Circuitos Elétricos no Concurso Petrobras: como cai na prova | Eng. Elétrica e Eletrônica

Hinzugefügt: 2026-06-07

[00:00:00]A few years ago, a dream began. A dream born from the desire to transform the lives of those seeking more than just a job. I've been on the same side as you. I've been unemployed, I've been undervalued, and I've worked in places where insecurity and uncertainty about the future were constant companions.

[00:00:16]IBGE estimates that nearly 1.5 million Brazilian music graduates are unemployed. In one year, this number increased [music] by 13%.

[00:00:25]The crisis in the job market has hit young people hard.

[00:00:30]Newly graduated professionals are struggling to find a job in their desired field.

[00:00:35]And then when I [musician] graduated from college, you know, I managed to finish college on time and I thought, like, man, now I'll be able to work, I'll be able to stop being supported by my mother. [music] The market was kind of a mess, I even managed to get an internship to finish my degree, but a job was impossible to find. [music] Everyone was either firing people or at least not hiring, right?

[00:01:00]I know what it's like to fight to achieve something better, something that truly brings stability and security. Imagine working in your field of study, earning a salary of over R$ 10,000 per month, and never having to look for a job again in your life. This was the change that happened to me, and I truly wish it would happen to you.

[00:01:19]A fresh start with stability, security, and the opportunity to grow in a solid career. Our journey began in 2017, and we quickly realized the impact we were making. Over 60% of those who passed the Navy entrance exam were our students. Back in 2019 we already had over 5,000 active students, and in 2023 we surpassed the mark of 1,000 approved students. A special mention goes to the Petrobras, CNU, Aeronautics, Army, and many other competitive exams. The success we've achieved so far is just the beginning. What I want, what we want, is to see you achieving great victories in whatever public career you desire. For candidate number 1, preparation goes beyond just studying. We create high- quality preparation materials specifically tailored to each competitive exam. We also offer all the necessary support for you to achieve your goal. And the goal, of course, is to transform lives. So that you not only get a job, but have a successful career with stability for the rest of your life. Join us.

[00:02:26]The transformation begins here with your decision to not stop until you reach your goal.

[00:02:32]And I managed to pass the English course at Cebrasco.

[00:02:36]But what would you do? Would you talk about us?

[00:02:39]Look, I highly recommend C01 for the following reason: you have a mentoring process there, a leader who will guide you, and you know you won't be walking alone. And that gives you very important emotional support. You know where to find the information and what to do to get approved, right? It's different. Oh, it's on YouTube, everything's on YouTube, man. You can learn how to do heart surgery on YouTube, you can learn everything there. It's just that, you need a step-by-step guide, and that's already been pretty thoroughly explained in the course, you know? So, there's no need to think that you have to look at the course as an investment, right?

[00:03:14]Short, medium, long term, depending on your preparation, whether you're making mistakes or not. And to invest in that, being sure, as you said, right? You will only fail if you have a health problem or if you give up due to necessity. And that approval changes everything, it really does. The guy goes, his son goes to a better school, he'll enroll him in a language course, he'll have the best health insurance. Uh-huh.

[00:03:36]And the certainty that every month changes there, that's also a very good thing to hear, right? A professional who's 40 years old, man, has never been late, [laughs] I've never had any trouble. If we extracted a lot, we extracted a little, we were never late. It drips there every month. I'm going to dominate the pinga.

[00:03:51]So this is quite an achievement.

[00:03:54]Thank you too, William, because you guys were the pioneers back in 2015. That little seed was planted, you know, it was watered, it grew and grew. I continued investing, I believed in the project, and, wow, thanks to you guys, this approval was a very important step forward. Thanks to you, Lauro, and the whole team at the former Marin and now Concurseiro 01. I'd get home, have dinner, then coffee, obviously, and then I'd start studying, right?

[00:04:24]Obviously, the main tool for all of this was the course I took, which was the Concurseiro 01 course. It helped me a lot because it has many practice questions and many mock exams. I don't know if it's the right methodology, but it works for me, which is basically just doing exercises, reading a lot of theory, I read what I see I'm getting wrong a lot, and you correct it based on the feedback from the questions there. That's it. I don't know if it's the best methodology. I think if you look at the courses they offer, they always have theory followed by exercises. But I did it this way, mainly because of the short amount of time.

[00:04:59]It looks like it's not going to work. I've always thought that to pass a competitive exam you needed to prepare in what way?

[00:05:05]Two years focused solely on that competition, right? No, it's possible in 4 months. The platform there helped me a lot with exercising. That's when I saw that theory, which wasn't very, very good.

[00:05:17]That's how I watched the classes, the video lessons. After studying for practice tests, I was doing really well, I don't know, out of 120 I was getting around 105 right. Dude, I'm saying that a job competition isn't something immediate; it takes time for the results to come. It won't happen on the first try, or the second, but you have to study hard to get there, and eventually you will. You don't need to be the smartest person in the world, or the best, but it 's about persistence, studying, and using the right tools, right? The classes and mock exams I had at Concurseiros were... That helps a lot, right? It's doable and worth it, right, man? It is worth it. And it's worth it.

[00:05:56]I'll say that the key difference was the correct setup of the study cycle, the table of topics that will be covered, right? How many questions per subject have you had in recent years, and also, as I always mention, the practice tests, for those who like soccer, right?

[00:06:11]We don't do friendly tests with the national team, the guy puts on A, puts on B, puts on C, the simulated test is for the real test. Oh, I'm doing badly, man. So, try it, change the way you study, change the way you're taking the test. It's the opportunity you have to make a mistake. If we hadn't actually done the practice test on the day of the actual exam, we wouldn't have succeeded. So, that's the rocky road ahead.

[00:06:36]So when I got to the announcement, I was like, "Okay, it's possible to get there, but how am I going to prepare?" I didn't know where to begin.

[00:06:45]And if you don't have support, if you don't have someone helping you structure your schedule, you know, understanding the call for proposals, it becomes very difficult.

[00:06:54]It becomes much more difficult to understand what you can choose not to study, rather than what you actually need to study, you know? To understand what you might not study, you have to understand how the exam board works, how the board asks questions.

[00:07:04]So, this focus on teaching materials has undoubtedly greatly increased my productivity. I recommend 100% how you guys did this for me, it helped me to do this curation. So, we're still before the official announcement, but you can already get yourself ready, get your house in order, and soon you'll be able to compete. You'll already be way ahead of the rest of the country who haven't done it.

[00:07:26]And I went looking for some support, some help, you know? And that's where I met the number one test-taker, right? And it was a blessing for me, because you were offering so many great things that would make my studies easier, right?

[00:07:40]I got the job, I worked in the Navy, and we used the preparatory course for competitive exams for the first time; I wasn't even a competitive exam candidate at the time.

[00:07:48]After that, I spent some time in the Navy and managed to pass the Petrobras entrance exam in 2022. And whenever people ask for recommendations, I always mention Concurseiro 01 (the number one exam taker), and some of my colleagues are even watching the live stream right now. Every week someone sees my testimonial and sends it to me, and I say, "Oh, you can trust the number 1 test-taker," he'll prepare you well, and you just have faith, just study.

[00:08:11]For those who don't know where to start, having that security of having a schedule there—I remember you already provide a suggested schedule—and having a mentor, that was also very interesting, right? We had a mentor who was already at Petrobras, right, who gave us tips on what might come up in the exams, who also had a lot of study experience, and having a group there was a great opportunity for us to exchange experiences and ask questions. Wow, that gives me so much comfort when I'm studying. So, uh, I highly recommend it.

[00:08:48]Hi everyone, how are you?

[00:08:51]Welcome, everyone. Good evening to all our fellow electrical engineers who are watching our live stream today, whose topic is well- worn, but if not the most important of all, then electrical circuits. Guys, leave your goodnight here, tell me how your audio is. The professor has a slight cold, okay?

[00:09:14]So maybe my voice isn't at its best. So, let me know if our audio is good, if everyone can hear us clearly.

[00:09:32]Thanks for the feedback, everyone. Class, today we're going to talk about a topic that those of us who have been here a long time will know, that we started these periodic live streams a few years ago, actually, this isn't something new. And we've already had many live streams discussing and answering questions about electrical circuits.

[00:09:52]Guys, this is a very serious conversation about Petrobras, okay?

[00:09:57]If you don't have a good understanding of electrical circuits, you urgently need to start studying this subject. And it is by far the most frequently tested topic on the Petrobras exam, prepared by CESGran Rio, as well as being the most frequently tested subject, with the highest number of questions, whether direct or indirect, in electrical engineering exams, okay everyone? So it's very, very important that we have a good understanding of electrical circuits. The CESGran Rio exam board's profile for this subject suggests an extremely practical approach, meaning solving electrical circuits, right? So we're being held accountable for this a lot, okay?

[00:10:42]So it's fundamental. You can't afford to get the electrical circuits question wrong during the exam if you want to be among the top scorers, folks, you really can't. It's our basic food, it's our rice and beans.

[00:10:55]So, that's why it's important for you to be in this class so we can practice all the typical questions, right? And the typical way this might be presented to you during the exam, okay?

[00:11:09]Following our same pattern, we'll be working through the problems, and as we solve them, if you have any questions, just let us know in the chat. Otherwise, we'll keep going, but feel free to interrupt me so we can teach you in the best way possible.

[00:11:28][cough] Sorry.

[00:11:31]Let's get started, everyone.

[00:11:41]The first issue we're going to address here is a matter of three-phase systems, okay? It's a subject within electrical circuits that you guys test a lot on, but I don't think the exam board tests it heavily. I think the most difficult part in terms of resolution or extension of resolution that we have in the electrical circuits section is direct current circuits, that's where it can present some difficulty in terms of extension, okay? Generally, three-phase systems are quicker for us to resolve issues. As long as we don't fall for any of the tricks from the betting shop, okay?

[00:12:20]So, our question states the following: in an installation, the power supply for loads consisting of impedances, namely ZAB, ZBC, and ZCA, is provided by a three-phase circuit with a three- wire connection, where the line voltage is 220 V and the phase sequence is CBA.

[00:12:46]We usually have ABC or ACB. That's the first thing the bank tries to confuse you about. It starts with reference A. The next letter is C and then B. So, what we actually have here is the sequence ACB, right? She just wanted to confuse you here, right?

[00:13:05]Do that inversion there. So, we have a line voltage of 220, and we have the sequence which is ACB. There, the loads are connected in a delta configuration, and we want the total power in watts. The data I have is that ZAB has 20 angles of 90º, ZBC has 44 angles of 60º, and ZCA has 22 angles of 0º. The line voltage reference is VBC with a 0° reference.

[00:13:36]Sine of 60 is approximately 0.8 and cosine of 60 is approximately 0.5.

[00:13:43]So, folks, first thing, we have to connect our loads in a delta configuration and observe the power developed based on that power supply.

[00:13:53]One important point we need to understand when dealing with three-phase circuits is whether the circuit is balanced or not, right? In this case, and what makes the issue a bit more difficult, this circuit is not balanced. Why? Because the impedances have different values. Since the impedances have different values, we consequently have an unbalanced circuit. In other words, I can't calculate my power from one phase and multiply it by three. I am required to calculate the power of each of the phases and then add them together. I can't multiply by three because, since the circuit is unbalanced, those values ​​per phase won't be equal. So, first thing to pay attention to here, folks. Be very careful about this. We are talking about a three-phase system that is unbalanced.

[00:14:47]So, here we have our connection [clearing throat] of our delta impedances. We have the line voltage and magnitude, which was given as 220 V.

[00:15:01]We can calculate the line and phase current to obtain the power in each of these phases.

[00:15:05]What am I going to do? I'm going to calculate the power in this phase, in this one, and in this one.

[00:15:11]Remember that it is the active power that is given in watts at the end. So, I'm going to add them up. The first thing I'm going to do here is calculate the phase current. Since I have the voltage between each of these phases here, I can easily calculate the currents within them. Simply put, this current here is the voltage divided by the impedance. And so on, just like in the other cases, I will calculate my current for each one, right?

[00:15:42]So, for phase AB, or the current between the impedances A and B, we have a voltage of 220. But what was given to us here? We were told that the VBC reference was 0º. VBC was 0º. I [cough] need the VAB reference and I need the VAC and CA reference, sorry. Putting VBC here was given as 0°, but the sequence is negative.

[00:16:22]The sequence is not ABC. The sequence was given as ACB, as we saw, right? And he told us that this 0º would usually be there in the VAB. But he showed up here at VBC, no problem. In the negative sequence, everyone is rotated relative to their reference point of -120º.

[00:16:44]Typically, the VAB is zero, so with -120 it will become -10.

[00:16:52]The typical VBC here for us is C [clearing throat] is -10.

[00:16:58]Rotated by -10, it becomes the zero that he had already given us in the reference frame. And the VCA is plus 120 minus -10, which will give us those [roaring] 240º there, folks. So we can now proceed with this calculation based on the reference that has been given to us here. So the current AB is 220 A at an angle of -10° and the impedance was given as 20 A at an angle of 90°. If I divide the magnitudes and subtract the phases, my current is 11 A at an angle of -210°.

[00:17:38]And that's how I'll do it for the other phases.

[00:17:42]The BC current is 220 at a 0º angle.

[00:17:46]divided by the impedance, which is 44, angle of 60º. So I'll divide the modules and subtract the phases. It will give a 5- degree angle of -60. And finally, in the AC current I will have 220 angle of 120º divided by 22 angle of 0º, which will give me 10 angle of 120º.

[00:18:12]amp. Note that the currents are completely different and are not 120º out of phase, but this proves that this circuit is unbalanced based on these currents right now. And we 'll realize that these phases here won't matter much when it comes time to do my calculations. What can I do? I will then be able to calculate my power values ​​here as the real part of the impedance times the square of the current magnitude.

[00:18:45]That's why I said that the final result for the current doesn't matter much, because how do we calculate the active power in each of the phases?

[00:18:55]It's the magnitude of the current squared times the actual portion of my impedance there. Thus, the real part of the impedance is nothing more than the magnitude and cosine of its angle. Therefore, for the impedance modulus, which is 20, and the cosine of the angle, which is 90, and the calculated current, 11², which equals 0. And we follow the same reasoning. Take the magnitude of the impedance, the cosine of its angle, and then multiply the square of the calculated current. For B it gives 550 and for C, following this procedure, it will give 2200.

[00:19:32]Therefore, the active power that the source supplies to the entire circuit is the power consumed in phase A plus that of phase B plus that of phase C, that is, 0 + 550 + 2200, which will give 2750, which is letter C. If the circuit were balanced, the power developed in A would be a value that would be repeated in B and C. In this way, I wouldn't need to calculate for the other phases, thus eliminating a good part of the problem. You would simply need to calculate for one phase and multiply by three. But since the circuit is not balanced, that's not true. We absolutely need to calculate separately for each phase and then add up our power at the end.

[00:20:21]So, we have resolved our first issue. If you have any questions, everyone, just let us know here, okay? But here are some important insights to pay attention to regarding three-phase circuits. First, check if it 's balanced or not. It's extremely important for us to realize this during the exam, especially since we're talking about an exam board with a very high level of difficulty, so they're not going to make it easy for us, okay?

[00:20:54]So, pay close attention to this point.

[00:20:58]Our next question, which is already a question about direct current circuits, because in terms of calculation it's easier for the exam board to ask us about that, is a typical example of what César Gran Rio likes to ask, okay? It's a matter of direct current circuitry, okay? which seems simple, but requires a certain insight and good skill when applying the problem-solving methods, whether it's Kirchhoff's Current Law (KCL) or Kirchhoff's Voltage Law (KVL), which I don't highly recommend for a public service exam because it can confuse you more than help you.

[00:21:33]Yes, in that case, using mesh analysis or nodal analysis tends to give you much faster answers right when you're developing the problem, okay? So let's go.

[00:21:47]The figure below represents a circuit.

[00:21:50]We're looking at it here; the question asks for a resistance equal to 1 ohm. What is the value of the constant K? So let's go.

[00:22:03]He wants to know what value of resistance would be perceived between these terminals if I came here and measured the equivalent resistance with an ohmmeter, for example.

[00:22:17]This circuit consists of three resistors, two of them with a value of 2 ohms and one with a value of 1 ohm. But if that were all, it would be too easy.

[00:22:27]We would create a set of parallel series. Our lives were settled.

[00:22:31]thus obtaining our equivalent resistance. However, in this case, we have a dependent source. Note that source V3 depends on the value of V2. The V3 source is nothing more than K x V2. So, since I have a dependent source here, I can't use a series and parallel combination to solve the problem, folks. That's not how it works when you have a dependent source. In fact, this is the worst possible scenario when trying to solve a direct current circuit with dependent sources, folks. So, professor, how can I calculate the equivalent resistance when I am dealing with a dependent source? First step, independent sources should be eliminated. The voltage source becomes a short circuit and the current source an open circuit.

[00:23:29]This step isn't necessary here, since we don't have independent sources, okay? But since we have a dependent source, to calculate the equivalent resistance between A and B, I will connect a voltage source between these terminals. Whatever amount you want, any amount will work. I'll typically choose 1 V, which is the most established one there. Once that's done, I 'll calculate the current flowing through that source. The ratio between the voltage and the current that has just been calculated will then give me the equivalent resistance that I can see through those terminals there. So, with that information, I'll be able to calculate K, since I've already been told that the equivalent resistance is 1 ohm. Let's stop talking and get down to business.

[00:24:23]So, the first thing we're using here is the ohmmeter method, which is simply associating a voltage source of 1 V and trying to calculate the current passing through it to obtain the equivalent resistance.

[00:24:40]So, my goal here is to calculate this current that I'm referring to here. Therefore, I will use the mesh chain method for this. I'm going to put our mesh I1 and our mesh I2 here. Write the equations, solve them, and then obtain a relationship there for I, right?

[00:25:03]For our first loop, loop number one, we have, starting here, 2i, continuing the path with another 2i. This resistance is shared, so I will have 1 x I1 - I2 between loop number 1 and loop number 2. So I have 1 x I1 - I2 plus the value of the dependent voltage source, which is K x V2. So, + K x V2 = 0.

[00:25:35]So, I have my expression. But we have a small problem.

[00:25:40]In our method, the mesh current method, I have to obtain I1 I2, in such a way that other variables should not appear. In this case, the dependent variable V2 appears, so I have to eliminate it by expressing it as a function of the mesh currents. In other words, we will express V2 as a function of I1 and also of I2. So let's do it.

[00:26:08]So we're going to observe that the voltage V2 is here with this specific polarity, right? And we realize that current one goes from its higher to its lower end, which is in the correct convention, the so-called passive convention. Voltage is not simply resistance times current. Therefore, our voltage V2 is nothing more than the resistance of 2 ohms multiplied by the current passing through it, which is I1. So V2 is twice in one. So I was able to replace the dependent variable in terms of the mesh currents. So I'll be able to substitute here in such a way as to have an expression that depends only on one, and then notice that I substituted here and arrived at an expression that is condensed: I1 5 + 2k - I2 = 0. This is for loop one.

[00:27:11]And now we're going to do mesh number two. In loop number two, starting here at our source, we then have -k V2. It goes into the minus and comes out in the plus, as we're seeing here. 1 x I2 - I1, as we see here, and the value of the source is 1 V + 1 = 0. We already know the relationship between V2 and I1. V2 is 2 x I1. So I'll replace it here again.

[00:27:44]And when rearranging our expression, I have I1 - 2k - 1 with + I2 = -1. So notice that I have two equations here and I have to obtain an expression as a function of k, which is the desired variable for this current here. We observe, right, that in this case the current i, which is the one I want to calculate, that is passing through my source, is nothing more than the opposite of I2. So I is - I2. So what am I going to do? I'm going to get the solution for the current as a function of K, right? In this case, it's very simple. Since I have two expressions, I just need to isolate the current in one, I1, and substitute it into the other. So I'll obtain the relationship of I2 as a function of K. Then it's just a matter of solving the system there. By doing this, I get 2 = -5 + 2k / 4. But that's not the current I want to calculate. The current I want to calculate is I, as it is shown here. Therefore, IG = - I2. So, that 's the negative side of it. This current I is 5 + 2k divided by 4. So, now I have the current value and I already had the voltage, which was 1 V.

[00:29:13]Therefore, my equivalent resistance is the source voltage, which was 1, divided by the calculated current.

[00:29:22]So, as we saw in the problem statement, this equivalent resistance of mine was 1 ohm. Therefore, the only variable that will remain is k. So I have to have 1 = 4/5 + 2k, right? When dividing fractions, I keep the first fraction and multiply by the inverse of the second. So, that's why it's 4/5 + 2k. If we solve this, we'll easily obtain the value of K there, just by using cross-multiplication or the rule of three, whichever way you find best to determine the value of K. So, for this equivalent resistance to be 1 ohm, it 's exactly -0.5.

[00:30:08]The correct answer is D. It's not a difficult question, far from it, okay, everyone? That's far from it, but it's an issue that ultimately requires a lot of calculation from us, doing a reverse process, which we can sometimes find difficult. And one point I want you to note is that when we have an electrical circuit that uses a dependent source, sit down because here comes a story, we need to be careful, we need to calculate the equivalent resistance. What do we need to do? Use the ommeter method. Connect a voltage source and calculate the current flowing through it. All good? So, pay close attention to that. Do you have any questions about this, everyone? While I'm reading the next question—we have a few that I've set aside here—you can ask any questions and then come back to the chat while I'm reading our next question, if we have any doubts.

[00:31:13][snoring] Just a reminder that all these questions I'm solving are CESGran Rio questions, okay, everyone? Well, these aren't questions that we didn't take from the exams.

[00:31:28]Yes, Adriano, the source can be 1V, 10V, 1000V, 15V, 14V, any value will always give exactly the same response. It can be any amount you want. 1 V is because it makes the calculations easier, okay?

[00:31:42]But it can be whatever amount you want. It will always lead you to the exact same answer.

[00:31:51][snoring] Guys, let's do one more question here. And this issue is quite tiresome. Why? When we talk about first-order and second- order circuits, they are also frequently tested at Petrobras and in general in competitive exams. And especially when we talk about initial conditions, that's where we encounter the main problems. What does this issue with CES Gran Rio tell us? The figure below represents a circuit operating in steady state and with direct current. At a given moment, t = 0, the switch S closes. So, this switch here, before it closes, right, at time t = 0, it remains open the entire time, it was in steady state, which are key conditions for the answer here, but then it closes exactly at t = 0, right?

[00:32:55]Then when t = 0 it closes.

[00:32:59]So we have the following: v0 of t and dv0 of t dt, v0 is the voltage, folks. which is found in this 2-ohm resistor here.

[00:33:10]DV0 is the rate of change of that voltage at time zero plus, which is immediately after closing that switch.

[00:33:20]And then he says how much each of these elements is worth. Everyone's v0 is worth 10, right? And it's the dv0 dt that changes. Look at the templates here, right? But unfortunately I can't get one without the other, right?

[00:33:36]Otherwise, I would just calculate the dv0 dt right away, okay? So he wants the initial conditions in this case at time t = 0.

[00:33:45]For that, folks, we're going to have to do two analyses: the initial conditions before this switch closes and how our circuit will be after the switch closes, okay? In this case, we have a resistor, which are the voltages across the resistors, but we also have a capacitor and an inductor, which have certain initial conditions when they operate in steady- state direct current.

[00:34:12]So let's separate our analysis here, folks. First, I'm going to analyze my circuit at zero minus, that is, when the switch is still open, right? When the switch is still open, my circuit looks like this.

[00:34:28]What happens? The key is open.

[00:34:31]Since we are in a steady-state direct current regime, the inductor behaves like a short circuit, while the capacitor behaves like an open circuit.

[00:34:42]So, the key was locked, it was here, right? It's closed here, it was open, okay? Our inductor, which was in this position, becomes short-circuited, as we can see here. And our capacitor here becomes an open circuit. This is at time zero minus, he wants it at time zero plus.

[00:35:04]But I need to do some analysis before that. Why? Because I know that at the instant zero minus and zero plus, two things are conserved before and after the switch is closed, which is precisely at that moment our current in the inductor and the voltage in the capacitor, which are variables that cannot change abruptly. Therefore, at time zero minus and at time zero plus we have this equality. So, I'm going to look for that equality at the negative zero and then I'm going to analyze it at the positive zero. Let's then calculate our currents in the inductor and the voltage across the capacitor under these conditions. Right off the bat, in our circuit, we can see that the voltage across our capacitor at -0 is 10V. Why, everyone? Note that the current will only flow here, since this is an open circuit. So, the voltage across that resistor is also the voltage across the capacitor, which is in the open circuit there.

[00:36:08]So, the voltage is resistance, which equals one, and the current equals 10. Therefore, the voltage across the capacitor at that instant is 10 V. And the current in the inductor, folks, is there any current? There isn't even a source. Therefore, the current is zero. Therefore, the inductor current is zero. But that's at the moment zero minus. But given these conditions that I discussed exhaustively in our theoretical class, in our material, the zero-time minus for these variables only applies to the voltage across the capacitor, and the current through the inductor is exactly the same. 0 men exactly the same in 0 more. Therefore, the voltage across the capacitor at +0 is 10 V and the current through the inductor at +0 is 0.

[00:36:58]Having obtained these variables, we now return to our circuit.

[00:37:03]But now in our circuit we're going to analyze the zero point, which is the moment the switch has just closed.

[00:37:12]My key locked. In that case, I am not on the permanent regime. I just transitioned the circuit. Therefore, the inductor reverts to being an inductor and the capacitor reverts to being a capacitor, and no longer a short circuit and an open circuit, respectively, since the steady-state condition has ceased. So, take a look at our circuit. Here was the key that ends up being locked. The inductor becomes an inductor again, and so does the capacitor. But I know that at the zero point, what else? I know that the current in the inductor is 0 amps. At that instant it is zero plus. So the current passing through here is zero. And I also know that the voltage across my capacitor right now is 10V. How cool is that? Knowing this, if I apply this circuit here, the voltage across the capacitor is equal to the voltage V0 itself, which is the voltage across the 2-ohm resistor; they are in parallel. So, that's why the voltage V0 at 0+ is exactly equal to the voltage across the capacitor at 0+, in this case, which is 10 V for us. That's why it was so easy that every answer key said it was 10, right? Now we have to find DV0 dt, which won't be difficult either; we just need to remember two expressions: that our current in the capacitor is C DV0 dt and the current in the inductor is LDT.

[00:38:48]These two expressions, derived from inductive and capacitive elements, will help us obtain the derivative of DV0 dt. To help us, let's remember the following. The current in the capacitor is C, derived from the voltage across the capacitor with respect to time. But what did we just see? The voltage across the capacitor is the voltage V0 itself.

[00:39:13]We saw that they were working in parallel.

[00:39:14]I just stated that right here. They were running parallel.

[00:39:19]Therefore, the current in the capacitor is C DV0 dt, as we have here. So, here is the element that we want to calculate. which is dv0 dt. So I have that this ratio, this derivation, is nothing more than the current in the capacitor divided by the capacitance.

[00:39:40]I have the capacitance value, but I need to track down the current in the capacitor at time zero plus. To do this, I'm going to go back to my circuit and, using those previously calculated initial conditions, obtain the current at that instant. So I have that the derivative of V0 with respect to time is the current in the capacitor at zero plus the capacitance. How much will this chain be worth? If I go back to my circuit, I can easily do it.

[00:40:11]Why? I want this current here, which is the current in the capacitor at zero plus.

[00:40:17]So, if I analyze this node here using DKC, I have this current coming towards me, right? I also have this other chain coming to me here, in this case, for the amount of 10, right? We just need to remember that the voltage here is 10 V divided by the resistance of 1. So, that's why the current here is 10. But I also have another current of 10 arriving at this node. I have the zero current previously calculated in the inductor, and I now have the current in this element, whose voltage is 10 V, calculated by the resistance, which is 2 ohms, which is 5. So, in this circuit, there are several current outputs. If I take DKC into account here, I've arbitrarily assumed that the outgoing currents are positive and the incoming currents are negative, so I arrive at my capacitor current as 10 - 0 - 5, which will give -5.

[00:41:25]Therefore, the ratio dv0/dt is -5, which is the current value divided by 4, which is -1.25 V. The answer is B.

[00:41:37]Interestingly, this sign is negative, as expected. Why? The negative sign indicates that the voltage is dropping, right? In this case, if we observe our circuit, after the switch closes, this inductor will become a short circuit again later on, when the steady state is established. And since it's in parallel with a short circuit with that resistor whose voltage is V0, the final voltage there will be zero.

[00:42:05]So it was to be expected that as time passed, this tension would decrease, as indicated by the negative sign. This is a strong indication for us here that we would have the right answer. So we can still use this theoretical analysis to indicate that we are correct. That was an issue from Petrobras, okay everyone? I know it 's not a trivial matter, we need to practice, but pay attention to this type of question, okay? Very important. Even worse, if after all that we still needed to solve some differential equation—and they usually include that—it would take even longer, which is already an extremely lengthy issue for us, okay?

[00:42:54]If you have any questions, let us know here, folks, otherwise we'll move on to our next question.

[00:43:02]Our next question, Petrobras or Transpeto, I don't remember exactly off the top of my head, but I know that CES Gran Rio, regarding Petrobras system exams, says the following: Consider a three-phase network supplying two balanced loads, both configured in a star connection.

[00:43:27]These loads are designated as loads C1 and C2, possessing impedance per phase, respectively. One is worth R + JX and the other R - JX. Since they are balanced, all phases assume this value.

[00:43:46]Charges C1 and C2 will be replaced by a single equivalent charge. So I have to establish the equivalence between C1 and C2. And this equivalent load C3 is configured in a delta configuration and dissipates the same power as the previous loads.

[00:44:05]He wants to know the values of R and X as a function of R and X of the equivalent impedance. In other words, I have two loads connected in a star configuration, and I want to create the equivalent of them, but now connected in a delta configuration. This is a piece of cake, guys. Whenever we have two three-phase loads connected in parallel, the tip is: convert them to delta if they are not already connected in that way, because I can only connect them in parallel if they are in delta. So, the configuration that is proposed here for us is the following. I have charge C1 with the value R + JX and I have another charge C2 with the value R - JX.

[00:44:57][snoring] First step for me to simplify and obtain a single delta load, I'm going to convert this circuit here and this circuit here to delta. How do I do y? Simply remember that Y-connected impedances are nothing more than delta-connected impedances divided by three when balanced. So, in this case, we're going to make use of that here. Since I want the impedances in a triangle configuration, if I do the reverse, then dividing is multiplying. Therefore, the impedances in a delta connection are nothing more than 3 times e.

[00:45:37]So, notice what I've done now?

[00:45:40]I converted the star connection to a delta connection, and all the loads were multiplied by three. So here I have 3 multiplying R + JX in each phase, and for another load 3 multiplying R - JX for the other phase here. Now I can connect each phase in parallel, obtaining equivalent impedance for each of them. So I'm going to draw a parallel between 3r + jx and 3r - jx. Making a parallel comparison is the product times the sum. Therefore, I will have here 3r + jx divided by 3 - jx, which is the product, divided by the sum 3r + jx and 3r - jx.

[00:46:36]In multiplication, 3 x 3 up there is 9.

[00:46:40]And now, using the distributive property, I have R x R, R², R x - JX - JXR, JX x R + JXR and JX x - JX + X², because remember that J² is -1, right? With that other negative, it became positive. And down here we'll see that this jx term will cancel out, leaving me with only 3 x 2r. If we simplify things here, we can cancel out this element. So we have this simplification here as well. So we have 3 multiplied by r² + x² divided by 2r. This is the value of our only equivalent impedance in a delta configuration, whose answer is E. Nothing supernatural about it. The key tip is that when connecting three-phase loads in parallel to perform the product-by- sum calculation, they must be connected in a delta configuration. That's a very interesting tip.

[00:47:50]For you guys. So, we have the correct answer here as letter E. If you have any questions, let us know, okay everyone? Don't be afraid to ask your questions. This is the right time, is n't it? You just can't make any mistakes on the test. Here you can make mistakes or get confused at any point.

[00:48:10]Let's ask another question. I personally love this question because it looks so easy, but it's actually quite difficult, folks. It 's not simple, even though it's very short, it does require a little bit of work. Let's see how we can address this issue.

[00:48:29]For the circuit below, a resistor RL was inserted, which is shown here, and its chosen value has a very famous name: maximum power transfer. Guys, this is a study that we do exhaustively in our material. The mass- power transfer theorem is invariably linked to the equivalent of TVN. So now you know the value and the subject matter we're going to have to calculate here, right? So he wants to know the value of RL for maximum power transfer to occur. Since this is going to happen, he wants to know how much power is being developed in that resistor. Everything alright? Let's go. The first thing we need to do, folks, is remember that for the load resistor, in this case the added resistor RL, to have maximum power transfer, its resistance must be exactly equal to the resistance of TVAN. So, our first step here is to calculate the resistance of TVNã. How do you obtain the resistance of TVN? So, see through this RL resistance here.

[00:49:53]The normal procedure is to disable the independent sources, in this case the 12V source.

[00:49:59]And our resistors will be, you know, a series and parallel combination there. But why is the question malicious? Because you can't do either a series or a parallel one, you can't get out of that.

[00:50:14]One cool way is to do the star-delta transformation, but it will be a lot of work because it's an unbalanced star-delta transformation and the formulas are very extensive, okay? So, in this case, the only way to obtain the equivalent resistance seen through RL, which will then give me my TVN resistance, which is exactly the value it should assume to have maximum power transfer, will be by again using the meter method. What am I going to do?

[00:50:47]I nullify independent sources. So, I disconnected the 12V source. And where that resistor was, which are the terminals that will be open, I connect the voltage source. In this case, I orbited 1 V and I want to calculate the current that is there. That's what I do. Note that here I have neutralized the voltage source, resulting in a short circuit, as expected. I connect the value of the 1V source to where I want to obtain the equivalent resistance, and then I go there and want to calculate this current here. So now I need to solve my circuit so that I can calculate this current. If I write out the mesh chains, and since we've practiced a few times, I'll be more direct: for mesh one we'll get the following expression: 24 and 1 - 4 and 2 - 23 = 1. For mesh 2, we'll have -4 and 1 + 18 and 2 - 10 and 3 = 0. And for mesh number 3, -21 - 102 + 53 = 0. Practice at home, okay? You will have access to this recording immediately after the class ends.

[00:52:01]Put it together, see if you're able to develop this method well, because it's fundamental to doing well on the test. Write out the three equations. Then you can calculate the current.

[00:52:14]We realize that the current I want to calculate is exactly my current number one. Note that current one flows in that direction, exactly the current I want to be flowing in the source. By solving using the mesh current method, I obtain that I1 is 0.0833, which is precisely the current passing through the source. So, the resistance of TVNã is equal to the resistance of my power supply divided by the current value, which is 12 ohms.

[00:52:53]So, now that I have this TVN resistance and I want to calculate the power, I already know that the load resistance must be equal to the TVN resistance. And in that case, considering that 12 ohms is the ideal voltage for maximum power transfer, I 've already solved that part. But since I also want the power developed, we have an expression for that. And the expression depends on the tension of TVNã and there goes more counting then. The issue is not that simple. So I want to obtain the TVN voltage, which is called the open-circuit voltage between my terminals here. Note that, as the theory already tells us, to calculate the TVN voltage, we don't have to cancel any source. I calculate the voltage across the open terminals and that's all.

[00:53:46]By doing this, I only have two mesh chains, I1 and IS, because this mesh here, obviously, is what? If it 's open, no current will pass through it.

[00:53:57]So, writing the mesh current for I1 and I2 and solving them, I get the value of I1 as -0.75 and I2 as -0.15.

[00:54:11]Therefore, with these currents, I can calculate the voltage of TVNã in the loop you want. In this case, a simpler loop to obtain here, the calculation of this TVN voltage is using this path here, which is then minus the TVN voltage 20 times my current I2 and four times my current I equals 0. By doing this and substituting our already calculated current data, my TVN voltage is 6 V. As the question wanted to complicate things and asked for the maximum power, this maximum power is calculated as the TVN voltage squared divided by 4, sorry, divided by the TVN resistance. So the calculated voltage was six and the calculated resistance was 12, okay? Its maximum power is 3/4 W. It's a very extensive question, because having to calculate the resistance of TVNã and then calculate the voltage during the exam is very complex. It gets quite lengthy when I explain it here, but whether we like it or not, the examiners end up asking us about it, okay everyone? How do we do that? By practicing a lot of problems, improving, and solving many circuits, you can solve the problem as quickly as possible and also learn the best methods when solving them. So this is going to be strictly up to you guys.

[00:55:49]Practice circuit problems a lot to gain agility and speed when solving them, okay?

[00:56:00]And to finish our lesson, I'm going to do one more question, which is also difficult, and which will deal with topics that are frequently tested, which we just saw earlier, namely the TVNAN theorem, which is closely related to the maximum power transfer theorem. And in this case, which is the worst-case scenario, we're going to address the issue of dependent sources again. Look, I've selected the most tricky questions we found in the material, okay?

[00:56:33]So, look what my question says. Consider the circuit below.

[00:56:38]This circuit has a 16V source.

[00:56:40]It only has resistors, with two open terminals, A and B. And it has a voltage-controlled current source. So you can see that I have a dependent current source, right?

[00:56:55]Because we have a rhombus, generally the rhombus indicates to us that it is a dependent source. And she is controlled by tension. It is controlled by the voltage between C and A, which is this voltage here. VCA is the voltage across the 4-ohm resistor. Here it's positive and here it's negative because of the sequence, right?

[00:57:15]First C, then A. And he wants to know which option contains the value of the TV's resistance. Again, this is a very straightforward question, very much in the style of CES Gran Hill, a very direct question: calculate this, give us the circuit and calculate it.

[00:57:32]In this case, we will obtain the resistance of TVN as the equivalent resistance between terminals a and b. Up to that point, it would be cute. If there were no dependent sources, what would the procedure be?

[00:57:47]It cancels the voltage source and makes the series parallel, if possible. We saw in the previous question that it wasn't, right? Even though it didn't have a dependent source, the arrangement was such that I couldn't simplify it. Then I had to use the meter method.

[00:58:04]In this case, I am forced to use the ommeter method. Why? Due to the existence of the dependent source. So I'm going to calculate the resistance between A and B.

[00:58:14]But what is the step size? I'm going to disable the 16V power supply there, thus creating a short circuit. And I'm going to connect a source of any value, in this case 1V, between A and B. And I'm going to calculate the current that will be passing through it. So the equivalent resistance, which is the actual resistance of TVNã, is nothing more than the value of the voltage source divided by my calculated current.

[00:58:41]So let's do it, everyone. So, here we have our ommeter method drawn. So, I have my voltage source set to 1 V. I want to calculate the current that will be passing through it here. To calculate this current, I have three loops, so we will use the mesh current method. But right away, when solving using the mesh current method, we learned, whether in college or in the material we prepared for you here in the course, that when a mesh shares a current source, as in this case, we have the formation of a superloop. For supermal, we write two expressions, one associated with the current source itself and the other associated with the path of our current, or our loop, in this case the sum of the voltages equals zero.

[00:59:39]So, let's observe here that we have Supermal, right? In this case, they merged [sighing] into one.

[00:59:51]The first expression associated with the current source. Note that I have the current in this direction here. I have current I1 in this direction here.

[01:00:02]So, I have I2, [rumbling] which agrees with the direction of the current source which is upwards, - I1, which is in the opposite direction, is equal to the value of the current source, which is 0.1 VAC.

[01:00:18]But here, this control variable, which is VCA, has to be a function of the loop currents. And then I realize that the AC voltage, which is this voltage here, is nothing more than 4 x I2. The voltage between C and A is the voltage across the 4- ohm resistor times the current passing through it, which is I2. So I have that VCA is 4 I2.

[01:00:45]If we substitute and rearrange, then I have that -i1 + 0.62 equals 0. First expression of supermalia. The second expression of supermal is using LKT here on our path. So I have the voltage across the 8-ohm resistor, which is 8 times I1, as shown here. Continuing, 4 times I2.

[01:01:10]Continuing, 4 x I2 - I3. And then we don't have anyone here anymore. That all equals zero. So here is the first expression, and rearranging the second expression, 8 and 1 + 8 and 2 - 4 and 3 = 0. I have three unknowns: one, two, and three. So I need three equations.

[01:01:35]The last equation will come precisely from our mesh number three. From our grid number 3, we will then have 4 and 3 - 2 + 1 = 0, as I wrote here.

[01:01:50]Organizing, we have our third expression. If I solve this system with the three equations, I get the currents I1, I2, and I3. My goal is not to obtain these currents, but to obtain this current here, Ai, which is exactly the opposite of the current I3, which is in the opposite direction. So, I is the opposite of 3, that is, 0.364.

[01:02:22]So now, my equivalent resistance between those terminals, which is nothing more than the value of the voltage source, which is 1, multiplied by my current, which is i, which is 1 divided by 0.364, giving 2.75 ohms. Our answer is shown here in option C. An extremely interesting question, right everyone? The level of expertise we're providing you with at Petrobras is exactly what we're preparing you for, okay? So [snoring] this is the type of circuit question you can expect during the exam, nothing easy is going to come, okay? So we saw that here, all the questions, if they weren't difficult, they were lengthy, okay?

[01:03:15]So we need to be very careful and attentive when studying circuits, because we can't afford to get these questions wrong.

[01:03:23]Now that we've covered the topics we set aside for you all, feel free to contact us with any questions you may have about the competition, the exam, the topics, or any other related subject you may have.

[01:03:40]You can contact us here in the chat and we'll respond promptly in this final stretch.

[01:03:47][snoring] [snoring] Thanks for the feedback, Adriano.

[01:04:33]Class, if we have no doubts, then we'll meet in the next class.

[01:04:37]Remember that in our material we have a series of questions about electrical circuits. I've already prepared another 200 for you guys, besides the questions from the 01 app, which is really good, okay? These are very good, even difficult, questions, which is what you will have to answer. So, from now on, study and practice so you can do well on those circuit questions, which can easily account for more than 10% of your Petrobras exam, okay? Have a great night, happy studying, and see you next time, everyone!

[01:05:13]A big hug.