What is the charge on the 5 F capacitor? (hint: be careful with series vs parallel)

Hey, guys. I hope you're able to solve this one on your own. If not, here's a little bit of help. What is the charge on the 5 farad capacitor? Unlike most of the problems that we've done, this is a targeted question which means we need to use our modified algorithm where we're going to collapse wanna want to address. Let's start with the upper loop because it's the more complicated one. In it, we have 2 pairs of the pair of 2 farad capacitors and the pair of 4 farad capacitors. Okay. These pairs might look like they're in parallel but they're actually each in series. If you look between the 2 farad capacitors, there is a direct connection. And between the 4 farad capacitor, there is also a direct connection. When you have a direct connection, that's gonna be series, not parallel. So we'll use our quick equation for capacitors in series because we only have 2 capacitors for each pair. For the green pair, that's going to be 2 times 2. 2 +2 is 4. That's 4 divided by 4, which is 1 farad. For the red, this is gonna be 4 times 4 over 4+4, which is 8. That's 16 over 8, which is 2 farads. Okay? So let's redraw this. We have our battery, we have one capacitor, another capacitor, and our 3rd capacitor, and then at the second loop which is untouched. Okay? So in the upper loop, the leftmost capacitor was the red pair, that's 1 farad. The middle was untouched, that was already 1 farad. The rightmost capacitor was the red pair which is now 2 farads. The bottom loop unaffected, that's 3 farads and 5 farads. Now the upper loop is 3 capacitors in parallel, the lower loop is 2 capacitors in parallel. We can address those very easily. We have our to add those, and that's 4. The lower loop was a 3, and a 3, and a 3, and have to add those and that's 4. The lower loop was a 3 and a 5. We just have to add those and that's 8. What we're left with is 2 capacitors in series so we can combine them into a single equivalent capacitor, and all we have to use is our shortcut equation given over there to the left. C equivalent is going to be 4 times 8 over 4 +8 which is 12 and that's about 2.7 farads. So we have an 8 volt battery and a 2.7 farad capacitor. What's the charge on that capacitor going to be? Well, q equivalent is gonna be 2.7 times 8, just c times v, which is about 21.6 coulombs. Now we have to work backwards. Okay? Let me give myself some breathing room here. So first, I'm gonna redraw the circuit directly above me. The circuit that immediately preceded, the one we were just on, those 2 capacitors, farad, the lower one is 8 farad, those 2 capacitors are in series. So they're gonna have the same charge as their equivalent capacitor which was 21.6 coulombs. So 21.6 coulombs 21.6 coulombs. Now we can find the voltage due to both of these if we wanted but notice all we're interested in is the 5 farad capacitor right? The 5 farad capacitor is part of this lower loop which is this lower equivalent capacitor. So we can actually just ignore the upper one. We We don't want anything to do with it. What's the voltage of the lower one then? Well, that's 21.6 divided by 8. It's the charge Okay? And now we can work backwards again. Okay? And now we can work backwards again. Where once again, we are completely ignoring that upper group. Okay. We don't care about that upper group. We know the voltage right here 2.7 7, of our equivalent capacitor and since these two capacitors right here are in parallel, they each carry that 2.7 volts. We have the 3 farad capacitor and the 5 farad capacitor, each at 2 point 7 volts. So our question was what is the charge on the 5 farad capacitor? Well we know enough to solve that now. The charge on the 5 farad capacitor is going to be the capacitance times the voltage which is just 13.5 coulombs. Okay. 13.5 coulombs. You don't have to work all the way backwards. For this problem we completely ignored this equivalent capacitor because it led to the upper loop that we did not care about. The capacitor we were interested in was in the lower loop so we found the equivalent capacitor, the total one up here, which we absolutely needed to do. But once we got past that, we started completely ignoring the upper loop because it had nothing to do with what we were trying to do. Alright? I hope this helped. Thanks for watching, guys.

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What is the charge on the 5 F capacitor? (hint: be careful with series vs parallel)

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