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3 . Determine la capacitancia equivalente entre los puntos A y B. Si una diferencia en potencial de 40.0 V es conectada entre los puntos A y B, halle la carga total...

Question

3 . Determine la capacitancia equivalente entre los puntos A y B. Si una diferencia en potencial de 40.0 V es conectada entre los puntos A y B, halle la carga total del circuito. Debe ir mostrando la reduccion del circuito mientras lo va a resolviendo_ 20 puntos 5.0 uF 24 uF4.0 pF12 pF.0 uF8.0 uF

3 . Determine la capacitancia equivalente entre los puntos A y B. Si una diferencia en potencial de 40.0 V es conectada entre los puntos A y B, halle la carga total del circuito. Debe ir mostrando la reduccion del circuito mientras lo va a resolviendo_ 20 puntos 5.0 uF 24 uF 4.0 pF 12 pF .0 uF 8.0 uF



Answers

In Fig. $25-28,$ a potential difference $V=100 \mathrm{V}$ is applied
across a capacitor arrangement with capacitances $C_{1}=10.0 \mu \mathrm{F}$ $C_{2}=5.00 \mu \mathrm{F},$ and $C_{3}=4.00 \mu \mathrm{F} .$ What are (a) charge $q_{3},$ (b) poten-
tial difference $V_{3},$ and $(\mathrm{c})$ stored energy $U_{3}$ for capacitor $3,$ (d) $q_{1}$ (e) $V_{1},$ and $(\mathrm{f}) U_{1}$ for capacitor $1,$ and $(\mathrm{g}) q_{2},(\mathrm{h}) V_{2},$ and $(\mathrm{i}) U_{2}$ for
capacitor 2$?$

How about it? So put those problems. We need Teoh. Find a C three, OK, And so now we're gonna be dio, um, solving to get to that. So we have C 1 12 There we go equals one. See wine, a plus, one seat here. Um, And where this one If we're gonna flip it over, that Cee Lo equals C one C two C one plus C two and here get for for four plus or equals 2.0 for you. Uh and we also have 01 seem to 123 we just see. Wanna see Teoh C one plus C two plus C three, which is, um, two plus Teoh four plus C three and then, um, have see equation is gonna be equal Teoh C four times, See 123 And, um Then we have a c 123 plus C four. And on this, we have eight times you 04 plus C three equals had a plus sea breeze. Bring this down. Okay, so now 1/2 times fusion Well squared. So we're gonna gonna solve the 2.9 time sentiment. The three jewels equals one class times um 16 plus a C three lauded by time, that plus C three equal times, 48 votes squared crime and, um, cool down. So we're gonna bring not over years. And, um, we're gonna bring night over their hero and waas. So this we're going to 0.517 You okay? Well, after we do all that on now, someone comes up by time plus C three. Okay, Mrs. People to 16 point C three. Okay. And, um, now we're gonna have on one of these sides. You, uh you. So we have 25.17 plus 2.517 C three equals 16. Okay. You, uh and then we have plus a you of C three. Ennis is also you see three. So now we have 5.483 You, uh I see three people's 9.17 You, uh, so we can say that C three iss 1.67 You you guys. Thank you.

In this problem, we're considering three capacitors connected to a potential difference. As shown in the circuit diagram. And we want to find the Charge potential difference and energy store for each of these three capacitors. So in order to do this, we're going to rely pretty heavily on the formula Q equals C. V. Which will give the charge on a capacitor if you know its capacitance and voltage and then U. Equals one half C C V squared which gives the potential energy stored in a capacitor. Again, if you know its capacitance and voltage. So four part A. Let's start with Q three, That's going to be capacitance C3. And in this case the voltage across the third capacitor will be equal to the voltage drop. The Uh so when you plug that in, you just get that the charge here is 400 micro combs. Now we want to find the voltage drop. We already said that V three is equal to V which was 100 volts. And for part C we now want to find the potential energy. You that's going to be one half C three times the squared. And again you can just go ahead and plug in the numbers for that and you would get two times 10 to the negative to jules for part D. We want to figure out the charge Q1 on capacitor one and in order to do that, we want to recognize that the charge on capacitor one will be equivalent to the charge on capacitor to because they are connected in series and that's going to be equal to the charge Q. That can be found by using the equivalent capacitance of those jew capacitors times the voltage drop across both capacitors. So the equivalent capacitance for capacitors one in two because they are connected in series is one, oversee one plus one, oversee two and the inverse of that. uh and you can now go ahead and plug in and get that. The charge on capacitor one should be 3.3 times 10 to the -4 columns. Now that we know the charge though, it's pretty easy to find the voltage V one, it's going to be the charge Q. One divided by the capacitance C. One. So using that charge that we found in part D. You should find that the voltage drop across capacitor one is 33.3V. And then for part F we want to find the Energy stored in Capacitor one. You which will be 1/2 C one V one squared again. We know what C. One is. We just found The one. So using that we get 5.6 times 10 to the negative three jewels of energy stored in capacitor one. Yeah, we're going to erase A B and C. Here to make space for capacitor to but it's going to be that same process yet again. So for part G. We already know what the charge is. We found that in part D. It's 3.3 times 10 to the negative. For school loans. For part H. To find the voltage, it's going to be charged on capacity to divided by its capacitance, Which leaves you with 66.7V. And for part I to find the energy stored in a capacitor to it's again 1/2 pass it. It's two times the voltage to that. We just found in part each Squared and you would get 1.1 times 10 to the negative to jules.

We have two capacities of capacities C1 equal to micro far And city equal to four microphone. So this is a scenario and put in some different is 300 volts. So we need to find that and calculate the total energy story the opposite of what we do that you will do nothing about half of cbs and you know that these two are parallel with the battery light. uh so this this is the craft that we look at starting 14 different about this too will be same. Or rather we can see that they can help the equivalent positions of the panel capitated when we see one prosciutto. Right. So you believe that half of seven placido that was two plus four into the r minus six for micro into three 100 square, do it with three into nine independently managed to So you 2.27 jewel will be the answer. Okay, thank you.

Hi there. So for this problem we are given the figure that is shown in here and it is a circuit with a potential difference be which is equal to 100 bulls And it is applied to across these capacitors and we are given the information about the capacities of each of these capacitors. So for C1 we have 10 negro for it's for the capacities to we have five micro for its and for the capacity three we have for micro for its and what we need to obtain for the first part of this problem is the charge of the capacity tour three which we called Q three. Now in order to obtain that, we know that the charge in the figured the voltage in in that in that capacity is simply the the the potential difference that is applied to these capacitors. So we just need to simply use our equation for the charge. We know that the charge is equal to the project between the capacitance of the capacitor times the potential difference. So we have these two values, we know that the capacities is for micro for it times I am 100 balls remember that micro means 10 to the minus sits. So substituting that into the calculator, we obtain that the charge Q three is four times 10 to the minus for columns. So this is a solution for the first part of this problem. Now, for part B we are asked about The potential difference in B three. So that is the potential difference in the capacitor three. But as you can see since this is in parallel to the other Kappas doors and at the same time with the voltage we will have that this potential difference is the same as the potential difference that is applied to all of these capacity tours which is 100 poles. So there is a solution for part be of this problem. Now for parsi we are asked about the stored energy for the capacitor three so we can call this you three that is stored energy. We know that the stored energy is equal to one half of the capacities times the voltage in that capacitor. Now in this case we will have that for the capacitor three we will have that that is one half of the capacities three of this capacitor and its voltage which is B three square. So we need to simply substitute all those values. We know that the capacities of this capacitor is for me, Crow for it's the Voltage is 100 poles. This to the squared. So plugging this into the calculator, we obtain that the energy stored on that capacitor is equal to two times 10 To the -2 jules. So this is a solution for part see of this problem. Now for party we are asked about The charge Q one In the capacity or C one. Now in days again we will have that the well from the figure We can see that the charge Q one In the capacitor, C1 and the charge Q two are the same. So we know that the capacity charge Q1 is equal to the capacitance to and in these times remember that in this case we will have um these two capacitors are in serious so we will have that that is the The effective capacitance of these two which is C1 time C2 over the sum of these two Because they are in serious times the potential difference. So we need to simply substitute all of those values that we know in here. So Q1 is equal to the capacitance one which is 10 me crow for its times the capacitance to which is fine micro for its and of course these are over the sum of these two which is faith in micro forests and these times 100 bowls. So plugging this into the calculator, we obtain that the charge Q1 is equal to 3.33 times 10 To the four the -4 cole lums. So there is a solution for part D of this problem. Now for party we are asked about the potential difference in the capacity tour want. Now in this case we know that they, the um potential difference in that capacity is its charge over its capacities. So we already obtained its charge which is 3.33 times 10 to the -4 columns and this over the capacities of C1 which is if I'm correct 10 micro forests, 10 micro for its so this will give us a potential difference in that capacity of 30 3.3 Bulls. So this is a solution for part E of this problem. Now for part F we are asked about The energy is torn in that capacity or one. So again we use the same equation that we are used before. We know that the energy store is equal to one half of the capacitance of that capacity or and its potential Difference. So substituting those values we obtain that the capacitance. Well, we know the capacitance is 10 micro forests and the bowl touch the potential difference that we obtain is 33.3 poles and this to the square. So substituting that into the calculator, we obtain value off 5.55 Times 10 to the -3 jewels. So this is a solution for part F of this problem. Now for part G, What we are asked is about the charge Q2 but we stayed in the part The of this problem that the charge Q2 Q1 is equal to the charge of the capacity or two because they are in series. So in here we just simply can obtain that the charge Q two is equal to the charge you want. That is equal to the value that we obtain that that is 3.33 3.33 Times 10 to the -4 columns. Now for the other part of this problem. Times age. Well getting here for that part of the problem, we need to calculate the potential difference. Mhm In the capacitor to now we know again that that is defined as the charge you two over the capacity is too. And we know those values. We know that the charge Q3 is 3.33 times 10 to the -4 colom's and these over the capacities to which in this case it is given of five negro for its Find Me Crow for its And this will give us a potential difference of 66.7 bulls. And finally, for part I of this problem, we, what we need to obtain is the energy a store and we know that that is one half of the its capacitance times the voltage on that capacity or the potential difference. So that is one half of the capacities that we know is finally grew for its And the potential difference which is 66.7 bowls and this to the square. So we obtained that the energy is stored on that capacitor Is equal to 1.11 times 10 to the -2 jewels. So this is a solution for this problem. Thank you


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