Concept: Solving Capacitor Circuits10m
Hey guys. In this video we're going to be doing what I call the second half of solving the circuit problems with multiple capacitors in them. Alright, let's get to it. In circuit problems you're going to need to find the charge and the voltage of the different capacitors that make up whatever your circuit is, okay? We saw last time there are two types of connections, you have series connections and parallel connections, we also know that the equivalent capacitance of multiple capacitors in series is given by this equation, and the equivalent capacitance of multiple capacitors in parallel is given by this equation, what we need to do is take the step further, each of these capacitors in series share a charge with each other, okay? So, I know that one of these capacitors has 1 Coulomb of charge then the others have to have 1 Coulomb of charge no matter what the charge has to be the same for each capacitor in series, not only that, but they share the charge with the equivalent capacitor as well, so if each of those three capacitors is 1 Coulomb of charge then the equivalent capacitor also has to have 1 Coulomb of charge, okay? This is incredibly important to know, for parallel connections each of these three capacitors shares a voltage with one another. So, if one of the capacitors has, let's say 2 volts of voltage, they all have to have 2 volts of voltage, not only that but they share the voltage with the equivalent capacitor as well. So, if each of these capacitors has 2 volts the equivalent capacitor has to be two volts as well, okay? This has to always be true. Alright, and it's very, very important to remember the rules for series and parallel connections, not only the equations to find the equivalent capacitance but to remember that series capacitors share charge, parallel capacitors share voltage, okay? So, what we have is basically an algorithm for solving these problems, a series of steps, the first is to collapse everything down into one equivalent capacitor, that was sort of what I call the first half of the problem and we did that in the last section, next, you're going to do one calculation to find the voltage and the charge on the equivalent capacitor, okay? Usually you're going to know one or the other, lastly you're going to work backwards. Alright, noting the both the voltage and the charge on each of the capacitors as you work backwards, okay? What do I mean by working backwards? That's what we're going to see in this example right here.
What is the charging voltage of each of the capacitors in the following circuit, step one, we have to collapse this down into a single equivalent resistor the first thing, we want to do is attack the loops, okay? We have one loop with a 2 and 1 farad capacitor that are in parallel, okay? So, once that loop is collapsed our circuit is going to look like this, this is the 6 farad capacitor, this is 10 volts, that has not been changed, okay? The loop was formed by two capacitors in parallel, that's easy enough you just add them 2 plus 1 is 3 farads, okay? Now, what we have are two capacitors in series, so this is going to be absolutely equivalent to a single capacitor connected to the battery, the battery is still 10 volts, we're going to use our shortcut equation because this is two capacitors in series and we'll say the equivalent capacitance is C1, C2 over C1, plus C2 which is 3 times 6 over 3 plus 6, which is 9, this is 18 over 9, which is 2, so this is 2 farads, okay? This is the goal of the first step of the algorithm, is to get to this point where we have a single capacitor connected to a single battery, the reason is, is we already know how to solve this problem, we know that the two share a voltage so the batteries is at 10 volts then the capacitor is at 10 volts as well, this means that the charge on the equivalent capacitor has to be the capacitance times the voltage, which is 2 times 10, which is 20 coulombs, okay? Now, there's going to be a lot of things to keep track of in this problem. So, what I want to do is draw a little table where we can keep track of our results, okay? In this table, we're going to have 4 capacitors, okay? I drew this table a little too far down because you guys can't see it there, that's better, we're going to add 4 capacitors, we're going to have I'll call this C, let's not bother with that, we'll say the 2 farad capacitor the 1 farad capacitor the 6 farad capacitor and the equivalent capacitor, and what we want to know, is what is the voltage and what is the charge on each, this is a really good way to keep track of our results and to make sure that we know if we're on the right track and what we're missing if we're missing anything, okay? So, we know the voltage on the equivalent capacitor is 10 volts and we know the charge on the equivalent capacitor is 20 coulombs, okay? Now, let's work backwards as I said. Well, this circuit is exactly equivalent to the circuit in the picture before it, right? All we're going to do is expand the last capacitor that we collapsed, okay? So, remember, we went from this upper left one here to this one and now we're going backwards, okay? Because these two capacitors here were formed in series, we know they have to have the same charge as the equivalent capacitor, okay? The equivalent capacitor had 20 coulombs of charge. So, each of these are going to have 20 coulombs of charge and this one's a 3 farad capacitor and this one's a 6 farad capacitor. So, boom right away our 6 farad capacitor has 20 coulombs of charge, that's another entry in our table. So, let's find the voltages for each of these, the voltage for the 3 farad capacitor is going to be the charge over the capacitance. So, it's going to be 20 coulombs over 3 which is about 6.7 volts, the voltage for the 6 farad capacitor is going to be 20 over 6, which is about 3.3 volts. So, now we know that the 6 farad capacitor is 3.3 volts of voltage, we're going to go back one more step into our original circuit, okay? We're already done with the 6 farad capacitor. So, we can ignore it, we know that the upper capacitor here is two farads and the lower capacitor is one farad but we also know that their equivalent capacitor had a voltage of 6.7 volts, that's important because these are formed in parallel. So, they share that equivalent, sorry, they share the voltage with that equivalent capacitor. So, each of these also has 6.7 volts of voltage, so right away, we can put 6.7 volts for each of these, all we have to do now, is find the charge for each of those and remember that charge is just capacitance times voltage. So, that's going to be 2 times 6.7 which is going to be 13.4 coulombs, and lastly for the 1 farad it's going to be 1 times 6.7, which is going to be 6.7 coulombs, those are the last two entries that we need, 13.4 coulombs, 6.7 coulombs, okay? And this whole process can take a while but if you set up this table and you go slowly collapsing everything down to one and then working backwards you'll always be able to fill that table with the right answers. Alright, that wraps up our discussion of how to solve these complex circuits, thanks for watching guys.
Problem: What is charge and voltage across each capacitor below?5m
Example: Find Charge of One Capacitor (Simple)4m
Hey guys. Let's do an example. What is the charge on this 3 farad capacitor below? And all the problems we've done so far, we've asked for the charge and the voltage on every capacitor in circuit. Now, however, what we're asking is a targeted question, what does it charge at a particular capacitor, okay? So, we need to modify our algorithm a little bit, the first two steps are going to be the same, collapse everything down into a single equivalent capacitor and find the charging voltage on the capacitor, but then we're not going to work backwards finding everything, we're only going to work backwards until we've answered our targeting question, okay? So, let's begin, first the 1 farad, 3 farad and the bottom ones are all on loops. So, we want to say that they're in parallel but these two are in series and we need to address that first, okay? So, this is going to stay at 1 farad capacitor, this is going to stay a 5 volt battery, this is going to stay a 3 farad capacitor, this one we're going to use our shortcut equation C equivalent is C1, C2 over C1 plus C2, which we can use because there are only two capacitors, this is 2 times 4 over 2 plus 4, which is 6, 8 over 6, which is 1.3 farads, okay? Now, we can collapse everything down here, because all the capacitors are actually in parallel, it doesn't matter about this battery, this battery right here, this whole branch, it doesn't matter that it's there, all those capacitors are still in parallel, because you can ignore that branch and you can think about this whole thing right here,as being a loop, ignoring that branch. So, those are in parallel, these are in parallel, they're all in parallel, okay? So, this is going to be a 5 volt battery connected to a single equivalent capacitor, which is just the sum, right? It's going to be 1 farad plus 3 farads plus 1.3 farads which is 5.3 farads, from there we can find that the charge on this, however, we don't actually need to go this far, we know that because this is a single capacitor connected to a single battery that it's going to have 5 volts of voltage because it was made up by a bunch of capacitors in parallel, each of those capacitors are going to carry the same 5 volts including our 3 farad capacitor, our 3 farad capacitor is at 5 volts. So, what's the charge on the 3 farad capacitor? it's just the capacitance times the voltage which is 3 times 5 which fifteen coulombs, okay? So, this is an example of where you can modify the algorithm to answer a targeted question. Alright, thanks for watching guys.
Problem: What is the voltage of the battery below?4m
Problem: What is the charge on the 5 F capacitor? (hint: be careful with series vs parallel)7m
Three capacitors C1 = 4µF, C2 = 8µF, and C3 = 6µF are connected across a 20V battery as shown.
Three capacitors C1 = 4μF, C2 =8μF, and C3 = 6μF are connected across a 20 V battery as shown.
a) Find the voltage drop across C1.
b) Find the voltage drop across C2.
c) Find the voltage drop across C3.
d) Find the charge on across C2.
e) Find the charge on across C3.
Find the charge out of the 10V battery in the circuit?
(a) 2.5 μC
(b) 48 μC
(c) 72 μC
(d) 96 μC
Three capacitors are connected to a 1 V battery as shown beloe. The capacitances are C1 = 3 F, C2 = 1 F and C3 = 5 F. Rank the capacitors according to the charge they store from largest to smallest.
a. Q 3 > Q 1 > Q 2
b. Q 1 > Q 2 = Q 3
c. Q 2 = Q 3 > Q 1
d. Q 1 = Q 2 = Q 3
e. Q 1 > Q 3 > Q 2
Consider the circuit shown in the sketch. When the capacitors have their final charges, what is the charge on C2?
Three capacitors are arranged as shown below. C1 = 3.0 x 10-6 F. C2 = 6.0 x 10-6 F and C3 = 5.0 x 10-6 F. When the capacitors have reached their final charges, the charge on C1 is q1 = 3.0 x 10-4 C. What is the voltage V2 across C2?
Consider the circuit of capacitors shown below. The equivalent capacitance of the circuit is 9.0 μF.
Determine the charge stored on the 4.0 μF capacitor. Assume that voltage across nodes A and B is 12 volts.
Three capacitors are connected in the circuit as shown in the figure.
(a) What is the equivalent capacitance in the circuit?
(b) What is the energy stored in each capacitor?
Four 4μF capacitors are connected into a square. A battery of unknown voltage is connected between the diagonal corners of the square. As a result, one of the 4μF capacitor is charged to 0.12μC.
a) Draw the schematic of the connection.
b) How much energy is stored in this 4 μF capacitor?
c) Find the total amount of charge provided by the battery.
d) Find the voltage of the battery.