In Fig. E24.20

Question

In Fig. E24.20

, C1 = 6.00 uF, C2 = 3.00 uF, and C3 = 5.00 uF. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge on C2 is 30.0 mC. (a) What are the charges on capacitors C1 and C3? (b) What is the applied voltage Vab?

Accepted Answer

Welcome back everyone. We are taking a look at three different capacitors arranged in the following arrangement here we are told that the capacitance for capacitor one is nine micro fare adds for capacitor two is six micro fare adds and four capacitor three is four micro fare adds. Now there is a voltage source connected across the combination of all of them and that is going to give a voltage per source and we are tasked with finding a couple of different things here. Right. We are tasked with finding a what? Q one and Q three, R and B. What the voltage per source actually is. Now one other bit of information that we are told is that the charge on our second capacitor is micro columns. So let's go ahead and start with part A here. Let's try to find the charge on capacitor number one. Our charge is going to be equal to the voltage uh times our capacitance. But what is V one here? Well, if you look C one and C two are arranged in parallel meaning V one is going to be equal to VT Now, V two of course is just Q two divided by C two. And we have those values. This will just be 40 micro columns divided by six micro fare adds, giving us a voltage of 6. volts or a potential difference of 6.67 volts. Now we can go ahead and calculate Q one, this is just going to be 6.67 times nine times 10 to the negative six, giving us 60 micro columns. Great. So now what about the charge of capacitor three? Well if you'll see capacitor three is in series with the parallel combination of C one and C two meaning Q three is just going to be Q one plus Q. To this of course is 40 micro columns plus 60 micro columns which is equal to 100 micro columns. Great. Moving on to part B here. Now we need to find the potential difference per that source. So this is going to be equal to Q total divided by our equal capacitance. Now be careful here, Q total does not necessarily mean adding up all three charges and you'll see why that is in just a second. I'm actually gonna scroll down just a little bit here let's go ahead and try to calculate what our equivalent equivalent capacitance is according to the arrangement here we have that are equivalent capacitance or one over equivalent capacitance will be one over the capacitance of C three plus the equivalent capacitance of capacitors one and two. Now the equivalent capacitance for capacitors one and two since they are in parallel is simply just C one plus C two, this is equal to nine plus six giving us 15 micro fare adds. Now going back down to this, we have we have the one over because as we know C three is in series with our parallel combination. So this is just going to be 1/4 micro fare adds plus 1/ micro fare adds. Giving us that are equivalent capacitance is 3.16 micro fare adds. Now what is our Q total here? Like I mentioned earlier, it's not just adding them all up. The queue total is going to be the charge of our equivalent capacitance. So in this case we are going to be adding the charges of our first capacitor and our second capacitor, this gives us 40 plus 60 giving us micro columns. Now we are ready to find the potential difference per source. This will be 100 micro columns divided by 3.16 micro fare adds, giving us 31.64 volts. So now we have found the charges on all of our capacitors and we have found the potential difference per source and this all corresponds to answer choice. C Thank you all so much for watching. Hope this video helped. We will see you all in the next one

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