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Exercise 28.AF.19each case below; cakculate ihe time constant then find values for the voltage acro5s the capacitor and voltage across the resistor when t = Part (a...

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Exercise 28.AF.19each case below; cakculate ihe time constant then find values for the voltage acro5s the capacitor and voltage across the resistor when t = Part (a) uses resistance and capacitance values that can be set wilh Ihe sliders, 50 you can use the animation verly your cakulations. Part (b) uses values ouiside the lider ranges each case, Ihe battery voltege is 10 V, (a} Let R 18 MQana 1,4 HF.Find Vz and Vc when 5.00(D) Lcz R 2.4 Mnand c -2.5 HF.Va and Vc when5.00Summary Incteased take M

Exercise 28.AF.19 each case below; cakculate ihe time constant then find values for the voltage acro5s the capacitor and voltage across the resistor when t = Part (a) uses resistance and capacitance values that can be set wilh Ihe sliders, 50 you can use the animation verly your cakulations. Part (b) uses values ouiside the lider ranges each case, Ihe battery voltege is 10 V, (a} Let R 18 MQana 1,4 HF. Find Vz and Vc when 5.00 (D) Lcz R 2.4 Mnand c -2.5 HF. Va and Vc when 5.00 Summary Incteased take More celectez Mott Samt tne same celef time for the capacitor charge incrcascd [akcs Sclcc: Mare Ine Same s2ec timne Ior Lne cdpacilor charac



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A $2.0-\mu \mathrm{F}$ capacitor is charged through a $30-\mathrm{M} \Omega$ resistor by a 45 $\mathrm{V}$ battery. Find $(a)$ the charge on the capacitor and $(b)$ the current through the resistor, both determined $83 \mathrm{~s}$ after the charging process starts. The time constant of the circuit is $R C=60 \mathrm{~s}$. Also, $$ q_{\infty}=V_{\infty} C=(45 \mathrm{~V})\left(2.0 \times 10^{-6} \mathrm{~F}\right)=9.0 \times 10^{-6} \mathrm{C} $$ (a) $q=q_{\infty}\left(1-e^{-t / R C}\right)=\left(9.0 \times 10^{-5} \mathrm{C}\right)\left(1-e^{-83 / 60}\right)$ But $\quad e^{-83 / 60}=e^{-1.383}=0.25$ Then substitution gives $$ q=\left(9.0 \times 10^{-5} \mathrm{C}\right)(1-0.25)=67 \mu \mathrm{C} $$ (b) $i=i_{0} e^{-t / R C}=\left(\frac{45 \mathrm{~V}}{30 \times 10^{6} \Omega}\right)\left(e^{-1.383}\right)=0.38 \mu \mathrm{A}$

Okay, so does Chapter 26. Problem 93. So it says an RC circuit contains a resistor are of 15 killer homes the the capacitor of 0.3 micro parents in a battery with an E m f nine volts. Starting a time equals zero when the battery is connected. Determined the charge Q on the capacitor in the current I and circuit from time equals t two. Time equals he time equals 02 time equals 10 milliseconds, and it wants to do it at 1.1 millisecond intervals for graphing capabilities. Cool, Cool made grass trimming the charge. Cute. I change with time. Um, from the grafts, find the time at which the charms. Okay, cool. So to figure out how in our C circuit you've always like this. We need to look back at the equations 26 6 a in 26 8 Okay, so what these show us is, they show us the charge as a function of time is given by C E. M f one minus B to the negative t over RC. They also show us the current as a function of time given as e M. F. Over our E to the minus t over RC. Okay, now, from this, we can immediately graft these with graphic capabilities. It's about the second part First for a second. So it says, find the time at which the charge of chains 63% of its final value. So if we want the charge to be 0.63 que final. Okay, well, what is Q Final? Just a question to final for charge. You can see as t goes to infinity. This approaches to this c times t m f. So this is 0.63 C times. Now we can display visit time 0.3 micro thirds times line votes That comes up to being 1.7 times 10 to the negative six schools. Mrs R. Q. I'm gonna call it P for an appointment. Find them a graph. Let's do the same thing for current. So I p is now. We wanted to be 37% of the final. Great. Well, again, the final current is given. Oh, sorry. As the initial current final charge with the initial Kurt, the initial current is given by looking at this time equals zero just absolutely over our. So this town become 0.37 slung over our 0.37 nine votes over 15 killer homes. And we get an I P. Of 2.22 times 10 to the negative for wimps. So these are the two points we want to figure out on our cards and current versus time plants. So from these equations are we have to do is get an array of T. If you want to use this and pipeline, you can do some pie dot lynn space 0 to 10. No seconds at a point. One nanosecond in between. This gives you an array, and then you can just evaluate this at each point and then plant it. So I'm gonna go ahead and sketch what the parts are gonna look like here. So it's straw this charge versus time slot. So this is time in seconds. 0 to 10. Mrs. Charge, uh, zero, when I say appears to 0.5 racoons. So we started zero then. This kind of follows this kind of powder. Okay. And if we want to find the time when we just find when this equals R Q p that we found before and we should see that happens at about 4.4 milliseconds. Okay, let's plot the other wrong here. So this is the current Oh, sorry. This is still time on the X axis. Of course, this is time in milliseconds from 0 to 10 milliseconds. And now this is current. This goes from 0 to 0.6. This is Miller Imps. Now this We start at 06 kind, uh, decays very slowly along this path and again we should see that I p what we found before, the point that we wanted to find happens at the same time, which is roughly 4.4 no service.

In this question were given an RC savior with resistance 135 home and capacitance 15 Michael which is 1.5. I'M stand by my six fatty. I'm sorry. It's micro fatter. Yes. Now. So obviously that I'm constant in this case is RC which is 1 25 and 1.5 cent. Apartment six, which is 0.88 music now this Now that you know this, let's go. Bye. In this question, the capacitor is charging. So we have a sister on the capacitor and they're connected to our batteries farces. I know the capacitor is charging. We know that the master is charging up the charge cause us you're not times one minus minus. Steve, I tell Thank you. Not is the final charge in maximum charge after a really long time. So do you want has given the question? We want to find out when the charge will be one minus one. By e times the maximum charge. So we want our charge to be one minus one by he was the maximum charge. So using these two equations equating these two we get, you know, dance one minus four minus t vital if you call the Cunard into one minus one by so we can see you're not every day 51 toget to get us. If I minded, Steve is one by which is minus one. Now, if you can see this, we can see that because the exponents must be seen. Which means he wiped out his one, which means Tiesto. And this happens after one that one time constant or B. You want to find out what the current will be? This phone. You know that What for charging capacitor. I goes us. I not even dynasty by Tel, that is it. Start off starts off with a maximum value on exponentially decreases as the charge as the capacitor getting starch. And when these evil hotel as we just saw I do not I'm skip are minus one That is I not buy it. Hence it falls after one. Thank you. After one constant, the current is won by e times the maximum current Just settle on 368 times the maximum current. Now for the next part we want to consider the case Window capacitor is discharging. So we have the your sister and we have the capacitor on nothing else in the secure and the kid Faster has some charge on So in this case for park Si we want to find out who ended it Charge will be one by e times the maximum charge. We know that for a discharging capacitor the charge expanding exponentially d case that is at any time he touched you will be the maximum charge I'm even dynasty white out hence using these two equations we write you know by E is equal to you not even dynasty right out Excuse us If our dynasty by tell is even minus one it means he is a good hotel The same as the last question It happens after one time constant within boxy we all want also want to find out when the current will be maximum current times one by now for discharging capacitor You bastard! You know that the current again d case as if our dynasty wayto Because as the charge gets defused, the pastor's world which falls and the current also falls. And even in this case, using these three questions you see I not evil dynasty right help is not equal minus one, which means he's able to tell which is the same for all four questions

Hi. In the given problem, the capacities of the capacitor Is given as C is equal to 1.50 micro Farage and its resistance Is given as R is equal 225 mm so that I'm constant of this. C. R circuit Will be given by Tao is equal to the product of capacities with the resistance. So this is 1.50 micro favorite, multiplied by 125 home sort Approximately. It comes out to be equal 288 micro second. No, in the first part of the problem using the equation of instantaneous charge stalled over a capacitor while it is being charged is given a skill is equal to Q. Not one minus erase the bar minus D. Bye joe. No, as for the given problem, the charge strolled over the period of the capacitor, instant in his charge Is given to be 1 -1 x eight times of the maximum charge. And putting this value of cube in the given equation we get 1 -1 by E. In two. Q not is equal to Q, not one minus irish par minus t by top. So canceling this cannot we can write it like one minus the rich par minus one is equal to one minus E. Raised to the par minus T by tao, Then canceling this one and compensating this negative sign here. Finally we get, each bar -1 is equal to irish bar minus T by tao. Hence comparing these two sides, selection side and right hand side. We Get to know T by Tao is actually equal to one. So finally the time in which the charge over the plates of the capacitor will become 1 -1 x eight times of the maximum charge. The time is actually equal to one time constant which is equal to 188 Micro 2nd. The answer for the first part of the problem. Now, in the second part of the problem, we have to find the current passing through the circuit at that much time in comparison to the maximum current passing through the circuit. So using the equation of current in a charging circuit in a charging cr circuit which is given as I is equal to I not one minus cherish the par minus t right out. And we have found in the first part of the problem that this tea, it's nothing but equal to tell. So using that I comes out to be equal to I not 1 -1 minus D by minus towel by toe, Which will come out to be one. So this is I is equal to I not 1- Irish Par -1. And this value comes out to be I not 1 -1 0.368 or finally the current passing through the capacitor at that much time is 0.63. two times of the maximum current passing through the circuit which is the answer for the second part of this problem. Now In the 3rd part of the problem now we have to use the equation of discharging cr circuit and that equation of discharging C. R. Circuit is Q. Is equal to Q. Not perish the par minus D by toe. And again it is given that in stan Tania's charges over the political the capacity remaining charge Should be equal to one x 8 times of the maximum charge. So putting this value of the charge in the previous situation we get when by eq not is equal to do not times erased. The par minus T by top canceling the skill note. And this can be written as he reached for -1 is going to flourish for minus city by tao. Again we get the time equals to tao means again this is 188 microsecond. Answer for that heard part of this problem. Finally, In the 4th part of the problem we have to find the time taken by The charge taken by the current passing through the circuit. two decade up to one by eight times means I is equal to one by eight times that maximum current passing through the circuit. So using the equation of current for discharging cr circuit I is equal to I not perish the par minus city By Tao. Here I is equal to one by E. Times I not is equal to I know times erased par minus city by tao. Again, canceling design. Not again areas by minus one is equal to erase by minus city by toe. So canceling this E a negative sign. Finally, we get T. Is equal to tower again, means hunter and 88 Micro 2nd, Which is answer for the 4th and the last part of the problem. Thank you.

Current as a function off time is given by I not exponential deal with dough. So Ah, yet how comes out to be I not over e e equals I don't These 0.7 AM beer Andi One of our ease viewpoint 368 it was zero foreign Zito 257 Ambien. Now hard into the figure we have, we can figure out what is the time constant. So the charge will be I not Times Tao equals I not ease 0.7 ampere multiplied by time, constant reading it from the figure It comes to be 0.6 second. So the total charge comes out to be four point toe. May they couldnt Now we can find the capitals, which is Q over. V equals 4.2. Medical them divided die nine village equals 0.47 Millie Fad It was 4 70 Micro Fattah


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