the-square-wave-voltage-shown-in-figure-24-31-is-applied-to-an-r-c-circuit-sketch-the-shape-of-the-i

Question

The "square-wave" voltage shown in FIGURE $24-31$ is applied to an $R C$ circuit. Sketch the shape of the instantaneous voltage across the capacitor, assuming the time constant
of the circuit is equal to the period of the applied voltage.

Luis Rios · Accepted Answer

for this problem were given a graph that shows the voltage Nick capacitor over time one is connected to a battery were also given the resistance, which is our is equal to 4000 700 Holmes, part A for us to find the voltage in this battery. Well, once the capacity reaches maximum charge, it will be the same voltage as the battery so it levels off at nine volts. Part B wants to find the time constant how the physical to our time, see and use the equation for the voltage in a capacitor when it&#x27;s charging. So v c musical to bolt is in the battery times one minus you to the minus t over R C from the graph. If you say at T equals toe one second, then we get the voltage in the capacity to be four months now with this information on you to do is not solved for our time seen so the viable side by the voltage in the battery and now we have we see well, everybody is equal to one minus e to the minus t over our see subtract the one we have the sea over absolutely bolting the battery like this one. Tu minus e to the minus t over our See Now if we just plug in the numbers for the voltage in the capacitor and bolted to the battery just to simplify a little bit, we have receiving her for the water by nine minus the one. So we go minus minus 59 Now we can plug that in, and if we get rid of the negative on the right side, we just have five nights positive. 5/9. Physical to E to the minus. G over. Arm. See. So now we take the natural log rhythm of both sides to get rid of that exponential. We have Ellen have all nine physical to L A and e to the minus T R c well, and of e Ellen of years one. So that was the way. So we&#x27;re left with Ellen. I have. Ninth is equal to minus t over our I see. That&#x27;s for looking for our time. See, you multiply both sides by our time sitting and divide by the Eleanor five or nine to get R C. Which is the time constant is equal to minus t over. Ellen five nights. So just plug in your T value, which we took as one. What&#x27;s minus one over Ellen by nights. So for a time, constant town just quickly R. C is people too. 1.71 seconds. Now, with that information, we can find the answer to part See, so we&#x27;re looking for is the capacitance. So we know R. C is equal to 1.71 So it&#x27;s all or C. We get C is equal to 1.71 that of my arm. And we know our that was given to us earlier. So it&#x27;s 1.71 divided by 4000 700. So for part, see, capacitance C is equal to 0.36 I was 10 to the minus six the Reds.

Doruk Isik · Answer

in this problem. We&#x27;re given a crap, and we&#x27;re told that graph displaced the voltage is a function of time and were asked to, um fine, we&#x27;ll change. One time is seconds. So we&#x27;re actually trying to fine we probably 20. Celestial attention, Leonard. Reverend Engine line function is they talked? Their function, So not times in lang seems to be passing through borns. This is three reasons for a few dozen Your horns. 10 free. So excesses tea and waxes we end. Hurry, Cool. So we&#x27;re gonna say your prime is equal to DVD T and that is that the deputy? So four months, Every year I buy the reman is Stan. We find this to be was Europe I wolves for second key growing tension legs. As you see ancient lines 82nd gets more and more presented. Two more horizontal tension line. Thanks. It lines are It means that slower there will

Carlos Henrique De Lima · Answer

So this problem We have a parallel plate with era A and suppression de and the potential various s A T square. And we want to know this basement current. So we know the capacitance from from the This is an area off the plates. So the car passed Owes is a given and then the athletic flux is also given because it is just a kowtow. The total charge divided by absolute zero. And this can vary the I C V over absolute zero No, the displacement currents, the cap Pastors, we can write us absolute zero a over the So this is a V over the flats. Then we can write used this expression Toe writes that special current This based on current it&#x27;s written us absolute zero defy a t There&#x27;s this absolute zero a over the They didn t off the a t square and then we take this The relatives of this becomes too absolute zero a did away the decided by B So there&#x27;s this the final result and we can simplify this in terms off the campus is if we write this absolute zero A and the in terms of the capacitance so This will be just to see 80 if you use the capacitance. But we want to write in terms of being Lee. So the assistance

Luis Mendoza · Answer

so into throwing we have on RLC all tragic Turin Circuit General Sisi, researchers of other assistance capacity or any doctor. And they get extra generator on India. Specific problem. The inductive generator is going to be zero. So a legal zero who is basically trickling to saying that we are ignoring opens now. We&#x27;re told that the voltage across the generator exploding for balls year devoted across the resistor is 45 balls here. Now ask what is the voltage across the capacitor? So things were in a serious RLC circuit. Voltages can related by the equation. He&#x27;s here always boulders in the generator br voltage into her sister bl boulders in the doctor of the Civilization, the capacitor bl zero because inducted zero I think it is a matter of solving for BC, which is to pull the genetic capacitor on the speeches thieves relationship here. So square root voltage of the generator squared minus voltage along the receipts. Two X squared and pulling numbers. In 45 24 we have pain bolted off 44 balls across the capacity

Sheh Lit Chang · Answer

this question were given this circuit, it has been connected for a long time. So, Josh, that part&#x27;s in this question, in part A Want to find the potential difference across the capacitor. Okay, So one thing we need to recognize is that, uh, when the switch and holding is connected for a long time, um, there&#x27;ll be no current slow in the capacity, so the card will be flowing, um, in the resistor. Okay. So, effectively, the diagram becomes you can kind of ignore, uh, presence of the capacity for the moment. Okay. Yeah. So you have this circuit diagram, and you can kind of in order capacitor one moment, okay? We just want to find a potential difference for the capacity. So what we need is, um, to find out potential at these two points. Okay, That would be the That would be the, uh, Let me take a difference. Then you can find a potential difference across the capacity. Okay, so So this is a parallel connection. Okay, so here we 10 words and what&#x27;s all right? And then you can kind of imagine that Ah, the current here will be to em. We can calculate the two ends. Who do you want them right, then, uh, after two for the left branch. Um, so after going through the one own resistor, the potential here will be a towards okay, and then he&#x27;ll be towards for the right branch. Okay, by just using the minus. I are. Okay, so here the potential difference. Um, based on that, neither potential difference across the capacitor hey is equal to eight. Minus two equals 26 words. Okay. And then this is the answer for party. Okay, don&#x27;t impress. Be, um the battery is disconnected from the circuit. A what time in the world. That&#x27;s the capacity discharged through 1 10 off its initial charge. Okay, so when you remove the cheat, Yeah. So when you remove the patchy, um, So when you remove the battery, the situation becomes like this. So you have ah, capacitor across to, uh, parallel network. So you have one own its own and then to own for them. Okay. Yeah. This is what happens when the battery is disconnected. Okay. In a new circuit, diagram becomes like this. So the equivalent resistance is 1/9 plus 1/6. When you take the inverse and gets 2.6 homes. Okay. And then you can calculate the, uh, time constant. So this is our equivalent time capacitance. So it&#x27;s 3.6 micro sha microseconds. Mhm a 3.6 times. One micro fair gives you treat my crew 3.6 microseconds. And then, if you want to calculate uh huh find a time interval for the discharging to take place. So the discharging equation, whether it will teach across the capacity will be be equal to be not e minus t over town. Okay. And then, um, we have b equals to 1 10 but the initial footage. So we have 0.1. It goes to e minus t o a towel and then take natural on both sides. And then how would just be minus a t? He is minus How natural law. 0.1. Um, complete this you get it? Point 29 microseconds. Okay, so this is the answer for part B. Okay? And that&#x27;s all for this question.

Problem

When a long copper wire of finite resistance is …

Problem 28 Easy Difficulty

The "square-wave" voltage shown in FIGURE $24-31$ is applied to an $R C$ circuit. Sketch the shape of the instantaneous voltage across the capacitor, assuming the time constant
of the circuit is equal to the period of the applied voltage.

Answer

$\therefore$ Voltage across capacitor for decreases form 0.588 $\mathrm{V}$ to 0.3566 $\mathrm{V}$ during the period $2.5 \mathrm{T}<\mathrm{t}<3 \mathrm{T}$

Related Courses

Physics

Related Topics

Discussion

Top Physics 102 Electricity and Magnetism Educators

Elyse G.

Christina K.

Liev B.

Zachary M.

Physics 102 Electricity and Magnetism Courses

Electric Charge

The Electric Force

Recommended Videos

Watch More Solved Questions in Chapter 24

Video Transcript

James S. Walker

Physics

Elyse G.

Christina K.

Liev B.

Zachary M.

Electric Charge

The Electric Force

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\therefore$ Voltage across capacitor for decreases form 0.588 $\mathrm{V}$ to 0.3566 $\mathrm{V}$ during the period $2.5 \mathrm{T}<\mathrm{t}<3 \mathrm{T}$

Physics

Elyse G.

Christina K.

Liev B.

Zachary M.

Electric Charge

The Electric Force

James S. WalkerPhysics

Elyse G.

Christina K.

Liev B.

Zachary M.

James S. Walker

Physics