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Remaining Time: hour; 53 minutes, 09 secondsQuestion Completion Status:QUESTIONCalculate Vo in this circuit R1=20,R2= 20,ZC =-j30,ZL=j1Q,V=2<600 VzV 22t0 fle und...

Question

Remaining Time: hour; 53 minutes, 09 secondsQuestion Completion Status:QUESTIONCalculate Vo in this circuit R1=20,R2= 20,ZC =-j30,ZL=j1Q,V=2<600 VzV 22t0 fle und -ubmwt. ( Yt? n Ui A#er5 () Sa"e all nnst'eru. Clc? Sc: 6t ' ubintAi DType here t0 search

Remaining Time: hour; 53 minutes, 09 seconds Question Completion Status: QUESTION Calculate Vo in this circuit R1=20,R2= 20,ZC =-j30,ZL=j1Q,V=2<600 V z V 22 t0 fle und -ubmwt. ( Yt? n Ui A#er5 () Sa"e all nnst'eru. Clc? Sc: 6t ' ubint Ai D Type here t0 search



Answers

Consider the rate $P_{1}$. at which energy is being stored in the $R-L$ circuit of $\mathrm{Fig} .30 .12 .$ Answer these questions, in terms of $\mathcal{E}, R,$ and $L$ as needed: (a) What is $P_{L}$ at $t=0$, just after the circuit is completed? (b) What is $P_{L}$ at $t \rightarrow \infty$, a long time after the circuit is completed? (c) What is $P_{t .}$ at the instant when $i=\mathcal{E} / 2 R,$ one-half the current"s final value?

To find the time the current will increase by 1/2. That made a difference between the maximum current and the actual. Do you think this question I won? Are you finding? And we can write a e one minus eat minus. I got the miners. It is here. Eat hot off with the opportunity is it was won by the worst any upon our lo 98968 in tennis to polar minus upon one. Oh, bigot! He is a coastal t want in tennis to bubble minus chickens.

69 to number 25. They give us an equation that talks about the current and it closed R l circuit, and they give us the equation. Isaac, Latvia V over R minus V over R E to the minus are over l t. And what they tell us on this question is that the steady state So over time, this current will reach a steady state, which is V. Cooper are and they're asking us is how long until it reaches 1/2 the steady state. So how much time until the current reaches half the steady state? So until I is equal to 1/2 the over our Okay, So in order to figure this out, we're just gonna have to set I equal toe one hat V over R. So that will give us equation of 1/2 V over R is equal to V over R minus. V over r times e to the minus are over l teeth. So I mean cancellation here. So you're gonna end up with 1/2 is equal to, and this is just minus. Excuse me. You're gonna with one minus e to the minus are over. L t And so if I subtract one from both sides of this equation so that gives me minus 1/2 Goes minus e to the minus are over l of tea. I can get rid of these minus signs, but that those could become plus now just take the natural log of both sides. So the natural log of 1/2 is equal to the natural log of e to the minus are over l t. And so this tells me that, um, you've got the natural log of one minus. The natural log of two is equal to minus are over. L t natural log of one is zero. So this tells me the natural log of two is our over l t so t therefore, is going to be l over our have is a natural log of to So that is the amount of time that must elapse before I reach 1/2 of the steady state. So the time is l over our and notice it doesn't depend upon the voltage. Okay, so it's just l over our times the natural log of to

It's Claire's. So when you right here. So we're gonna write the standard first order differential equation for current I as a function of tea, which is time. So we have tea I over g t plus our over tee times I which is equal to be over l Then when we saw for I we get V over r times one minus e to the negative are divided by l t. So we get 1/2 times be over our, which is equal to V over r times one minus e to the negative are divided but oh, terms t are multiplying Both sides by r over v to get 1/2 is equal to one minus e to the negative are over l a times t Then we get 1/2 is equal to eat the negative are over at all times t then Ln of one minus Ellen of two is equal to l n of e to the negative are over all times t we're just taking the longer of them and we get t to be equal to go over our l and of two seconds. Yeah,

Okay, we got another application problem here. However, we strip away the application, we end up with an equation for I. Not the imaginary unit but the current here in our circuit I equals 4.42 times either the negative 66.7 tee times. Sine 2 2016. Okay. We are asked to find the derivative with respect to times to the I. D. Team. So we'll break this into two functions that are multiplied together, which means we're gonna need a product rule. Okay, we'll just get into that. Okay. We should be very comfortable product rule at this point. So take the derivative of the first function need the second function alone. Okay, derivative. The first function will be 4.42 times negative 66.7. He did the negative 66.17. Okay. Plus leave the first function alone ugly numbers here but we're not afraid of that. Times the derivative of the second function derivative. Second function 2 26 co sign. Right that below. A little bit more space. 2 26. Co sign to 26 T. Okay. We can simplify this down a little bit. We can take this E to the negative 6.67 T. Out in front. And then what are we left with? 226. Times 4.42 I'm gonna about to play those two together. First. Yeah. Positive term up there around that. To 999. So 999 co sign to 2060. Hey minus we've got this 667 times 4.42 We'll do that calculation next. About 295 minus 295. Sign 2 26. T. Mm. That's not derivative.


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