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Provide rlc transirinn starcz and inrcrcdiarcw and f MAJOR pnc Jucrs for thc followin? transtormation Ut thcrc (Cactinn starc VRIOCHJUH MannKnT-cuiolCH;SNaTH} (a el...

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Provide rlc transirinn starcz and inrcrcdiarcw and f MAJOR pnc Jucrs for thc followin? transtormation Ut thcrc (Cactinn starc VRIOCHJUH MannKnT-cuiolCH;SNaTH} (a elller ,AcCtaCH OHProrutufijcu Mtlhiiym Tor €

Provide rlc transirinn starcz and inrcrcdiarcw and f MAJOR pnc Jucrs for thc followin? transtormation Ut thcrc (Cactinn starc VR IO CHJUH Mann KnT-cuiol CH;SNa TH} (a elller , AcCta CH OH Prorut ufijcu Mtlhiiym Tor €



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Determine the steady-state current in the RLC circuit that has $R=\frac{3}{2} \Omega, L=\frac{1}{2} \mathrm{H}, C=\frac{2}{3} \mathrm{F},$ and $E(t)=13 \cos 3 t \mathrm{V}$

Okay, so to determine the study state current, Um, first, we note that the charge follows the form of the square Q over DT squared, plus our over l D Q. The tea. Ah, plus, oh, one over L C. Q is equal to one over l e of t like so now, ah, to determine the steady state charge, then we're going to use a trial solution. Why a P course I Q p of tea. This is gonna be able to a not ah, and then this right hands are sorry. Right hand side has co signed three t is going to be a not ah co sign three t plus B not sign three t. So then now the, uh, for this term here, we need to find q p prime. This is equal to three. Uh, negative three a. Not sign three t and then minus three b not cosigned. Three t I'm gonna are sorry. Plus three b not I'm gonna write it so that the co sign term is in front, so I'll move this over here. So that's like this and then get rid of the plus over here and then Q p double prime. Now, T, this is gonna be able to negative nine b not sign three T minus nine. A not, uh, cosigned three t. So again, I'm going to move this. Ah, so that this co signs are all on front. Now, uhm, I'm going to plug these into this here to get our steady state charge to solve for a not and be not so I'm multiplying. Ah, this by our over l so are over l. That's three halves divided by 1/2. The twos will cancel out, so we just get three. So we're going to multiply this by three here, so this becomes a nine, and this also becomes a nine. So you have nine and nine next. This one over Elsie. So it's 1/1 half times to over three. So these council out and then the three can come up to the top, so that's also to be multiplied by three here. So we're gonna multiply this by three and then multiply this by three as well. Now we're gonna add all these terms so that all the coastline terms will just add. So here we have a three a and a negative. Nine A. So that's gonna become a negative six. A. Not and then we also have a plus nine B. Not now. This is going to be equal to the, um the coefficient here, one over l E to the t. So that's 13 cosigned three t and then 1/1 half. So the two comes out, so he becomes two times 13. So it's gonna be equal to 26 here next. Ah, the sign terms. So the A there's only a negative nine. A not. And then we have a three B not in a negative 90 knots. That's minus six. Be not. And that's gonna be able to zero because we have no sign term here less. Ah, so this, like, plus zero sign of three t. Right? So from this equation here, we can get that a not Okay, let's see. We move this six b not over here is equal to negative nine A. Not then divide by negative nine. So we get negative 2/3. So that becomes negative. 2/3 be not next. We can substitute this into here, so we have negative six times negative. 2/3. That becomes ah, just positive for So we have four plus nine. Be sorry for be not plus nine B not 0 to 26. This becomes 13 Be not equals 26 Divide by 13 On both sides we get be not listen to be equal to to So that means a not is gonna be good too negative for over three. So that means our steady state charge is going to be equal to so for q p of tea is equal to so negative 4/3 co sign three t plus and then ah to sign three t now to find our study, stay current our city state current e i p a t It's gonna be able to the derivative of this. Okay, that's he or ah. So we're going to take the derivative of this now to find are, um, steady state. Ah, charge. So taking the drive of this. Okay, so we get i p of tea, it's going to be equal to So it's three times this and then times negative. So it's gonna be four sign three t and then now derivative of this. So bring out the three that becomes plus six, uh, cosigned three t like so? So this is going to be our study state current

So the party, The resident's frequency oven R L C circuit. We would be resident's frequency this of equaling one over to pi times the square root of the induct INTs, times the capacitance. And so this would be equaling one over to pie multiplied by the square root of 1.25 times 10 to the negative sixth Henry's multiplied by 40 7.0 times, 10 to the negative ninth Farage's and we find it. Then the resonant frequency of the R L C circuit is equaling 6.57 times 10 to the fifth hurts. Now, this would be your final answer for party. For Part B, however, we want to find we want to solve really required capacitance, given that we have a certain resonance frequency and so we can say that again. The resonant frequency is one over to pie, but over two pi times the square root of Elsie and so solving for the capacitance we have, this is equaling one over four pi squared, multiplied by the induct INTs, al times the resonant frequency squared and so the capacitance would be equaling. One over four pi squared, multiplied by 1.25 times 10 to the negative six. Henry's multiplied by 1.33 times 10 to the sixth hurts quantity squared. And this is giving us 1.15 times 10 to negative eighth ferrets. So this would be our final answer for Part B. That is the end of the solution. Thank you for watching.

In this problem, I'm writing the reactions. So just look at it carefully. OK. FECN six came food, FT CN six plus six. He S CN k s c N will react to form, get to the f e SCN six and now I'm going to write the second reaction. They just look at it carefully Or FECL three. That's pretty careful. F E CN six, then it will give 54 F e CN six plus two old KCL. So option age correct here.

Okay, so this problem, um, for we need to first find the solution here. So to do so Well, we need to first, uh, find the differential equation. That's going to be the formula. So D squared que the t squared and then plus, and then we'll have our over. L are over l d Q over DT plus and then one over l c que is equal to one over l e o T. Okay, so then now we need to find our plug in our information here. So we'll have first the square Q over D T squared. Plus, and then our is 3/1 half, so the two will come up to the top and it's become six. So we'll get six the Q over DT plus and then this L c. So it's 1/1 half times 1/5 again, That's 1/1 10th. And this tank can come up to the top now. So we get 10-Q is equal to and then 1/1 half times two cosine, Omega t. So that's gonna become four, uh, cosigned. Oh, my God, He like So So first we need to find our particular solution. So we need to solve the auxiliary equation. P of our is he with you? R squared plus six r plus 10 is equal to zero. So then now, um, if we try to factor 10 that's gonna be one in 10 or to end five, which both don't add up to six. Ah, so we're gonna need to use quadratic equations that's gonna be our is equal to negative. Be so negative six plus or minus. And then we'll have b squared minus four a. C We just four times one and then times 10 all over two times. One to a so out front here will have a negative three sort of negative six divided by two then plus or minus. This is 36 minus 40 which becomes square root of negative four. So that's two I. And then we have a divided by two here, so there's gonna b plus or minus I. So now our, um complimentary solution Q c of tea. It's gonna be equal to e to the negative three t the men times see one, uh, co sign of tea plus C two. Sign of t like so next we need to find our, um, our particular solution. So since on the right side, our route here on the right side is just omega or B is ableto omega, right? So since we don't have an a part, there's no e to the negative or either to something um, we can just do czar trial solution Q of p of tea is in the vehicle to a not and then we'll have cosigned omega T and then plus B not sine omega t hips sine omega t so Q p prime it's gonna be equal to. Then we'll do negative a not omega sine omega t and then plus B not Omega. Uh, Omega co sign will make a t. I'm gonna rearrange this so that then the co sign is outfront. So we'll get this here and their next I'm going to do que p double prime tea. This is gonna be negative. Be not Omega squared sine omega t minus a not omega squared cosigned omega T. And again I'm going to rearrange it so that the coastline is outfront. Okay, All right, there we go. Next. Ah, I need to plug these into our left hand side here. So I'm gonna multiply first, Uh, this by 10. So if I multiply that by 10 So let me ask you teach this a little over. Ah, move this over as well. So multiply this by 10. Okay, then I'm gonna have a 10 out front here, then a 10 out front here and then when it works by this by six. So I'm gonna get a six out front here and then also a six out front here. And then Now I'm just going to add down the the line here. So all the coastline terms, So if we notice these, um, costar in terms we're going to have on a not, um, times 10 minus omega squared and then plus a six b, not Omega. So these cosigned terms here are going to be equal to the coastline term here, so we're gonna set that equal to four. Next, we're going to set the sign terms equal to zero. So that's gonna be this this and this. So we get Ah, negative six a. Not Omega of in plus also 10 minus omega squared. Be not is equal to zero. So first we can solve for a not so a not it's going to be equal to so we'll take, Be not remove this over to the other side. Right, So we get six a. Not Omega. So we'll just buy this by six omega. So it's gonna be 10 minus omega squared, divided by six omega. So let's plug this now into here. So we'll get, uh, 10 minus omega squared squared over six Omega plus six Be not Omega. Also, this is times be not here. I forgot that is able to four. So I'm going to multiply this by six omega Oops! Six omega so that we get six omega over six Omega, um and then factor out to be not so what this is going to be is we have 10 minus omega squared, squared plus 36 omega squared over six omega and then be not is able to four. So that means be not is going to be equal to, um right down the denominator first. Okay. And then on the top here we have 24 omega clips 24 mega. Then a not is going to be, um that should be will be not here. Then we multiply this by this. So if you do that, this is gonna cancel, and we'll have a four. They're so we'll just have this times four divided by this. So a not this kind of equal to, um let me just move some of this up a little, Okay? So have room to write it here. Okay, so four times 10 minus omega squared over and then 10 minus omega squared, squared, plus 36 omega squared. Next, I'm going to, um, right that over to the right part here. Okay. So que of see no or a q of t It's gonna be r E to the negative three t time. See one sign or see one co sign T plus C to sign O t. And then plus, and then a not Is this part here? I'm gonna factor out this here. Okay. From both of the terms. So we get, um, on the top, we just get Okay, So we have four times 10 minus omega squared and then, uh, cosigned Oh, my God. Tea. And then plus, and then the be not the numerator. Hey, it is going to be four Ah, 23 24 omega and then times sign Omega T, and then this is all over here. So we have 10 minus omega squared, squared, plus 36 Omega squared. All right. Okay. Next, what we're going to do is find the amplitude of this. So the amplitude of this is going to be equal to the, um The total amplitude is going to be a not squared, plus B not squared, and then the square root of that. So that's going to, um, So we'll take. First of all, we're gonna take, um, this here. We're gonna take it out front here. So let me do this in blue. So we're going to take, um, this out of these two terms first, right? So we just get ah, four times 10 minus omega squared, squared. So that's actually gonna be a 16 out here. Ln plus 24 squared is 576 and then Omega squared. Okay. And then now, right then we divided by and sorry. That's a square root of that divided by 10 Linus omega squared squared plus 36 omega squared. Next, we can actually factor out a 16 from both these terms, so we're just gonna get a four out front. So this is a we're gonna get a four out front and then times a square root of you have 10 minus omega squared, squared plus 36 omega squared and then divided by, uh, 10 minus omega, squared, squared, and then plus 36 omega squared. So this now is equal to this bottom part here, so we can actually rewrite it as just we can take the square root down. So if we had x square root of X divided by X If we multiply by square root of X on top and bottom, we're gonna get X divided by X times square root of X. So this is just going to be equal to one over square root of this. So we'll have 10 10 minus omega squared, squared, plus 36 Omega squared. Okay. Um, all right, we're taking the square root of all that, and then we have lips four divided by that Now to maximize a, uh, we need to minimize this bottom part here. So, men, this here now to minimize that, um, it's the same as minimizing uh, the square of this. So we're gonna just minimize Ah, 10 minus omega squared, squared plus 36 Omega squared 36 Omega squared. Now, to do that, we need to take the derivative with respect to W. So the derivative respected w Okay, let me do this in the different color here, that's going to be two times 10 minus omega squared and then times a negative to omega. And then plus 36 times two is 72 and then Omega. And we said that able to zero. So let's simplify this. We have a four omega, um, negative for omega times. 10 minus omega squared and then plus plus 72 omega. We can, uh, equals zero. And then now we can divide both sides by a negative for Omega. So this becomes 18. So we'll actually just get, um, negative 10 minus omega squared. Plus 18 is equal to zero. Okay, so we get 18 is equal to 10 minus omega squared. If we move this over to the right hand side. So we get so we'll move the Omega squared over here and the 18 over here, we get that omega squared, it's actually gonna be equal to okay. Sorry. First, uh, to solve this uh, let's go back to this step here instead of dividing by four omega, I'm going to Ah, factor out a four Omega. Okay, so I'm gonna factor out of four omega. So now I get 70 are Sorry. Um, 72 divided by four is I get 18 and then minus 10 minus omega squared is equal to zero. So this part can simplify into omega squared and then plus eight. But then we have also four omega. This is equal to zero. This part can never be equal to zero. So that means we have that Omega must be equal to zero. So So that's just gonna be, um, homemade equals zero last night. Yes. So this is our solution for any Omega. And then now our solution for, um if you plug in zero for Omega. This is the salute, the current that maximizes. Um well, this is the omega that maximizes the current, so, um, we'll just leave it at that. That omega equals zero is the value that maximizes the altitude.


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