitors,
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always being careful to use their specific values,
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we will get that the voltages on the capacitors
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are 2 volts through the 96-far capacitor,
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12 volts on the 16-far capacitor
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and 4 volts on the 48-Far Capacitor.
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The real reason we went through this is,
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because I wanted to show you something nice.
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If you collect the tensions that exist
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on each of the capacitors, you get 24 volts,
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just like the value of the battery.
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This is no coincidence.
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If you collect the voltages on the components
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in each such chain from one line, the sum of the voltages
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it will always be equal to the battery voltage.
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And this principle allows us to find the formula,
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which we used for equivalent capacity
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of sequential capacitors.
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To find the formula, let's say
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that we have three capacitors with capacities C1, C2 and C3,
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connected in series to a battery with a voltage of V.
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Now we know that if we collect the voltage across each capacitor,
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this will add up to the battery voltage.
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Using the capacity formula,
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we can see that the voltage across a single capacitor
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will be the charge of this capacitor
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divided by its capacity.
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The voltage across each capacitor
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will be Q / C1, Q / C2 and Q / C3, respectively.
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I didn't record Q1, Q3 or Q3 because, remember,
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all charges on the series capacitors
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they will be the same.
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These voltages should add to the battery voltage.
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I can post a general Q article,
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because it has it in every member on the left.
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And now I will divide each side by Q.
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I did this because, look what we got
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on the right side of this equation.
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Battery voltage divided by stored charge
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is equal to 1 on the equivalent capacity,
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because Q / V equals equivalent capacity.
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And here it is.
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This is the formula we used,
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and it comes from here.
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We found it from the fact that the tensions
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on these successive capacitors
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should collect the battery voltage.

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