If you're coming from the previous article, take a deep breath, because we're going to continue with important concepts. Take your time to slow down and make sure you understand what's being explained. If you have any questions, leave a comment and we'll try to help you out or improve the explanation. By the way, you might want to review everything mentioned in the second article in this series. Do you remember series-parallel associations? Short circuits or open circuits? If the answer is "phew...", you definitely need to go review it.
In this article, we'll land the previously discussed concepts and apply them to basic circuits. We'll look at the basic elements of electrical circuits and how they're represented. Then, we'll use them to provide practical examples of Ohm's Law. Finally, we'll test your understanding of the concepts with some slightly complicated examples. Let's get started!
Circuit elements
So far, we've covered circuits by talking about light bulbs and sockets, but it's clear we can't get very far like this. A circuit uses a series of symbols to represent components whose purpose is to describe the circuit's behavior. There are a ton of different symbols, but most aren't commonly used. We won't cover them all here; I'd rather you focus on the important stuff.
Voltage sources
A voltage source is an ideal element that guarentees a potential difference between its terminals. I've emphasized "guarantees" because that's its main characteristic. No matter what we connect to it, it will provide the voltage it is set to. To do so, it will generate the necessary current. In real life, ideal voltage sources obviously don't exist. Depending on the specific situation, voltage sources represent different thing. For example, a voltage source could be a power outlet, a AAA battery, a cell phone charger, a hydroelectric power plant... We'll see how they're made much later. In any case, when you see a source in a schematic, you should implicitly understand that this voltage source is suitable for ensuring the potential difference in that application. There are different ways to represent them in circuits. In the following image, I show you some common ones.
In my experience, the first two representations are used interchangeably. When I draw by hand, I use the middle representation because I'm terrible at circles, even if the voltage source isn't a battery, and it's still understandable. The representation on the right is quite common to explain that voltage is alternating (i.e., it varies over time). In addition to these symbols, there are others used to represent specific types of voltage sources, but as I said, these cover 99% of the cases.
Voltage sources are elements that provide electrical power to the circuit.
Current sources
A current source is an ideal element that guarentees a current flows through it. To do so, it will generate the voltage it needs. As in the previous case, this is also an ideal element. In real life, there aren't many examples of human-scale current sources; however, at the circuit level, they are everywhere; we'll get to that later. The best-known element that acts as a current source is the solar panel, which, for a specific operating point (lighting, temperature), provides a current regardless of the load, as long as it doesn't exceed its maximum capacity. With current sources, there's less creativity, and almost everyone draws them like this:
I'm not saying it's the only way, but it is the most common. Current sources are also elements that provide electrical power to the circuit.
Resistors
The symbols we'll see below are used to represent an electrical component called a resistor , or anything else that behaves like one. For example, traditional light bulbs behave like resistors (not LEDs), so they're drawn as such. Since you already know what Ohm's Law is, you can imagine what a resistor is. It's a component that guarantees a specific voltage-current relationship across it. There's a lot of language abuse with these elements. It's common to see the words resistor and resistance interchanged. The former is the component, and the latter is the physical magnitude. In fact, we're so lazy that we sometimes just say the letter "R" 😆in phrases like "that R is 3 KOhm" or even "that R is 3 Ks." There are two typical symbols for Rs: the American one, and the one from the rest of the world. They look like this:
There are many more symbols, even for components you've heard of: diodes, transistors, transformers, relays, inductors... Everything will come in due time. Now it's time to solidify Ohm's Law, so there's no need to complicate things further for now.
Unlike voltage sources, resistors are elements that dissipate power.
Electrical connection
We've seen some element symbols, but we haven't discussed how to connect them to each other. In circuit theory, we assume that elements are touching each other with nothing in between. When I say nothing, I mean nothing: not even a millimeter. This isn't possible in reality, but the error introduced into the calculations is negligible in most cases. This is because, to connect elements to each other, we actually use conductors of various types (see the first part of this series if you haven't already) that are almost ideal conductors. That is, their resistance is very low. Because it's so low, for "normal" currents, there's no voltage drop. In other words, the entire connection between elements is at the same potential and behaves like a single point. Therefore, when you see a schematic (circuit diagram), you can assume that the connections between elements are a single, infinitely small point, without having to worry about how that's physically implemented. In my experience, this concept seems simple, but you'll see that it can lead to some confusion when interpreting schematics.
Examples with Ohm's Law
Circuit with voltage source
Finally, you now have enough elements to analyze simple circuits. In this section, we'll look at some examples. We'll go to the most basic circuit possible, consisting of a voltage source and a resistor. In fact, the circuit we're going to look at is the one Georg Ohm himself used to study his famous Law. In his case, the voltage source was the thermocouple, and the resistor was the wire he used. If you don't know which thermocouple I'm talking about, see the previous entry in the series. You can see this circuit drawn below.
To analyze a circuit, we must first look at it and think (never forget thinking). What do we see here? On the left, there is a 5V voltage source. This means that, no matter what happens in the universe, there will be 5V more on the positive terminal than on the negative terminal. What else is there? On the right, we see a 1KΩ resistor. This value tells us that 1000V (or 1KV) is needed for 1 Ampere to pass through the resistor. More generally, it tells us that the ratio of voltage across the terminals of the resistor to the current passing through it is 1000. Come on, let's keep looking and thinking, there's more. Between the source and the resistor, there are two "wires". One connects the positive terminal of the source to the top terminal of the resistor. There is another wire below that connects the other terminals. There is one last important element: the ground symbol . This symbol marks what we consider 0V in our circuit. We already reflected on this in the previous entry in the series, so what I'm about to say shouldn't surprise you: we could draw that symbol elsewhere in the circuit and the behavior would be the same. If the phrase doesn't make sense to you, review it again or ask in the comments. We've just seen the circuit, but we haven't finished thinking. The "wires" make the circuit closed. This means that a current can exist in it. In this circuit, the current could flow in two directions: clockwise and counterclockwise. We've established, by convention, that it runs from the positive terminal to the negative terminal. For now, let's not question this convention. The path the current follows is usually drawn. In this circuit, it's obvious, but there are more complex cases. Now we have to follow the current leaving the source and see what elements it encounters until it returns to it. In this case, it only encounters the resistor. We mark the voltage that will fall on the resistor by placing a "+" where the current enters and a "-" where it exits. We use these symbols to explain that the voltage will increase or decrease as the current flows in the same direction. Now we just need to numerically determine the current flowing and the voltage across the R. It's not too mysterious, using Ohm's law:
That is, if you apply 5V to a 1K resistor, you force a current of 5mA. Now, what's the voltage across R1? Well, again, we can use Ohm's Law.
Obviously, this gives us the source voltage. This is because there is only one element in this circuit. If there were more elements in series, the source voltage would be divided among them. Another way to see this is to realize that the source and the resistor are in parallel (totally lacking in rigor: they share wires on the top and bottom terminals). The elements in parallel are always at the same voltage, as we saw in the second part of this series.
This example may have seemed very bland to you, and it's true, but it's a necessary step. Be careful not to get complacent; it all seems so easy until, suddenly, you don't even know how to begin analyzing. Let's test ourselves with the following example.
Circuit with current source
Take a look at the following circuit. You'll see that it's very similar to the previous one. It has two elements, yes, but one is new! There's a current source, which we've called I2. Why I2 and not I1? Because we're going to use I1 for the current flowing through the circuit. That is, I2 is the name of the "current source" component, while I1 is the name of the current flowing through the entire circuit.
In this circuit, I1 equals the current generated by I2. This is because there is only one loop . I've used a new word, although it's not a new concept. In electronics, we say that a loop is a closed circuit through which a current can flow. Let's continue studying the circuit. Remember that a current source imposes a current passing through it regardless of what's connected to it. In this case, the source imposes 2.5 amps. The current passing through the source passes entirely through R1 since it's in the same loop. What voltage will be dropped across R1 by this current? Ohm's law to the rescue:
Pretty simple, right? I'm going to ask you a question so you can assess whether you know the answer before reading on. What voltage is dropped across source I2? Here, the answer may seem obvious to you, or you may think something like, "What a tricky question..." We need to think about the topic a little. Up until now, if we wanted to know the voltage dropped across an element, we used Ohm's Law. How do we apply Ohm's Law to a current source? We know the current, fine, but... what about the resistance? It's clear that Ohm's Law doesn't seem to fit to solve this problem. We have to resort to other ideas. In the previous example, the voltage source, we saw that 5V was dropped across the resistor. We didn't need to ask ourselves, as we are doing now for the current source example, what the voltage was dropped across the voltage source. There was no need to because the answer was obvious. However, a very important subtlety has been hidden: the voltage generated by the source, 5V, added to the voltage dropped across the resistor gives 0V. Note that the voltage of the source is generated (positive, voltage increases), while the 5V of the resistor drops (negative, voltage decreases). The fact that the sum of the circuit voltages gives 0V is not a special case of this example. This phenomenon is called Kirchoff's Voltage Law , and it basically states that the sum of voltages along a circuit equals 0V. In more human terms: if you start adding the voltages from a point, and you arrive back at that point, the sum of the voltages will give you 0V, regardless of the elements you pass through and how long or short the path is. This makes quite sense given what voltage is. If you go back to the third part of the series and refresh the voltage section, you'll see that it makes sense that if a charge moves through the electric field starting from a point but returning to that same point, the total work is 0 (as long as the field doesn't change), since the field forces will sometimes make it difficult for it to move, but they will help it return. Well, I've thrown rigor to the ground. If it helps you understand it, fine. If not, forget what I said. Well, then, with all this, we can now answer the question we were asking: What voltage does I2 drop ? Well, -10V because 10V drop across the resistor , and the sum of the voltages across the mesh must be 0V! Why with a negative sign? Because I used the verb "to frop". If I had used the verb "to generate", I would have said 10V, positive. Then, of course, I would have had to count the volts across the resistor in negative.
Here, reasonably, you're probably going to be 💩🤪jinxing everything. It seems like I'm trying to confuse you with wordplay and symbols. I did this intentionally 😆so you can see that the important thing is the sum of voltages from 0V across a circuit. Obviously, to avoid going crazy and to understand each other, we always assume that voltage increases across sources and decreases across passive elements. However, in complex circuits with many sources and components, you won't know at first either the directions of the currents or the voltages at the nodes, so don't be afraid to assign signs to voltage drops. As long as you maintain the same criteria throughout the entire circuit, you'll end up with the correct answer. We'll see more on this later.
Phew... That circuit was quite dense. It wasn't actually complex, but I took the opportunity to sneak in a very important concept. If you're not clear on the whole Kirchoff voltage thing, don't worry, it'll be very common, and you'll eventually get the hang of it.
Voltage divider circuit
You're about to learn about the most used circuit in the world (a made-up statistic). Before looking at how it's sized, let's consider its usefulness. Normally, in a circuit, we have access to one or more power supplies. However, to keep circuits simple, small, and inexpensive, we try to reduce their number as much as possible. On the other hand, in several parts of the circuit, we'll need to "use" voltages other than our source. This is where the voltage divider comes into play. This circuit allows us to generate any voltage lower than that of the source by appropriately selecting the resistors that form the divider (R4 and R3 in the example below). However, you must keep in mind an important limitation: the divider voltage (Vo) will be very inaccurate if the current draw hanging from this voltage is large relative to the divider current. We'll see why later. In the world of electronics, the verb "to hang" refers to circuits that are powered by a certain voltage. For example, in the circuit below, R4 and R3 hang from V4. Let's take a look:
Our goal is to calculate Vo, or the voltage divider's output. I'm going to go faster than the previous circuits because I assume you've already understood them, as well as the series/parallel resistor association from the second part of this series. If not, review again.
First, I see that source V4 is 10V. Due to Kirchoff's voltage law, these 10V must be shared between R4 and R3. I also see that the source and the two Rs form a loop through which current I1 flows. Okay, we don't need anything else. Let's calculate the current and voltage across the Rs using Ohm's law a few times.
1. I simplify R4 and R3 into a series resistor. This gives me an R of 3kOhm.
2. I calculate the mesh current I1 with Ohm's Law cleared for current: 10V between 3K = 3.33 milliAmperes.
3. Okay, now I can calculate Vo using Ohm's Law for voltage. I'll use the mesh current multiplied by R3. This is 2kOhm * 3.33mA = 6.66V . Look how easy it is! You now have the divider output.
4. It's not necessary, but we could calculate the voltage across R4. There are two ways:
a. By Kirchoff's voltage law: if 10V must be distributed between the 2 Rs, and R3 takes 6.66V, R4 is left with 3.34V.
b. By Ohm's law: I multiply the mesh current by R4 and I should get the same: I1*R4 = 3.33mA *R4 = 3.33V. It has been different due to rounding, if I had used infinite decimals it would have been the same.
Fantastic, you've seen how easy it is to work with voltage dividers. BUT WAIT!! You want it easier?! Well, it can be. Since the voltage divider is so widely used, there's a shortcut formula:
This equation has one good thing and one bad thing: the bad thing is that it seems like a bit of a rocket science if you don't spend a couple of minutes substituting variables and checking that it's actually correct. The good thing is that it's very clear how adjusting the Rs allows you to choose the output voltage. Let's calculate the divisor using the equation:
This equation is remembered with the following phrase: "the input voltage times the opposite R divided by the sum." Repeat again: "the input voltage times the opposite R divided by the sum." Once again: "the input voltage times the opposite R divided by the sum." You'll know it by heart. Of course, at Alternatronics, we've thought of everything, and you have a voltage divider calculator at your disposal. Check it out! But come back, we're not done.
We have one last important thing to note about the limitations of voltage dividers: we've said that voltage dividers are only effective if the current drawn at the divider's output is small compared to the mesh current. Take a look at the following image:
We've added another resistor to the circuit called R_L. The "L" for "Load" is often used. This resistor isn't actually a physical resistor, but rather a representation of all the circuits that hang from the voltage divider output (Vo). These circuits are in parallel with R3. Therefore, since Vo depends on the voltage dropped across R3, Vo will change if we put something in parallel with R3. If the consumption through R_L is very low (large R), the effect will hardly be noticeable. If the consumption is very high (small R_L), the parallel R3||R_L will output something close to R_L, and Vo will change. If the last few lines seemed unclear to you, I recommend reviewing them and leaving a comment asking anything that isn't clear. I'll give you another explanation (or rather, another way of looking at the same thing) about why the divider voltage drops if we load it. When solving the circuit, we assumed that the current through R4 and R3 was the same. If we connect something between Vo and ground, this is no longer the case. Now, the current passing through R4 is divided between R3 and the new load. If the current going to the new load is very, very small, the error will be small.
Circuit with two loops
Let's move on to the last example circuit. It's a very simple circuit and will serve to introduce another of Kirchhoff's Laws: the Law of Currents. It's very simple and intuitive. This Law states that the sum of the currents entering a node is equal to the sum of the currents exiting the node. I'm sure you've found this quite intuitive, but there's a subtlety. How do you know which currents are entering and which are exiting if you haven't yet solved the circuit? The answer is that it doesn't matter; what you need to do is establish a criterion and always maintain it. For example, in the diagram below, we've assumed that the current passing through the source (Iv5) enters the node, and that the currents passing through the Rs exit from it. We could actually have changed the direction of any of those currents, and the problem would be solved the same. What we would see is that the voltages would also change sign, and the effect would be compensated. It's very important that you fully understand what I've just explained. I challenge you to solve this circuit by assigning a different direction to the currents. If you don't know how to do it, ask below. Come on, let's figure it out.
We have a source (V5) and two resistors (R6, R5). The resistors are in parallel, so we already know that the same voltage is dropped across them. Obviously, the voltage dropped is that of the source, i.e., 12V. What is the current through each of them? Easy, Ohm's Law applied to each resistor.
You see how easy this is. All that's left to do is determine the current flowing through source V5. We use Kirchoff's current law. To do this, we look at the direction of the currents we've determined and see that there's one current entering (Iv5) and two currents exiting (I_R6 and I_R5). Therefore, if the current entering is equal to the current exiting, and the currents exiting are the R currents we calculated, the current entering must necessarily be the sum of those two. Therefore, 18 mA.
Final review
In this post, we've covered several important topics. It's been quite dense, so congratulations if you've made it this far. The concepts you've learned form the foundation of electronics. From them, you can create a lot of rehashes that are useful in many cases. We'll cover some of them later; I prefer not to introduce them yet because the basics can take you anywhere, and it's good not to get distracted by too many things. Let's make a list of the elements you've learned and that you should remember:
- You have seen the basic elements that make up a circuit: voltage sources, current sources, and resistors.
- You have learned to apply Ohm's Law to determine voltages and currents in a circuit.
- You've seen how a voltage divider works, which is one of the most commonly used basic circuits.
- You've learned Kirchoff's Laws: the Voltage Law tells us that the sum of voltages along a mesh is always 0 V. The Current Law states that the currents entering a node are equal to the currents exiting it.
All of this, combined with the concepts from previous posts (series-parallel connections, the concepts of voltage, current, and charge, learning how to draw electrical schematics, etc.), gives you the tools to begin to understand how electronics "works." There are still quite a few important aspects missing, which we'll cover in Alternatronics, but if you've fully grasped what you've covered so far, you're already well on your way.
Before closing this introductory series, move on to the next entry. It'll be a short quiz on unusual circuit situations that will help you strengthen your intuition.
- Conductors, insulators and semiconductors
- Electrical circuits
- Ohm's Law and other important stuff
- Circuit theory and Ohm's Law
- Testing circuit intuition
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