Understanding Kirchhoff's Voltage Law Basics

In this video, we're going to go over Kirchhoff's Voltage Law, abbreviated KVL. Now, his law states that if you have a closed circuit, the voltages around that circuit must add up to zero. Now, some voltages are positive, and others have a negative value. So you need to determine which ones are positive and which ones will be negative. Now let's say if we have a resistor and let's say this is at 20 volts and this is at 10 volts. You know that current naturally flows from the direction of high potential to low potential in a resistor. So it flows from positive to negative. Now, a resistor consumes energy, and any time the current flows from a high electric potential to a low electric potential, the energy of the charges that are flowing in that circuit is actually being reduced. Energy is being transferred. from the charges to the resistor. So therefore because the resistor decreases the energy of the charges, we're going to assign a negative voltage to it because it reduced the energy per unit charge. And that's what a volt is. One volt is 1 joule per coulomb. So voltage is basically the energy change per unit charge, which is basically per 1 coulomb. Now a battery does something different. In a battery, current can flow from low potential to high potential. So this could be 0 volts and this could be 12. And when you see this, whenever the current is flowing from a low potential to a high potential, that means that the energy of the charges is increasing. And so this is going to create a positive voltage because the battery is transferring energy to the charges. And so it increases the voltage of the charges, but a resistor will decrease the voltage of the charges. Now let's see if you remember. So let's say if we have a resistor, and a current is flowing in that direction. Will we assign a positive voltage or a negative voltage to the resistor? So keep this in mind, a resistor always creates a voltage drop. It always consumes energy from the charges. So always assign a negative V for resistor. Now what if we have a battery that's listed this way, and there's a current flowing in this direction. This battery, is it increasing the energy of the charges, or is it decreasing the energy of the charges? And compare it to this situation. So what if we reverse the polarity? And let's say the current is still flowing in the same direction. So which one should we assign a positive voltage, and which one should we assign a negative voltage? Now keep this in mind, anytime the current flows from, let me put that in blue, a high potential to a low potential, the energy of the charges is being reduced. So it's going to be a negative voltage. Now anytime it flows in the opposite direction, that is from a low potential to a high potential, then... the voltage is increasing, the energy of the charges is increasing. So looking at this diagram, the current is flowing from positive to negative. So it's going from a high potential to a low potential. So this battery reduces the energy of the charges. Now for this one, it's going from a low potential to a high potential. So it's increasing the energy of the charges. So we should assign a positive V to this value and a negative V to this battery. Now let's work on some practice problems. So let's say if we have a battery and it's connected to four resistors in the circuit. This is the positive terminal and this is the negative terminal. And let's say it's a 12-volt battery. And this is going to be 8 ohms, 10 ohms. and 12 ohms. So this is R1, R2, R3. So write an expression that highlights Kirchhoff's voltage law. Show that the sum of the voltages around the circuit must add to zero. Now the current is going to leave the positive terminal of the battery. So it's going to be flowing in this direction. So as it flows through R1, the voltage will decrease. But the battery, notice that the current flows from the negative terminal to the positive terminal of the battery. So the battery supplies energy to the circuit. So I'm going to put positive VB for the voltage of the battery. and then negative V1, that's the voltage drop across R1, and R2 also creates a voltage drop, and R3 will also create a voltage drop. And based on Kirchhoff's voltage law, the sum of all the voltages in a circuit must add up to zero. Now let's use that equation to calculate the current in a circuit. So the voltage of the battery is 12 volts. Now we know that V is equal to IR. And so the current is the same in the circuit because there's only one path for the current to flow. And so what we have is a series circuit. So V1 is going to be I times R1. R1 is 8 ohms, so we can say that V1 is 8 times I. V2 is I times R2, or I times 10, so that's going to be 10 times I. V3 is going to be 12 times the current that flows through it, which is I. So now we can add these like terms. Negative 8 minus 10, that's negative 18, minus 12, that's negative 30. So this is going to be negative 30i. Now let's move this to that side, so we have 12v is equal to 30i, and then let's divide both sides by 30. So the current that flows in this circuit is 12 volts divided by 30 ohms, and that's equal to 0.4 amps. Now let's say the electric potential at this point is 0 volts. Let's calculate the electric potential at these points. So what is the electric potential at point A, B, and C? So notice that the potential difference of the battery is 12 volts. And so since this is a positive sign, A has to be higher than this point. So at point A, the voltage is, or the potential is 12 volts. Now what is the potential at point B? So we need to calculate the voltage drop across R1. So we can use V is equal to R1. I want so we have the current which is point four amps and let's multiply by R one which is eight so point four times eight is three point two so the voltage is going to be it's going to drop by three point two so if we take 12 volts and subtract it by three point two this will give us a potential of eight point eight volts at this point Now if you want to show the work, here's what you can do. So we're going to start with this equation, Ohm's Law. The potential difference is equal to the current flowing through the resistor times the resistor itself. Now the change in the potential is VB minus VA. and with a resistor we know that there's a voltage drop so we need to put a negative sign so this is going to be negative i times r1 so to calculate the potential at point b it's going to be the potential at point a minus i times times R1. So the potential at A is 12, and then subtract that by a current of 0.4 times a resistance of 8, and that will give you 8.8. Now let's do the same thing for point C. So we can start with this equation. Now this is going to be VC minus VB and that's equal to negative IR2. So the potential at C is going to be the potential at B minus IR2. So at B it's 8.8 and the current is 0.4 and R2 is 10. So if you type in 8.8 minus 0.4 times 10, this will give you 4.8 volts. And so that's the potential at C. Now if you take the current, which is 0.4 amps, and multiply by R3, which is 12 ohms, you should get the difference between these two values. So 0.4 times 12. is 4.8 and that's the potential difference across R3. And so that's how you can solve this particular circuit using KVL, Kirchhoff's voltage law. Now let's work on another example. Let's say if we have this circuit. This is going to be a 50 ohm resistor, which we'll call R1. And R2 is a 30 ohm resistor. So here we have a 12 volt battery. And here we have an 8 volt battery. Go ahead and use Kirchhoff's Voltage Law to calculate the current in the circuit. Now the first thing we need to determine is the direction of the current in the circuit, because we have two batteries. Now the 12 volt battery wants to send current in this direction, so that is the clockwise direction. And the 8 volt battery also wants to send current in the clockwise direction. So therefore, these two batteries, they don't oppose each other, they support each other, because they want to send current in the same direction. the same direction. So now that we know the direction of the current, let's set up the equation. So let's call this VB1 and VB2. So the current flows from low potential to high potential for VB1. So therefore, VB1 increases the energy of the charges. So we're going to assign it a positive voltage. The resistor consumes energy from the circuit. So we're going to assign that a negative voltage. Now current flows from the negative terminal to the positive terminal. So therefore, the battery increases the energy of the circuit. So that's going to be plus VB2. And then this resistor will consume energy from the circuit. So that's going to be negative V2. So VB1 is 12. We can replace V1 with I times R1, so that's I times 50, or simply 50I. VB2 is positive 8, and V2, that's going to be I times 30, or 30I. Now let's combine like terms. 12 plus 8 is 20. And then negative 50i minus 30i, that's negative 80i. So moving this term to that side, we have 20 is equal to 80i. So if we divide both sides by 80, 20 divided by 80 will give us a current of 0.25 amps. So now that we have the current flowing in a circuit, let's calculate the potential at every point. So let's say this is point A, B, C, and D. And let's say that the potential at point A is 0 volts. Calculate the electric potential at every other point. Now we can see that the voltage across the battery is 12. And the plus side indicates that the top side is higher than the bottom side. So the potential at point B has to be 12 volts. Now the resistor, R1, is going to drop the voltage. To calculate the voltage drop, it's simply I times R1. So the current is 0.25 amps, multiplied by a resistance of 50 ohms. And so the voltage drop there is 12.5 volts. So 12 minus 12.5, this is going to be negative 0.5 volts at point C. Now you could set it up the way we did in the last problem. You could say that the voltage drop is I times R1, both a negative sign, because it decreases the voltage. And VC minus VB is equal to negative I times R1. So you could say that VC is VB minus I times R1. So VB is 12, I is 0.25, and the resistance is 50 ohms. So 12 minus 0.25 times 50, this will give you negative 0.5 volts. So that's the potential at point C. Now what about the potential at point D? So we have a battery that's 8 volts, and this is the positive side, which means that this side is going to be 8 volts higher than that side. So if we take negative 0.5, and then add 8 volts to it, this will give us 7.5. So that's the potential at point D. Now let's confirm that A is indeed 0. So let's use this equation. The voltage drop across R2 is going to be I times R2. And so VA minus VD, because we're going from D to A, that's equal to this value. So VA is going to be VD minus... IR2. So Vd is 7.5, the current is 0.25, and R2 is 30. So if you type in 7.5 minus 0.25 times 30, this will give you zero, the potential at 0.8. And so it goes to show that Kirchhoff's voltage law does work. The sum of all the voltages in the closed circuit will always add up to zero. let's try another example but one with a lot more stuff in it so this is going to be a 50 volt battery here are the positive terminals and here are the negative terminals This is going to be 30 ohms, we'll call it R1. And then this is going to be 70 ohms, which we'll call R2. And this is going to be a 10 volt battery. And also, this is going to be a 20 volt battery. First we need to determine the direction of the current. So let's call this battery 1, battery 2, and battery 3. So battery 1 wants to create a current that flows in this direction, away from the positive terminal. Battery 3 wants to create a current in this direction. and that is a counterclockwise current. This is a clockwise current. So those two batteries oppose each other. And this battery wants to generate a counterclockwise current. So these two batteries are in support of each other and they're against this one. However, 50 is greater than the sum total of 10 and 20. So therefore, the current is going to be in this direction. Now let's go ahead and calculate the current. So this battery will create a positive contribution to the circuit. It's going to increase the energy of the circuit, so we're going to say it's positive 50. Now the current is flowing in a different direction for this battery. It's going from high potential to low potential. So this battery will create a voltage drop. So therefore, we're going to assign it a negative value. And here, the current flows from high potential to low potential. But here, it flows from low to high, so that's a positive contribution. But here, since it flows from high to low, it's going to be a negative contribution. So this battery will decrease the energy of the circuit. Now, let's write an equation. So the first battery will add 50 volts or 50 joules per coulomb of charge to the circuit. R1 is going to consume energy from the circuit, so the voltage drop will be the current that flows through it times R1. And R1 is 30, so this is going to be 30 times I. Now this battery reduces the energy of the circuit, so that's going to be negative 20 volts. And this one also reduces the energy of the circuit, so that's going to be negative 10 volts. And this one also consumes energy from the circuit, so that's going to be negative 70 times. I this is supposed to be r2 and so all of this is equal to 0 so now let's combine like terms so we have 50 minus 20 minus 10 so that's equal to positive 20 and then we have negative 30 minus 70 so that's going to be negative 100 I and now let's move this to that side So 20 V is equal to 100 times I so the current is going to be 20 divided by 100 And so in this example, we have a current of 0.2 amps. Now let's call this point A. B, C, D, and E. Let's calculate the potential at each point. And let's say that A is at 0 volts. So B has to be 50 volts higher. Because that battery provides energy to the circuit. So this is going to be at 50 volts. Now let's calculate the voltage drop across the 30 ohm resistor. So it's going to be a current of 0.2 amps times 30 ohms. So the voltage drop across this resistor is negative 6. So if the potential is 50 at B, then at point C, it's going to be 6 volts less. It's going to be 44 volts. Now this battery will reduce the energy of the circuit by 20 joules per coulomb, or by 20 volts. So the potential at D is going to be 44 minus 20, which is 24 volts. Now this battery also reduces the energy of the circuit, so it's going to decrease the energy by 10 volts. So now the voltage is going to be 14 volts. To calculate the voltage drop across R2 is going to be the current multiplied by R2 so negative 0.2 times 70 That's negative 14. So that's the voltage drop across this resistor, and 14 minus 14 will bring us back to 0. So now you know how to calculate the electric potential at every point in a circuit. The key to understanding how to solve these types of problems is to know which device will increase the energy of the circuit and which device will decrease the energy of the circuit. So resistors always consume energy. So they will always decrease. the energy of the circuit. So therefore, they will always have a voltage drop. So you should always apply a negative V value to it. Now a battery can either increase or decrease the energy of the circuit. And for this, you need to look at the direction of the current. So, if the current is flowing from a low potential to a high potential, then the battery is increasing the energy of the circuit. So, in that case, we need to apply a positive voltage. to this battery because it's adding voltage to the circuit. Now if the current is flowing from a high voltage or high potential to a low potential then this battery is reducing the energy of the circuit and so we need to apply a negative voltage to it. So hopefully you understand that concept and it's going to help you to solve more complicated problems than the ones that we covered in this video. This is simply a basic introduction.

Now let's say if we have a resistor and let's say this is at 20 volts and this is at 10 volts. You know that current naturally flows from the direction of high potential to low potential in a resistor. So it flows from positive to negative.

Now, a resistor consumes energy, and any time the current flows from a high electric potential to a low electric potential, the energy of the charges that are flowing in that circuit is actually being reduced. Energy is being transferred. from the charges to the resistor. So therefore because the resistor decreases the energy of the charges, we're going to assign a negative voltage to it because it reduced the energy per unit charge.

And that's what a volt is. One volt is 1 joule per coulomb. So voltage is basically the energy change per unit charge, which is basically per 1 coulomb. Now a battery does something different.

In a battery, current can flow from low potential to high potential. So this could be 0 volts and this could be 12. And when you see this, whenever the current is flowing from a low potential to a high potential, that means that the energy of the charges is increasing. And so this is going to create a positive voltage because the battery is transferring energy to the charges.

And so it increases the voltage of the charges, but a resistor will decrease the voltage of the charges. Now let's see if you remember. So let's say if we have a resistor, and a current is flowing in that direction. Will we assign a positive voltage or a negative voltage to the resistor?

So keep this in mind, a resistor always creates a voltage drop. It always consumes energy from the charges. So always assign a negative V for resistor.

Now what if we have a battery that's listed this way, and there's a current flowing in this direction. This battery, is it increasing the energy of the charges, or is it decreasing the energy of the charges? And compare it to this situation. So what if we reverse the polarity? And let's say the current is still flowing in the same direction.

So which one should we assign a positive voltage, and which one should we assign a negative voltage? Now keep this in mind, anytime the current flows from, let me put that in blue, a high potential to a low potential, the energy of the charges is being reduced. So it's going to be a negative voltage. Now anytime it flows in the opposite direction, that is from a low potential to a high potential, then...

the voltage is increasing, the energy of the charges is increasing. So looking at this diagram, the current is flowing from positive to negative. So it's going from a high potential to a low potential. So this battery reduces the energy of the charges. Now for this one, it's going from a low potential to a high potential.

So it's increasing the energy of the charges. So we should assign a positive V to this value and a negative V to this battery. Now let's work on some practice problems.

So let's say if we have a battery and it's connected to four resistors in the circuit. This is the positive terminal and this is the negative terminal. And let's say it's a 12-volt battery.

And this is going to be 8 ohms, 10 ohms. and 12 ohms. So this is R1, R2, R3.

So write an expression that highlights Kirchhoff's voltage law. Show that the sum of the voltages around the circuit must add to zero. Now the current is going to leave the positive terminal of the battery. So it's going to be flowing in this direction. So as it flows through R1, the voltage will decrease.

But the battery, notice that the current flows from the negative terminal to the positive terminal of the battery. So the battery supplies energy to the circuit. So I'm going to put positive VB for the voltage of the battery. and then negative V1, that's the voltage drop across R1, and R2 also creates a voltage drop, and R3 will also create a voltage drop. And based on Kirchhoff's voltage law, the sum of all the voltages in a circuit must add up to zero.

Now let's use that equation to calculate the current in a circuit. So the voltage of the battery is 12 volts. Now we know that V is equal to IR.

And so the current is the same in the circuit because there's only one path for the current to flow. And so what we have is a series circuit. So V1 is going to be I times R1.

R1 is 8 ohms, so we can say that V1 is 8 times I. V2 is I times R2, or I times 10, so that's going to be 10 times I. V3 is going to be 12 times the current that flows through it, which is I.

So now we can add these like terms. Negative 8 minus 10, that's negative 18, minus 12, that's negative 30. So this is going to be negative 30i. Now let's move this to that side, so we have 12v is equal to 30i, and then let's divide both sides by 30. So the current that flows in this circuit is 12 volts divided by 30 ohms, and that's equal to 0.4 amps. Now let's say the electric potential at this point is 0 volts. Let's calculate the electric potential at these points.

So what is the electric potential at point A, B, and C? So notice that the potential difference of the battery is 12 volts. And so since this is a positive sign, A has to be higher than this point.

So at point A, the voltage is, or the potential is 12 volts. Now what is the potential at point B? So we need to calculate the voltage drop across R1. So we can use V is equal to R1. I want so we have the current which is point four amps and let's multiply by R one which is eight so point four times eight is three point two so the voltage is going to be it's going to drop by three point two so if we take 12 volts and subtract it by three point two this will give us a potential of eight point eight volts at this point Now if you want to show the work, here's what you can do.

So we're going to start with this equation, Ohm's Law. The potential difference is equal to the current flowing through the resistor times the resistor itself. Now the change in the potential is VB minus VA. and with a resistor we know that there's a voltage drop so we need to put a negative sign so this is going to be negative i times r1 so to calculate the potential at point b it's going to be the potential at point a minus i times times R1.

So the potential at A is 12, and then subtract that by a current of 0.4 times a resistance of 8, and that will give you 8.8. Now let's do the same thing for point C. So we can start with this equation. Now this is going to be VC minus VB and that's equal to negative IR2. So the potential at C is going to be the potential at B minus IR2.

So at B it's 8.8 and the current is 0.4 and R2 is 10. So if you type in 8.8 minus 0.4 times 10, this will give you 4.8 volts. And so that's the potential at C. Now if you take the current, which is 0.4 amps, and multiply by R3, which is 12 ohms, you should get the difference between these two values.

So 0.4 times 12. is 4.8 and that's the potential difference across R3. And so that's how you can solve this particular circuit using KVL, Kirchhoff's voltage law. Now let's work on another example.

Let's say if we have this circuit. This is going to be a 50 ohm resistor, which we'll call R1. And R2 is a 30 ohm resistor.

So here we have a 12 volt battery. And here we have an 8 volt battery. Go ahead and use Kirchhoff's Voltage Law to calculate the current in the circuit.

Now the first thing we need to determine is the direction of the current in the circuit, because we have two batteries. Now the 12 volt battery wants to send current in this direction, so that is the clockwise direction. And the 8 volt battery also wants to send current in the clockwise direction.

So therefore, these two batteries, they don't oppose each other, they support each other, because they want to send current in the same direction. the same direction. So now that we know the direction of the current, let's set up the equation.

So let's call this VB1 and VB2. So the current flows from low potential to high potential for VB1. So therefore, VB1 increases the energy of the charges. So we're going to assign it a positive voltage.

The resistor consumes energy from the circuit. So we're going to assign that a negative voltage. Now current flows from the negative terminal to the positive terminal. So therefore, the battery increases the energy of the circuit.

So that's going to be plus VB2. And then this resistor will consume energy from the circuit. So that's going to be negative V2. So VB1 is 12. We can replace V1 with I times R1, so that's I times 50, or simply 50I.

VB2 is positive 8, and V2, that's going to be I times 30, or 30I. Now let's combine like terms. 12 plus 8 is 20. And then negative 50i minus 30i, that's negative 80i. So moving this term to that side, we have 20 is equal to 80i.

So if we divide both sides by 80, 20 divided by 80 will give us a current of 0.25 amps. So now that we have the current flowing in a circuit, let's calculate the potential at every point. So let's say this is point A, B, C, and D. And let's say that the potential at point A is 0 volts.

Calculate the electric potential at every other point. Now we can see that the voltage across the battery is 12. And the plus side indicates that the top side is higher than the bottom side. So the potential at point B has to be 12 volts. Now the resistor, R1, is going to drop the voltage.

To calculate the voltage drop, it's simply I times R1. So the current is 0.25 amps, multiplied by a resistance of 50 ohms. And so the voltage drop there is 12.5 volts. So 12 minus 12.5, this is going to be negative 0.5 volts at point C.

Now you could set it up the way we did in the last problem. You could say that the voltage drop is I times R1, both a negative sign, because it decreases the voltage. And VC minus VB is equal to negative I times R1.

So you could say that VC is VB minus I times R1. So VB is 12, I is 0.25, and the resistance is 50 ohms. So 12 minus 0.25 times 50, this will give you negative 0.5 volts. So that's the potential at point C.

Now what about the potential at point D? So we have a battery that's 8 volts, and this is the positive side, which means that this side is going to be 8 volts higher than that side. So if we take negative 0.5, and then add 8 volts to it, this will give us 7.5. So that's the potential at point D. Now let's confirm that A is indeed 0. So let's use this equation.

The voltage drop across R2 is going to be I times R2. And so VA minus VD, because we're going from D to A, that's equal to this value. So VA is going to be VD minus... IR2. So Vd is 7.5, the current is 0.25, and R2 is 30. So if you type in 7.5 minus 0.25 times 30, this will give you zero, the potential at 0.8.

And so it goes to show that Kirchhoff's voltage law does work. The sum of all the voltages in the closed circuit will always add up to zero. let's try another example but one with a lot more stuff in it so this is going to be a 50 volt battery here are the positive terminals and here are the negative terminals This is going to be 30 ohms, we'll call it R1. And then this is going to be 70 ohms, which we'll call R2.

And this is going to be a 10 volt battery. And also, this is going to be a 20 volt battery. First we need to determine the direction of the current.

So let's call this battery 1, battery 2, and battery 3. So battery 1 wants to create a current that flows in this direction, away from the positive terminal. Battery 3 wants to create a current in this direction. and that is a counterclockwise current.

This is a clockwise current. So those two batteries oppose each other. And this battery wants to generate a counterclockwise current. So these two batteries are in support of each other and they're against this one. However, 50 is greater than the sum total of 10 and 20. So therefore, the current is going to be in this direction.

Now let's go ahead and calculate the current. So this battery will create a positive contribution to the circuit. It's going to increase the energy of the circuit, so we're going to say it's positive 50. Now the current is flowing in a different direction for this battery. It's going from high potential to low potential.

So this battery will create a voltage drop. So therefore, we're going to assign it a negative value. And here, the current flows from high potential to low potential. But here, it flows from low to high, so that's a positive contribution. But here, since it flows from high to low, it's going to be a negative contribution.

So this battery will decrease the energy of the circuit. Now, let's write an equation. So the first battery will add 50 volts or 50 joules per coulomb of charge to the circuit. R1 is going to consume energy from the circuit, so the voltage drop will be the current that flows through it times R1.

And R1 is 30, so this is going to be 30 times I. Now this battery reduces the energy of the circuit, so that's going to be negative 20 volts. And this one also reduces the energy of the circuit, so that's going to be negative 10 volts.

And this one also consumes energy from the circuit, so that's going to be negative 70 times. I this is supposed to be r2 and so all of this is equal to 0 so now let's combine like terms so we have 50 minus 20 minus 10 so that's equal to positive 20 and then we have negative 30 minus 70 so that's going to be negative 100 I and now let's move this to that side So 20 V is equal to 100 times I so the current is going to be 20 divided by 100 And so in this example, we have a current of 0.2 amps. Now let's call this point A. B, C, D, and E.

Let's calculate the potential at each point. And let's say that A is at 0 volts. So B has to be 50 volts higher.

Because that battery provides energy to the circuit. So this is going to be at 50 volts. Now let's calculate the voltage drop across the 30 ohm resistor.

So it's going to be a current of 0.2 amps times 30 ohms. So the voltage drop across this resistor is negative 6. So if the potential is 50 at B, then at point C, it's going to be 6 volts less. It's going to be 44 volts.

Now this battery will reduce the energy of the circuit by 20 joules per coulomb, or by 20 volts. So the potential at D is going to be 44 minus 20, which is 24 volts. Now this battery also reduces the energy of the circuit, so it's going to decrease the energy by 10 volts.

So now the voltage is going to be 14 volts. To calculate the voltage drop across R2 is going to be the current multiplied by R2 so negative 0.2 times 70 That's negative 14. So that's the voltage drop across this resistor, and 14 minus 14 will bring us back to 0. So now you know how to calculate the electric potential at every point in a circuit. The key to understanding how to solve these types of problems is to know which device will increase the energy of the circuit and which device will decrease the energy of the circuit.

So resistors always consume energy. So they will always decrease. the energy of the circuit. So therefore, they will always have a voltage drop.

So you should always apply a negative V value to it. Now a battery can either increase or decrease the energy of the circuit. And for this, you need to look at the direction of the current.

So, if the current is flowing from a low potential to a high potential, then the battery is increasing the energy of the circuit. So, in that case, we need to apply a positive voltage. to this battery because it's adding voltage to the circuit.

Now if the current is flowing from a high voltage or high potential to a low potential then this battery is reducing the energy of the circuit and so we need to apply a negative voltage to it. So hopefully you understand that concept and it's going to help you to solve more complicated problems than the ones that we covered in this video. This is simply a basic introduction.

Transcript for:Understanding Kirchhoff's Voltage Law Basics

Transcript for:
Understanding Kirchhoff's Voltage Law Basics