Suppose in Fig. 24–27 that C₁ = C₃ = 8.0...

Question

Suppose in Fig. 24–27 that C₁ = C₃ = 8.0μF, C₂ = C₄ = 16μF, and Q₃ = 21μC. Determine the voltage across each capacitor.

Accepted Answer

Welcome back. Everyone in this problem, consider a circuit where four capacitors are connected in series and parallel. The first pair consists of capacitors, C one and C two with values C one equals 12 mills and C two equals 24. Miller. The second pair consists of capacitors, C three and C four with value C three equals 12 miller and C four equals 36 milly. It is known that the charge on C three is 24 milli coulombs. Calculate the charge on each capacitor. Here we have a diagram showing our setup. And for our answer traces, A says Q one and Q two are 15 mill Coomes while Q three and Q four are 24 mill Coomes B says they are 21 mill Coomes and 24 mill Coomes respectively. C 15 and 20 mile Coomes respectively, and D 21 and 20 mile Coomes respectively. Now, if we're going to figure out the charge, there are a few things that we note here. OK. Now notice that in this setup, the capacitors are grouped in pairs where C one and C two are connected in series to form the first pair. While C three and C four are connected in series to form the second pair. And then the two pairs are connected in parallel. Now, if we use what we already know that Q three equals 24 milli columns for capacitors connected in series, the charge on each capacitor is the same. So that must mean then that Q one is going to be equal to Q two and Q three must be equal to Q four. So since Q three is 24 Milli Coomes, this tells us then that Q four is also going to be 24 mile Coomes. So we figured out the charge on our two capacitors here. OK. Now we need to figure out Q one and Q two. What do we know about charge and capacitance? Well, recall, OK, that the charge Q is equal to the capacitance multiplied by the voltage. In this case, then since we know Q one equals Q two, then their charge is going to be equal to the equi equivalent capacitance for our first pair. So let's call that equivalent capacitance for the first pair one multiplied by the voltage for or sorry multiply, yes, multiplied by the voltage across C one and C two. If we can find both of those, then we should be able to figure out Q one and Q two. Now let's start with the equivalent capacitance. OK. How can we solve for that? Well, OK, let's, let's make a note here. OK. We know that because coc one and C two are in series, OK? Then that means the equivalent capacitance are the reciprocal of the equivalent capacitance is going to be equal to the sum of the reciprocal of both capacitors. OK. So in this case, that's gonna be equal to, we know that C one is 12 mills, C two is 24 millirads. OK? Now, when we add both of these, this means that one, well, the reciprocal of the equivalent capacitance equals three divided by 24 mills, which tells us then that the equivalent capacitance for our first pair is equal to 24 mills divided by three or eight mills. So we have the first part of our problem next. OK? We need to figure out the voltage across the leg. OK? Or the voltage across C one and C two. Well, if we go back to our problem here, remember we already know the charge for AC three and C four. OK. And we also know that since they are collected in parallel, then the voltage across C three and C four is also going to be the same voltage across C one and C two. OK. So in that case, if we can find a voltage across C three and C four, we can use it in the charge for C one and C two. OK. Now how can we figure out what that voltage is? Well, from C three and C four OK. Using the same thing we know about charge, remember we know that charge equals the capacitance multiplied by the voltage. So that means then in this case, the voltage is going to be equal to the charge, which we already know is Q three and Q four divided by the equivalent capacitance in our second pair. No, the equivalent capacitance in our second period. That's one, the reciprocal of the equivalent capacitance. Sorry. In the same way, we found it earlier is going to be equal to one divided by 12 mills plus one divided by 36 mills. Now, when we add both of these, we get it to be one divided by nine miller adds, which means then that their equivalent capacitance for the second pair is going to be nine miller. Thus, that tells us then that the voltage V is going to be equal to 24 mile Coombs divided by nine mills, which is going to be equal to 2.67 volts. No, this is really important. Now, we have the equivalent capacitance for our first pair and we have the voltage OK, across CC one and C, well across C three and C four, which is the same voltage across C one and C two. Then this means then this means that the charge QQ one which is Q two is going to be equal to the equivalent capacitance eight mills multiplied by our voltage of 2.67 volts which equals 21.36 mile coons. Or if we round up to two significant figures, approximately 21 mile coulombs. Therefore, the charge on capacitors, one and two are 21 milli coons and the charge on capacitors three and four are 24 milli coons. If we look back at our answer choices. OK. That tells us then that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Suppose in Fig. 24–27 that C₁ = C₃ = 8.0μF, C₂ = C₄ = 16μF, and Q₃ = 21μC. Determine the voltage across each capacitor.

Key Concepts

Capacitance

Series and Parallel Circuits

Voltage Division

Watch next

Suppose in Fig. 24–27 that C1 = C3 = 8.0μF, C2 = C4 = 16μF, and Q3 = 21μC. Determine the voltage across each capacitor.

Key Concepts

Capacitance

Series and Parallel Circuits

Voltage Division

Watch next

Suppose in Fig. 24–27 that C₁ = C₃ = 8.0μF, C₂ = C₄ = 16μF, and Q₃ = 21μC. Determine the voltage across each capacitor.