Having to deal with asingle capacitor hooked up to a battery isn’tall that difficult, but when you havemultiple capacitors, parties typically getmuch, so much better flustered. There’s all kindsof different ways to hook up multiple capacitors. But if capacitorsare connected one after the other inthis way, we call them capacitorshooked up in series. So say you were takinga test, and on the test it asked you to find the chargeon the leftmost capacitor.What some people mighttry to do is this. Since capacitance is thecharge divided by the voltage, they might plugin the capacitance of the leftmost capacitor, which is 4 farads, plug in the voltage of thebattery, which is 9 volts. Solving for the charge, they’dget that the leftmost capacitor collects 36 coulombs, whichis totally the wrong response. To try and figure outwhy and to figure out how to properly deal withthis type of scenario, let’s look at what’s actuallygoing on in this example. When the battery’s hookedup, a negative charge will start to flow fromthe right side of capacitor 3, which makes a negativecharge get deposited on the left of capacitor 1. This makes anegative charge flow from the rightside of capacitor 1 on to the left sideof capacitor 2. And that makes anegative charge flow from the rightside of capacitor 2 on to the left sideof capacitor 3. Indicts willcontinue doing this. And it’s important tonote something here. Because of the mode thecharging process undertakings, all of the capacitors heremust have the same amount of price stored under them. It’s got to be that route. Looking at how thesecapacitors charge up, there’s just nowhereelse for the charge to go but on to the nextcapacitor in the line. This is actually good bulletin. This meant that forcapacitors in series, the charge storedon every capacitor is going to be the same. So if you find the chargeon one of the capacitors, you’ve found the chargeon all of the capacitors.But how do we figure outwhat that extent of bill is going to be? Well, there’s atrick we can use when dealing withsituations like this. We can imagine replacingour three capacitors with precisely a singleequivalent capacitor. If we pick the right valuefor this single capacitor, then it will accumulate thesame quantity of price as each of the threecapacitors in series will. The conclude this isuseful is because we know how to deal witha single capacitor. We call this imaginarysingle capacitor that’s replacingmultiple capacitors the “equivalent capacitor.” It’s called theequivalent capacitor because its effecton the circuit is, well, equivalentto the sum total effect that the individual capacitorshave on the circuit.And it turns out thatthere’s a handy formula that lets you determine theequivalent capacitance. The formula to find theequivalent capacitance of capacitors hooked upin series looks like this. 1 over the equivalentcapacitance is going to equal 1 overthe first capacitance plus 1 over the secondcapacitance plus 1 over the third capacitance. And “if youre having” more capacitorsthat were in that same series, you would justcontinue on this way until you’ve includedall of the contributions from all of the capacitors. We’ll prove where this formulacomes from in a minute, but for now, let’s justget used to using it and meet what we can figure out. Using the valuesfrom our sample, we get that 1 over theequivalent capacitance is going to be 1 over 4farads plus 1 over 12 farads plus 1 over 6 farads, which equals 0.5. But be careful. You’re not done hitherto. We crave the equivalentcapacitance , not 1 over the equivalentcapacitance. So there is a requirement to make 1 over thisvalue of 0.5 that we concluded. And if we do that, we get thatthe equivalent capacitance for this series ofcapacitors is 2 farads. Now that we’ve reduced ourcomplicated multiple capacitor question into a singlecapacitor difficulty, we can solve for the chargestored on this equivalent capacitor.We can use the formulacapacitance equals charge per voltage and plug inthe value of the equivalent capacitance. And we can plug in thevoltage of the battery now because the voltage acrossa single charged-up capacitor is going to be the same as thevoltage of the artillery that charged it up. Solving for the charge, weget that the charge stored on this equivalentcapacitor is 18 coulombs. But we weren’t trying to findthe charge on the equivalent capacitor. We were trying tofind the charge on the leftmost capacitor. But that’s easy nowbecause the charge on each of the individualcapacitors in series is going to be the same asthe charge on the equivalent capacitor.So since service charges onthe equivalent capacitor was 18 coulombs, service charges on each of the individualcapacitors in series “re gonna be all” 18 coulombs. This process can beconfusing to beings, so let’s try another example. This time, let’s say you hadfour capacitors hooked up in series to a 24 -volt battery. The layout ofthese capacitors appears a little differentfrom the last example, but all of these capacitorsare still in series because they’re hooked upone right after the other. In other terms, thecharge has no choice but to flow directlyfrom one capacitor straight to the next capacitor. So these capacitors are stillconsidered to be in series. Let’s try to figureout the charge that’s going to be stored onthe 16 -farad capacitor. We’ll use the sameprocess as before. First we imagine replacingthe four capacitors with a singleequivalent capacitor. We’ll use the formulato find the equivalent capacitance ofcapacitors in series. Plugging in our values, wefind that 1 over the equivalent capacitance is goingto equal 0.125. Be careful. We still have to take1 over this quality to get that the equivalentcapacitance for this circuit is going to be 8 farads.Now that we know theequivalent capacitance, we can use theformula capacitance equals charge per voltage. We can plug in the value ofthe equivalent capacitance, 8 farads. And since we have asingle capacitor now, the voltage acrossthat capacitor is going to be the same as thevoltage of the artillery, which is 24 volts. So we is my finding that our imaginaryequivalent capacitor would collect a chargeof 192 coulombs. This means thatthe charge on each of the individual capacitors isalso going to be 192 coulombs. And this pays us ouranswer, that the charge on the 16 -farad capacitoris going to be 192 coulombs. In information, we can go even further. Now that we know thecharge on each capacitor, we can solve forthe voltage that’s going to exist across eachof the individual capacitors. We’ll again use thefact that capacitance is the charge per voltage. If we plug in the valuesfor capacitor one, we’ll plug in acapacitance of 32 farads.The blame that capacitorone supermarkets is 192 coulombs. So we can solve for thevoltage across capacitor 1, and we get 6 volts. If we were to do thesame computation for each of the other three capacitors, always being careful that we use their particularvalues, we’ll come that the voltagesacross the capacitors are 2 volts across the9 6-farad capacitor, 12 volts across the1 6-fard capacitor, and 4 volts across the4 8-farad capacitor. Now, the real reason Ihad us “re going through” this is because I wanted toshow you something neat.If you make sense the voltagesthat exist across each of the capacitors, you’ll come 24 volts, the same as thevalue of the artillery. This is no coincidence. If you make sense the voltagesacross the components in any single-loop circuit likethis, resources in the amount of the voltages is always going to equalthe voltage of the battery. And this principlewill actually make us derive the formula we’vebeen using for the equivalent capacitance ofseries capacitors. To deduce thisformula, let’s say we’ve got three capacitors withcapacitances of C1, C2, and C3 hooked up in series to abattery of voltage V.We now know that if we add up thevoltage across each capacitor, it’s got to add up to thevoltage of the artillery. Using the formulafor capacitance, we can see that the voltageacross an individual capacitor is going to be the chargeon that capacitor divided by its capacitance. So the voltageacross each capacitor is going to be Q over C1, Q over C2, and Q over C3, respectively. I didn’t write Q1, Q2, or Q3 because remember, all the charges oncapacitors in series are going to be the same. These voltages have to add upto the voltage of the battery. I can pull out acommon factor of Q because it’s in eachterm on the left.And now I’m going todivide the two sides by Q. I did that becauselook at what we’ve got on the right-handside of this equation. The voltage across the batterydivided by the charge accumulated is just equal to 1 overthe equivalent capacitance, because Q over V is equal tothe equivalent capacitance. And here it is. This is the formulawe’ve been using, and this is where it comes from. It’s derived from thefact that the voltages across thesecapacitors in series have to add up to thevoltage of the battery ..