Essential DC Circuit Formulas and Principles

In this video, we're going to go over a list of common electrical formulas that you'll encounter when dealing with electricity and DC circuits. So let's draw a simple circuit, one with a battery and a resistor. So here is the positive terminal of the battery. This is the negative terminal. And we're going to have a current flowing from the positive terminal to the negative terminal of the battery. And of course, that is conventional current. We know that electron flow. in, electron flows in the opposite direction. Now the first formula you need to be familiar with is Ohm's Law, which relates voltage to current and resistance. V equals IR. V is the voltage measured in volts, I is the current measured in amps, and resistance is measured in ohms. Now the next equation that you need to be familiar with is power. The power dissipated by a resistor is going to be equal to the voltage across that resistor multiplied by the current flowing through the resistor. You can also use this formula as well, I squared R and also V squared over R. So that's how you can calculate the power dissipated by a resistor or even the power delivered by a battery. You can use any one of those formulas to get the answer. Now, the next formula you need to be familiar with is this. The electrical work done is equal to power multiplied by time. You can also think about work as energy. being transferred. So the amount of electrical energy being transferred to a resistor is power multiplied by time. Now we said that voltage is measured in the unit's volt, current, is measured in amps, resistance is measured in ohms, but power is measured in watts. Power, electrical power, is the rate at which electrical energy is being transferred. So power is energy divided by time. So it tells you how fast energy is being transferred to a device. One horsepower is equal to 746 watts. So those are some details that you want to keep in mind. One watt is one joule per second. So if you have a resistor that is dissipating 50 watts of electrical energy, or rather 50 watts of power, what that means is that it's converting 50 joules of electrical energy into heat every second. So a resistor that's dissipating 100 watts of power, what it's really doing is it's converting 100 joules of electrical energy into heat per second. So in 2 seconds, it will have converted 200 joules. electrical energy into heat. In three seconds, 300 joules and so forth. So that is the basic idea of electrical power. It is the rate at which electrical energy is being transferred from one device to another per second. Now there are other ways in which you can calculate electrical energy. So in addition to using this formula, electrical energy is equal to the charge that's being transferred times the voltage. Electric charge is equal to the current multiplied by the time, which means current is basically the rate at which charge is flowing per unit time. So what this means is that if you have a current, an electric current of 1 amp, what's really happening is that you have 1 coulomb of charge flowing through a circuit or through a wire per second. An electric current of 10 amps means that you have 10 coulombs of charge flowing through a wire every second. So electric current is the rate at which electric charge is flowing. And electric charge is basically the quantity of charges that you have in any given material. Now, if we replace Q with IT, we get this formula. Electrical energy is equal to voltage multiplied by the current multiplied by the time. Which you can also get that if you replace power with voltage times current. So those are some common formulas that you want to be familiar with if you're dealing with simple circuits, and you want to use Ohm's Law, and you want to calculate power or electrical energy. Now let's talk about the formulas that you need to know when dealing with a series circuit. So let's say we have a battery, and we have three resistors connected in series. So what we have here is the voltage of the battery. We'll call this resistor 1. This is going to be resistor 2 and resistor 3. Now here are some things you need to know. When dealing with a series circuit, The current flowing from the battery, we'll call it I-T, that current is the same current that's flowing through resistor 1. The current that flows through resistor 1 is called I-1. flows through resistor 2, well that current is called I2. The current flowing through resistor 3 is called I3. Now in a series circuit, because it's only one path for the current to flow, the current is the same. I1, I2, I3, they're all equal to each other and are equal to the current flowing from the battery. The total resistance of three resistors in series, it's equal to the sum of those resistors. Now, as regards to voltage, If you want to calculate the voltage across resistor 1, it's equal to the current flowing through that resistor times the resistance. So back to Ohm's law, V equals IR. If you want to calculate the voltage drop across resistor 2, it's equal to I2 times R2. And V3 is going to be I3 times R3. Now the voltage of the battery is going to be equal to the sum of all the three voltage drops, or the three voltages across the three resistors. So VB is going to be equal to V1 plus V2 plus V3. This has to do with Kirchhoff's voltage law, which states that the sum of all the voltages in a loop will add a zero. So if you were to move V1, V2, and V3 to the other side of the equation, you'll get this. Positive VB. The reason why it's positive is because the battery adds energy to the circuit. And then minus V1. V2 V3 the reason why they're negative is because they subtract energy from the circuit they absorb energy from it so they decrease the voltage whereas the battery increases the voltage of the circuit so we have a positive for VB but a negative for V1 V2 and V3 at least that's the way I like to think of it but this has to do with Kirchhoff's voltage law But it is a lot easier to see it this way though. So VB is going to be the sum of V1, V2, V3 when dealing with a series circuit. Now let's move on to a parallel circuit. So we're going to have three resistors connected in parallel to each other, which means they're connected across from each other. VB is going to be the voltage of the battery and the current flowing from it. We'll call it IT the total current So in a series circuit we had just one path for the current to flow in a parallel circuit there are multiple paths for the current to flow. So this is R1, R2, R3, and the current flowing through R1, we'll call that I1. This is going to be I2 and I3. Now, in a series circuit, VB was the sum of V1, V2, V3. In a parallel circuit, VB is going to be equal to V1, V2, and V3, because each resistor is connected across the same battery so therefore the same voltage will be applied to each of those resistors. Now in both series and parallel circuits V1 is still equal to I1 times R1 so the voltage resistor 1 is equal to the current that's flowing through it times the resistance of resistor 1. V2 is going to be equal to I2 times R2 just as before and the same is true for V3. Now, in a series circuit, the total resistance was the sum of the three resistors. In a parallel circuit, the reciprocal is true. So, 1 divided by the total resistance, or the equivalent resistance, is equal to 1 over R1 plus 1 over R2 plus 1 over R3. Now, in a series circuit, because there was only one path for the current to flow, The total current was equal to I1, which was equal to I2, which was equal to I3. For a parallel circuit, it's going to be different. The total current is going to be the sum of I1, I2, and I3. And let's talk about that. Because we talked about Kirchhoff's voltage law, it makes sense for us to talk about Kirchhoff's current law. So here we have the total current. And here it's going to break off into I1. And let's call this IB. Actually, let's call it IA. So if we focus on this point here, this junction, based on Kirchhoff's current law, the total current that is entering that junction must equal the total current that is leaving that junction. So I-T, that's the current that's entering that point, and I-1 and I-A, that's the current that is leaving that point. Now if we focus on this point here. Ia is flowing into that point I 2 is going towards resistor 2 and I 3 goes towards resistor 3 so we have Ia entering that junction I 2 and I 3 they're leaving that point. That's the current that's leaving that point So if we replace Ia with I2 and I3, we get that the total current is equal to I1 plus I2 plus I3. So this formula arises from Kirchhoff's current law. So let me give you another example. So let's say we have 20 amps of current flowing to this point. We'll call it point A. And let's say we have 18 amps of current going here. We have 13 amps of current going this way. And we have 25 amps of current going that way. What is the current flowing in this branch? And is it flowing towards point A or away from point A? What would you say? Well first, let's give the currents variables. Let's call this I1, I2, I3, and I4. And this will be I5. Now the sum of the currents entering and leaving that junction should be equal to the sum of the currents entering and leaving that junction. equal to zero, which is the same as saying the amount of currents entering that junction and the amount of currents leaving will be equal to each other. So we're going to use Kirchhoff's current law to calculate I5. So the current that is flowing into the junction, let's assign a positive value to those currents. The currents that are leaving that point, let's assign a negative value to it. So I1 and I2 is going to be positive because the current is flowing towards point A. And I3 and I4 will be negative because the current is leaving from point A. So we're going to have I1 plus I2 minus I3 minus I4. I5, we don't know if it's entering or leaving, but we're going to put a plus towards I5. So when we solve I5, if we get a positive answer, that means the current is flowing to point A. If we get a negative answer, that means the current is flowing away from point A. So I1 is 20, I2 is 18, I3 is 13, I4 is 25, and let's calculate I5. So 20 plus 18 is 38, and 13 and 25, that's also 38. So it looks like I5 is 0. But if we were to change this, let's say if we made it 29, so we can get a value for I5. What would I5 be if I4 is 29? Negative 13 minus 29, that's going to be negative 42. So this would be negative 4, which means I5 will be positive 4. So if I4 is 29 amps, I5 is going to be 4 amps. Because it's positive, current is flowing towards 0.8. Notice that if we were to add up these values, What this tells us is that we have 42 amps of current flowing to point A, and if we add up these values, we would have 42 amps of current flowing away from point A, which is how it should be. The amount of electrical current that is flowing to a point should equal the amount of current that is flowing away from that point. So that's the basic idea behind Kirchhoff's Current Law. So that's basically it for this video. I just want to give you a list of common electrical formulas that you'll use when solving DC circuits or when dealing with electricity in general. Thanks for watching. By the way, for those of you who want more example problems, feel free to check out the links in the description section below. I'm going to be posting more content as well as videos that are related to this video that you're currently watching, so feel free to take a look at that when you get a chance.

And we're going to have a current flowing from the positive terminal to the negative terminal of the battery. And of course, that is conventional current. We know that electron flow. in, electron flows in the opposite direction. Now the first formula you need to be familiar with is Ohm's Law, which relates voltage to current and resistance.

V equals IR. V is the voltage measured in volts, I is the current measured in amps, and resistance is measured in ohms. Now the next equation that you need to be familiar with is power. The power dissipated by a resistor is going to be equal to the voltage across that resistor multiplied by the current flowing through the resistor.

You can also use this formula as well, I squared R and also V squared over R. So that's how you can calculate the power dissipated by a resistor or even the power delivered by a battery. You can use any one of those formulas to get the answer. Now, the next formula you need to be familiar with is this.

The electrical work done is equal to power multiplied by time. You can also think about work as energy. being transferred.

So the amount of electrical energy being transferred to a resistor is power multiplied by time. Now we said that voltage is measured in the unit's volt, current, is measured in amps, resistance is measured in ohms, but power is measured in watts. Power, electrical power, is the rate at which electrical energy is being transferred.

So power is energy divided by time. So it tells you how fast energy is being transferred to a device. One horsepower is equal to 746 watts.

So those are some details that you want to keep in mind. One watt is one joule per second. So if you have a resistor that is dissipating 50 watts of electrical energy, or rather 50 watts of power, what that means is that it's converting 50 joules of electrical energy into heat every second.

So a resistor that's dissipating 100 watts of power, what it's really doing is it's converting 100 joules of electrical energy into heat per second. So in 2 seconds, it will have converted 200 joules. electrical energy into heat. In three seconds, 300 joules and so forth. So that is the basic idea of electrical power.

It is the rate at which electrical energy is being transferred from one device to another per second. Now there are other ways in which you can calculate electrical energy. So in addition to using this formula, electrical energy is equal to the charge that's being transferred times the voltage. Electric charge is equal to the current multiplied by the time, which means current is basically the rate at which charge is flowing per unit time. So what this means is that if you have a current, an electric current of 1 amp, what's really happening is that you have 1 coulomb of charge flowing through a circuit or through a wire per second.

An electric current of 10 amps means that you have 10 coulombs of charge flowing through a wire every second. So electric current is the rate at which electric charge is flowing. And electric charge is basically the quantity of charges that you have in any given material. Now, if we replace Q with IT, we get this formula. Electrical energy is equal to voltage multiplied by the current multiplied by the time.

Which you can also get that if you replace power with voltage times current. So those are some common formulas that you want to be familiar with if you're dealing with simple circuits, and you want to use Ohm's Law, and you want to calculate power or electrical energy. Now let's talk about the formulas that you need to know when dealing with a series circuit. So let's say we have a battery, and we have three resistors connected in series. So what we have here is the voltage of the battery.

We'll call this resistor 1. This is going to be resistor 2 and resistor 3. Now here are some things you need to know. When dealing with a series circuit, The current flowing from the battery, we'll call it I-T, that current is the same current that's flowing through resistor 1. The current that flows through resistor 1 is called I-1. flows through resistor 2, well that current is called I2. The current flowing through resistor 3 is called I3.

Now in a series circuit, because it's only one path for the current to flow, the current is the same. I1, I2, I3, they're all equal to each other and are equal to the current flowing from the battery. The total resistance of three resistors in series, it's equal to the sum of those resistors.

Now, as regards to voltage, If you want to calculate the voltage across resistor 1, it's equal to the current flowing through that resistor times the resistance. So back to Ohm's law, V equals IR. If you want to calculate the voltage drop across resistor 2, it's equal to I2 times R2.

And V3 is going to be I3 times R3. Now the voltage of the battery is going to be equal to the sum of all the three voltage drops, or the three voltages across the three resistors. So VB is going to be equal to V1 plus V2 plus V3. This has to do with Kirchhoff's voltage law, which states that the sum of all the voltages in a loop will add a zero.

So if you were to move V1, V2, and V3 to the other side of the equation, you'll get this. Positive VB. The reason why it's positive is because the battery adds energy to the circuit.

And then minus V1. V2 V3 the reason why they're negative is because they subtract energy from the circuit they absorb energy from it so they decrease the voltage whereas the battery increases the voltage of the circuit so we have a positive for VB but a negative for V1 V2 and V3 at least that's the way I like to think of it but this has to do with Kirchhoff's voltage law But it is a lot easier to see it this way though. So VB is going to be the sum of V1, V2, V3 when dealing with a series circuit. Now let's move on to a parallel circuit. So we're going to have three resistors connected in parallel to each other, which means they're connected across from each other.

VB is going to be the voltage of the battery and the current flowing from it. We'll call it IT the total current So in a series circuit we had just one path for the current to flow in a parallel circuit there are multiple paths for the current to flow. So this is R1, R2, R3, and the current flowing through R1, we'll call that I1.

This is going to be I2 and I3. Now, in a series circuit, VB was the sum of V1, V2, V3. In a parallel circuit, VB is going to be equal to V1, V2, and V3, because each resistor is connected across the same battery so therefore the same voltage will be applied to each of those resistors. Now in both series and parallel circuits V1 is still equal to I1 times R1 so the voltage resistor 1 is equal to the current that's flowing through it times the resistance of resistor 1. V2 is going to be equal to I2 times R2 just as before and the same is true for V3. Now, in a series circuit, the total resistance was the sum of the three resistors.

In a parallel circuit, the reciprocal is true. So, 1 divided by the total resistance, or the equivalent resistance, is equal to 1 over R1 plus 1 over R2 plus 1 over R3. Now, in a series circuit, because there was only one path for the current to flow, The total current was equal to I1, which was equal to I2, which was equal to I3.

For a parallel circuit, it's going to be different. The total current is going to be the sum of I1, I2, and I3. And let's talk about that.

Because we talked about Kirchhoff's voltage law, it makes sense for us to talk about Kirchhoff's current law. So here we have the total current. And here it's going to break off into I1. And let's call this IB. Actually, let's call it IA.

So if we focus on this point here, this junction, based on Kirchhoff's current law, the total current that is entering that junction must equal the total current that is leaving that junction. So I-T, that's the current that's entering that point, and I-1 and I-A, that's the current that is leaving that point. Now if we focus on this point here. Ia is flowing into that point I 2 is going towards resistor 2 and I 3 goes towards resistor 3 so we have Ia entering that junction I 2 and I 3 they're leaving that point.

That's the current that's leaving that point So if we replace Ia with I2 and I3, we get that the total current is equal to I1 plus I2 plus I3. So this formula arises from Kirchhoff's current law. So let me give you another example.

So let's say we have 20 amps of current flowing to this point. We'll call it point A. And let's say we have 18 amps of current going here.

We have 13 amps of current going this way. And we have 25 amps of current going that way. What is the current flowing in this branch? And is it flowing towards point A or away from point A?

What would you say? Well first, let's give the currents variables. Let's call this I1, I2, I3, and I4. And this will be I5. Now the sum of the currents entering and leaving that junction should be equal to the sum of the currents entering and leaving that junction.

equal to zero, which is the same as saying the amount of currents entering that junction and the amount of currents leaving will be equal to each other. So we're going to use Kirchhoff's current law to calculate I5. So the current that is flowing into the junction, let's assign a positive value to those currents. The currents that are leaving that point, let's assign a negative value to it.

So I1 and I2 is going to be positive because the current is flowing towards point A. And I3 and I4 will be negative because the current is leaving from point A. So we're going to have I1 plus I2 minus I3 minus I4. I5, we don't know if it's entering or leaving, but we're going to put a plus towards I5. So when we solve I5, if we get a positive answer, that means the current is flowing to point A.

If we get a negative answer, that means the current is flowing away from point A. So I1 is 20, I2 is 18, I3 is 13, I4 is 25, and let's calculate I5. So 20 plus 18 is 38, and 13 and 25, that's also 38. So it looks like I5 is 0. But if we were to change this, let's say if we made it 29, so we can get a value for I5. What would I5 be if I4 is 29?

Negative 13 minus 29, that's going to be negative 42. So this would be negative 4, which means I5 will be positive 4. So if I4 is 29 amps, I5 is going to be 4 amps. Because it's positive, current is flowing towards 0.8. Notice that if we were to add up these values, What this tells us is that we have 42 amps of current flowing to point A, and if we add up these values, we would have 42 amps of current flowing away from point A, which is how it should be. The amount of electrical current that is flowing to a point should equal the amount of current that is flowing away from that point. So that's the basic idea behind Kirchhoff's Current Law.

So that's basically it for this video. I just want to give you a list of common electrical formulas that you'll use when solving DC circuits or when dealing with electricity in general. Thanks for watching.

By the way, for those of you who want more example problems, feel free to check out the links in the description section below. I'm going to be posting more content as well as videos that are related to this video that you're currently watching, so feel free to take a look at that when you get a chance.

Transcript for:Essential DC Circuit Formulas and Principles

Transcript for:
Essential DC Circuit Formulas and Principles