Capacitors in parallel allow the charge a choice of capacitors. Potential difference is the same with multiple parallel capacitors but the charge adds. Like resistance in series, adding capacitors in parallel increases effective capacitance. The formula for determining effective capacitance is effective capacitance = capacitance 1 + capacitance 2.
Alright so let's talk about adding capacitors in parallel. Now as we all know parallel means that the current has a choice. Now when we're talking about capacitors instead of talking about the current having a choice we talk about the charge having a choice. So we'll write a parallel combination of capacitors just like this, so if I start charging this combination by sending some charge in here, that charge can go in 2 different branches. So it can charge up this capacitor or it could charge up that one. Alright so the potential difference is going to be the same like always is the case, with parallel combinations but the charge is going to add. So the idea is that if I want just one effective capacitor c parallel, to represent this parallel combination the charge that I should put on c parallel is going to be the sum of these 2 charges. Because as I try to charge c parallel all the charge that I put either on c1 or on c2 went onto cp.
Alright so charge adds, potential difference is the same. Alright so let's look and see what happens, q equals q1 plus q2 so since q is equal to capacitance times potential difference we write c parallel delta v equals c1 delta v plus c2 delta v potential difference is the same so delta v cancels. And that gives us a wonderfully simple formula where adding capacitors in parallel, c parallel just equals c1 plus c2. So that means all I have to do is just add which is always nice. It's just like resistors in series, so if I add more capacitors in parallel then the capacitance is going to increase. And this was actually used a lot in the early construction of computers; you would put a bunch of capacitors in parallel and then you'd be able to get a much bigger capacitance that you could get any other way. So we're going to add these 2 capacitors in parallel and I want to determine what is the total charge if the potential difference across these capacitors is 6 volts.
Alright let's do it 2 different ways, if the potential difference is 6 volts then it must be the potential difference across the 3 Farad capacitor so that means that the charge held on the 3 Farad capacitor is going to be potential difference times capacitance 18 joules. Alright and the charge held on to 2 Farad capacitor is going to be again capacitance times potential difference 2 times 6, 12 joules. So the total amount, I'm sorry not joules, Coulomb's so the total amount of charge that I'm going to hold is 18 plus 12 which is 30 Coulomb's alright now let's do it by adding them in parallel first. Well I add them in parallel then I say you know what I really just want to consider one capacitor with a capacitance of 5 Farads. If the potential difference across it is 6 volts much simpler. Alright now one other thing to think about when you've got capacitors in parallel the largest capacitor will store the most energy and that's because capacitors in parallel delta v is the same. So we say energy equals one half c delta v squared, this is constant so the bigger c biggest energy and that's the way that capacitors add in parallel.
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