In this video, we're going to talk about how to use Kirchhoff's junction rule and loop rule to calculate the current in a complex circuit. So let's start with this problem. We have a resistor in series with two other resistors that are parallel to each other. And let's say this resistor has a value of 3 ohms, and let's call it R1. This is going to be R2, and it has a value of 4 ohms.
And R3 has a value of... 12 ohms. So if you want to try this problem, feel free. Go ahead and pause the video if you want to work it out.
So calculate the current flowing through each resistor using Kirchhoff's rules. So let's say that the current that's flowing through the first resistor, we're going to call it I1, and the current flowing through the second resistor, let's call it I2, and the current flowing through the third resistor, we're going to call it I3. Now using Kirchhoff's Junction Rule, or his current law, Let's focus on this junction here.
So notice that we have a current I1 which flows into the junction. And I2 leaves the junction. I3 also leaves it to go this way.
Now according to this junction rule, the current that enters the junction is equal to the total current that leaves the junction. So the only current that's entering the junction is I1. The other two. currents are leaving the junction so I1 has to equal the sum of I2 plus I3. So that's one equation that we have so far and we're going to use it later.
Now the next thing that we need to talk about is Kirchhoff's voltage law which is important when using the loop rule. So that law states that the sum of all the voltages around the loop or around a closed circle must add to zero. Now when going going around the loop you need to determine which elements will create a positive voltage contribution to the circuit and which one will create a voltage drop or negative contribution to the voltage of the circuit so anytime you travel from a high potential to a low potential there's going to be a voltage drop. And it makes sense because you're going from high to low. Now, if you're going from a low potential to, let's say, a high potential, if you're traveling in that direction, then the voltage is increasing.
So, there should be a voltage lift. Now, here's a question for you. Let's say the current is flowing in this direction. And let's say as you're traveling around the circuit, you're going through the loop and it's in the same direction as the current. Will this be associated with a positive voltage or a negative voltage?
Now keep in mind, current flows from high potential to low potential. So we're going in this direction. We're going from a positive to negative. So therefore, this will be associated with a voltage drop.
So when using the loop, loop rule if you're going in a direction of the current that's associated with a voltage drop. Now if you're going opposite to the direction of the current, then you're traveling from a low potential to a high potential, which is what we have in this situation. so that's going to be associated with an increase in voltage or voltage lift because you're going towards a higher potential.
If you're going towards a lower potential, that's associated with a voltage drop. And that makes sense. So if you're traveling towards a negative sign, you should be associated with a voltage drop.
If you're traveling towards, let's say, a positive sign, we're going in that direction, that should be associated with a positive voltage or a voltage lift. So now let's say if you have a resistor And this part is positive and this part is negative. So if you're traveling in this direction using the loop rule, should you apply a positive voltage contribution or negative voltage in the voltage law equation? Now we know that current flows from high potential to low potential.
So notice that we're going against the current and we're traveling towards the positive side. So this is going to be associated with a voltage lift. Now let's say if we have a resistor and the current is flowing in this direction. And using the loop rule, we're traveling in this direction. Will this be associated with a voltage drop or a voltage lift?
Now we know that current flows from positive to negative. So we're traveling towards a low potential. So this will be associated with a voltage drop.
Now let's focus on a battery. So let's say this is positive and this is negative. And I'm going to draw another battery in this direction. So let's say if we're using the loop rule and we're going in this direction.
Should we apply a positive voltage or a negative voltage in the voltage law equation? Is this a voltage lift or a voltage drop? Now we're going towards the positive sign, so this is going to be associated with a voltage lift.
Now what if we're going in this direction, using the loop rule? So notice that we're going towards the negative sign, and so that should be associated with a voltage drop. So hopefully that makes sense, and understanding this will be very useful when solving these types of problems.
Because if you get, let's say, a negative sign wrong, the whole question... is gone. It's going to be very hard to find the right answer. All it takes is one single mistake to make a long problem and one big headache.
So now that we finished with all the sign conventions, let's go back to this problem. So let's focus on the first loop. We'll call this loop one. Now we're going in this direction. So we're going towards the positive side of that battery.
So we're going from from low potential to high potential. So that's going to be associated with a voltage lift. So that's going to be positive 24 volts. Now keep in mind, when using Kirchhoff's Voltage Law, the sum of all the voltages must add up to zero.
Now some of these voltages will be positive, some will be negative. Now when you're moving across a resistor, the voltage drop will be the current times the resistor. So just keep that in mind as well. Now current flows from high potential to low potential. So then this is going to be the positive side of the resistor, and that's the negative side.
And this is going to be the positive side of R2, and that's the negative side of R2, because current is flowing in that direction. And the same is true for R3. So now using loop rule, we're going in this direction through R1.
So we're going in a direction of the current towards a lower potential. So because we're going towards a lower potential, that's going to be a voltage drop. So the voltage drop is going to be a voltage drop. drop across R1 is V1 and V1 is equal to the current that flows through it times R1 and R1 is 3 so then this voltage drop is simply the current times 3 ohms which we can write as 3i1.
Now as we follow the direction of the loop we're going through R2 towards a low potential so that's going to be another voltage drop and so that's going to be minus i2 times the resistance of 4 so we could say 4i2 and so now we've completed loop 1 so we're going to set this equal to 0. Now what I'm going to do is take these two terms and move it to this side. So we have the equation 24 is equal to 3I1 plus 4I2. Now let's move on to the second loop. So we're going to go in this direction.
That's going to be loop 2. So initially, we're going up through R2 towards the positive sign. We're going against the current. And so since we're going from negative to positive, that's going to be a voltage lift and so the voltage lift is going to be I 2 times the resistance of 4 so it's 4 I 2 and then following the loop we're going in this direction through R 3 so we're going towards the negative sign we're going from positive to negative so that's going to be a voltage drop So the current is I3, the resistance is 12, so that's going to be 12I3. And so we have a negative sign for a voltage drop, but a positive sign for a voltage lift, and this is equal to 0. Now, notice the two equations that we have. We have two equations and we have three variables.
So in order to solve a system with three variables, you need three equations. And that's why this equation is important. So now let's combine those three equations. So what I'm going to do in this case is I'm going to replace I1 with I2 plus I3 because It's equal to that.
So 24 is equal to 3 times I2 plus I3 plus 4I2. And so 24 is equal to 3I2 plus 3I3 plus 4I2. And now we can combine like terms.
So far we have 24 is equal to 7i2 plus 3i3. Now notice that these two equations have the same variables, i2 and i3. So I'm going to solve by elimination, which means that I want to change this to negative 3i3, so that it can cancel with positive 3i3.
I'm going to multiply this equation by 1 fourth, which is equivalent to dividing everything by 4. Zero divided by 4 is zero, and then 4I2 divided by 4 is 1I2, and then negative 12I3 divided by 4 is negative 3I3. Now let's add these two equations. So these will cancel.
24 plus 0 is 24, and 7 plus 1 is 8. So now let's divide both sides by 8, and 24 divided by 8 is 3. So I2 is equal to 3 amps. That's the first answer. Now, let's find a second answer.
So, let's use this equation here to calculate I3 using I2. So, we have 0 is equal to I2 minus 3I3. So, I'm going to move this to that side.
So, 3I3 is equal to I2. So, 3I3 is equal to... I2 is equal to 3 amps, so let's replace it with 3. And then let's divide both sides by 3. So 3 divided by 3 is 1. Therefore, I3 is equal to 1 amp.
And so that's the answer for that. And now we can calculate I1. I1 is simply the sum of I2 and I3. So I1 is equal to 3 plus 1, so that's 4 amps. So now we have all the answers that we need.
So let's analyze the circuit to make sure that everything makes sense. So let's redraw the original circuit. So that's the positive terminal.
This is the negative terminal. And this is 24 volts. This was 3 ohms.
And we have a current I1 flowing through it. So that's a current of 4 amps. And then this resistor is 4 ohms, and I2 was flowing through that, so I2 is 3 amps. And then this resistor is 12 ohms, and I3 was flowing through that, which is a current of 1 amp. And it makes sense.
If 4 amps is flowing into this junction, 4 amps should leave it. We have 3 going here and 1 going there. So the numbers add up.
Now let's make sure that the potentials add up as well. So let's assume that this is 0 volts. So that's the potential anywhere along this portion of the wire. Now, here we have a difference of 24 volts across that battery.
So the potential at, let's say, this point is 24 volts. It's 24 volts higher than this point. Now, what is the potential at, let's say, point A? Now let's calculate the voltage drop across this resistor.
V equals IR, so it's the current that flows through it times the resistance. 4 times 3 gives us a voltage drop of 12. And 24 minus 12 tells us that the potential at this point is 12 volts. Now if we take 12 volts, because...
12 minus 0 gives us a voltage of 12 volts across this resistor. 12 volts divided by 4 ohms will give us the current of 3 amps. And 12 volts divided by 12 ohms will give us the current that flows through that resistor, which is 1 amp.
So all of the numbers make sense. So it's good to check your work to see if you have the right answer. Now let's try another problem.
If you want to, you can pause the video and work on it. So this time we're going to have a second battery in this problem. And so here's the positive terminal and here is the negative terminal.
And let's say on this side we have a 30 volt battery and on this side a 10 volt battery. So because the voltage of this battery is higher, chances are that the current will flow in this direction as opposed to that direction. Now, even if you use the wrong arrow and guess it incorrectly, the current will simply be negative instead of positive. So if you get a negative current, it just means you have to reverse the arrow, and then it's going to be positive.
So it really doesn't matter what direction you choose the current to be. You can still get the right answer. Now we're going to say that this resistor is a 2 ohm resistor, we'll call it R1. And this is going to be a 5 ohm resistor, which we'll call R2. And this will be a 3 ohm resistor, and that's R3.
Now once again, this current will be I1, and the current flowing through this resistor we're going to call I2. Now like before, we used to call this I3, but this time we're going to do something different. Now in this junction, we have I1 flowing in, I3 leaving the junction, and I2 also leaving it.
So based on Kirchhoff's current law, or junction rule, the current that enters... into the junction must equal the current that leaves the junction. So I1, which enters the junction, must add up to I2 and I3.
So if we wish to solve for I3, I3 is simply the difference between I1 and I2. Now why is this important? Instead of writing I3 as the current flowing through this branch, we could express it as I1 minus I2. And so instead of dealing with three variables, I1, I2, and I3, we can deal with two variables, I1 and I2.
Now let's say if we have two junctions. Let's say this is junction 1 and junction 2. Now we can say this is current 1, current 2, current 3, current 4, and current 5. Now you're going to need five equations to solve these five missing variables. However, we could simplify the process if we do it this way.
So I3 is going to be the difference between I1 and I2. So we could simply write it as I1 minus I2. Now we're going to have a new I3, which is different from this one.
So now the current that goes in this direction is going to be this current minus this one. So it's I1 minus I2 minus I3. And instead of having five different currents, we only have three different variables. So we have I1, I2, I3, instead of I1 all the way to I5. And so for more complicated examples, you can make the...
the process a lot easier if instead of writing this as I3, you can write it as I1 minus I2. And so from this point on, that's what I'm going to do for any future problems that I solve in this video. So if you don't get this, pause the video, rewind, and review this topic one more time until you understand it.
Now let's go ahead and finish this problem. So let's start with loop 1. So initially we're going in this direction towards the positive part of the battery. So that's going to be a voltage lift of 30. And based on the direction of the current, this is going to be positive, and this is going to be negative. So remember, the current flows from high potential to low potential. So now as we travel this way towards the resistor, we're going in a direction of I1 towards the lower potential.
So that's going to be a voltage drop of 2 times I1. So I1 times the resistance of 2. And now following the loop, we're going to go in this direction. That's in the direction of the current towards the low potential. So that's another voltage drop of I2 times the resistance of 5. So that's going to be negative 5I2.
And so that's equal to 0. Now I'm going to take these two terms and move it to the other side. So now I have this equation, 30 is equal to 2i1 plus 5i2. Now let's save this equation for later.
Now let's focus on loop 2. So here we're going against I2 towards a higher potential. So that's going to be a voltage lift of 5 times I2. So we can write that as positive 5I2. Next, as we follow the loop, we need to go through this battery towards the higher potential.
So that's another voltage lift of positive 10. And then we need to travel in this direction, that is in the direction of this current as it goes this way, towards the lower potential. So that's going to be a voltage drop. And so that's going to be negative 3 times this current, which is I1 minus I2. So because we have a single branch, the current that flows in this branch equals the current that flows through this resistor. And that's I1 minus I2, which would have been I3.
Now all of this is equal to 0, so keep that in mind. Now let's distribute the negative 3. So 0 is equal to 5i2 plus 10 minus 3i1 plus 3i2. And so let's go ahead and combine like terms. And so we have 0 is equal to 10 minus 3i1. And 5 plus 3 is 8. So that's going to be plus 8i2.
And I'm going to take these two and move it to the other side. So now I have positive 3i1 minus 8i2, that's equal to 10. So this is the second formula that we need. Now let's do some algebra.
So let's try to get rid of I1. The least common multiple of 2 and 3 is 6. So I'm going to multiply this by positive 3, and this equation by negative 2. So 3 times 30 is 90. 2i1 times 3 is 6i1 and 5i2 times 3 is 15i2 Now, negative 2 times 10 is negative 20 and then 3i1 times negative 2 2 that's negative 6 I 1 and negative 8 I 2 times negative 2 is positive 16 I 2 so now let's add up these two equations these will cancel 90 plus negative 20 is positive 70 and 15 plus 16 is 31 so I 2 is equal to 70 divided by 31 so I 2 is 2.258 Now, let's calculate I1. Let's use this formula.
So 3I1 minus 8I2 or 8 times 2.258 is equal to 10. Negative 8 times 2.258 that's negative 18.064 So let's add 18.064 to both sides So this is going to be 3i1 that's going to be equal to 28.064 and so I1 is 28.064 divided by 3 and so it's about 9.355 amps so now that we have I1 let's calculate the potential everywhere along the circuit so I'm going to redraw the circuit at this point and let's see if these two answers make sense So let's call this point A, point B, point C, which is the same as that point, and point D. Now let's say that point A is... the ground.
Let's say it's at zero volts. So given the potential at point A, calculate the potential at B, C, and D. Now we know this is 30, and this is 10, and this is a two ohm resistor, this is a five ohm resistor, and this is a three ohm resistor.
So go ahead and try it. So going from A to B, we have a voltage lift of 30 volts. So potential at B is going to be 30 volts. Now going from B to C, we know that the current flowing through the 2 ohm resistor is I1, and that's a current of 9.355 amps. So to calculate the voltage drop of this resistor, it's the current times the resistance.
So let's take 9.355 multiplied by 2. So that gives us a voltage drop of 18.71 volts. So we know this is positive and this is negative. So as we go from B to C, the voltage is going to decrease by 18.71. So 30 minus 18.71, that means that the electric potential at point C is 11.29 volts. Now as we go from, let's say, C to A, the potential drops to zero.
So we can calculate the current in the 5 ohm resistor by taking the voltage across the 5 ohm resistor, which is the potential difference between these two. points so that's 11.29 and then if we divided by the resistance of 5 ohms that will give us a current of 2.258 amps which is an agreement with this answer I2. So that means that we're on the right track. Now let's move from C to D. So as we travel in this direction, we're going towards the positive terminal, and so that's a voltage lift of 10 volts.
So it's going to be 11.29 plus 10, so the potential at D is 21.29 volts. Now, if we go from D to A, the potential difference, or the voltage, across the 3 ohm resistor is 21.29 volts. So if we take 21.29 and divide it by 3, that will give us the current flowing in this resistor, which is 7.09 and then 6 repeating, so we can round that to 7.097 volts.
I mean amps. Now the current flowing through the 3 ohm resistor is the same as I1 minus I2. So we could confirm this answer by subtracting these two currents. So if you take 9.355 and subtract it by 2, This will give you the same answer of 7.097 amps.
And so the numbers are all in agreement with each other. So we know these answers have to be correct. And so that's it for this problem. Let's move on to the next one. Now let's work on a more complicated example.
So feel free to try it if you want to. So this is going to be a 20 volt battery. This is going to be a 3-ohm resistor, and this is going to be a 2-ohm resistor. And then we have a 4-ohm resistor and a 12-volt battery.
And then this is going to be an 18-volt battery, a 10-ohm resistor, and a 6-volt battery, and a 6-ohm resistor. So go ahead and calculate. the current flowing through the circuit.
And we're going to say that the potential at this point is 0 volts. So relative to that point, calculate the potential at every other point in this circuit. Go ahead and try it.
Now the first thing I like to do is try to predict the direction of the currents in this circuit. So let's focus on this loop. Now the 20 volt battery wants to shoot out a current in that direction. And so that current wants to travel in this direction, in the clockwise direction. And the 12 volt battery wants to shoot out a current in this direction, which is going counterclockwise.
However this battery is stronger, so therefore, comparing those two circuits, I believe the net current in this part will be going in that direction. Now looking at this loop, the 12 volt battery wants to send a current going in this direction, and these two batteries, they add up to make a 24 volt battery. The combined effect of those two batteries wants to send the current going in this direction, as opposed to that direction. And this voltage wants to send the current going through this direction. So therefore, it makes sense that the current in this branch is going in that direction, towards the left.
Because the 24 volt battery wants to send current in this direction, which is greater than the 12 volt battery. And a 20 volt battery also wants to send current in that direction, which is greater than the 12 volt battery. So I believe there's going to be a net current going in this direction.
Now, if we compare the current flowing in this circuit, the 20 volt battery wants to send a current going in this direction. The 24 volt battery wants to send a current going in this direction. So I think there's going to be a current flow in this direction.
Now, let's assume that we have a current going in this direction, which we'll call I1. And there's a current going in this direction, which we'll call I2. Now, we could say that there's a current going in this direction, which is I1 minus I2. If we're wrong, we're going to get a negative answer for this result, which means that the current is going this way.
And it's going to be I2 minus I1 instead. But we can solve it this way. So we could define the current as going in that direction, and then change it later. It doesn't matter how you define it, as long as the math makes sense. So I'm going to define the current goal in this way, which is going to be I1 minus I2.
And let's see what happens in this example. Now before we begin, there's something else that we need to talk about. Perhaps you've noticed that there's two junctions, and we really don't need to take into consideration this junction. Because the current that flows through the 2-ohm resistor is the same current that flows in that part of the circuit.
So this is also I1. And this is I2. And the current that flows here is the difference between I1 and I2. So we get the same currents in that junction, so we don't have to worry about it. So that's why we only need to take into consideration one of the two highlighted junctions.
So now let's focus on the top loop. So let's define the loop going in that direction. We'll call it loop 1. I'm going to use a different color. So in loop 1, we're going towards a higher potential.
And so that's going to be a voltage lift of 20. And I1 is going in this direction. So therefore, this is going to be positive, negative, positive, negative. We're assuming that I1 goes in that direction.
And the current always flows from a high potential to a low potential. So as we flow through the 3-ohm resistor, as we follow the loop, we're going in the direction of the current towards a negative potential or lower potential. So that's going to be a voltage drop across the 3-ohm resistor. And that's going to be negative 3 times I1.
Now, as we follow the loop through the 2-ohm resistor, we're still going in the direction of I1. So that's another voltage drop, which is negative 2 times I1. And then we're going in this direction, through the 12-volt battery, but towards the negative terminal.
So we're going from positive to negative, so that's going to be a voltage drop of 12. And then we're going to travel through the 4 ohm resistor from positive to negative. And so the current in that branch is I2 as opposed to I1. And that's a voltage drop, so that's going to be minus 4 I1.
And so we've covered all the elements in loop 1. So now let's combine like terms. So we have 20 minus 12, which is 8. And then we have negative 3i1 minus 2i1, and that's negative 5i1. This is supposed to be i2.
I don't know why I have i1 here, because it was i2 times 4. So then this is going to be minus 4i2. Now let's take these terms and move it to this side. So now we have this equation.
8 is equal to 5i1 plus... For I2 now let's focus on the next loop so this is going to be loop 2 So let's start with this direction So we're going towards the negative terminal of the battery, so from positive to negative. That's a voltage drop of 18 volts. And then we're following this resistor from negative to positive.
So we're going from low to high potential against I2. So that's a voltage lift of 4 times I2. And then as we follow loop 2, we're traveling through the 12-volt battery from negative to positive, or towards the higher potential. So that's a voltage lift of 12 volts. And then we're going to follow through this resistor, which is in the direction of the 12-volt.
this current which we define it to be that way so we're going from positive to negative in the direction of the current so that's a voltage drop and that's gonna be negative 6 times I am 1 minus I 2 next we're going to go through this 6 volt battery from positive to negative so we're going towards the lower potential and that's going to be a voltage drop of 6 And then since the current is going in this direction, this is going to be the positive terminal of the resistor and the negative terminal of the resistor. So we're going from high potential to low potential in the direction of the current. So that's a voltage drop of 10 times I1 minus I2. And all of this is equal to 0. Now let's simplify.
So we have negative 18 plus 12, which is negative 6, plus another negative 6, so that's negative 12, and then plus 4i2. Now let's distribute, so that's going to be negative 6i1 plus 6i2, and then let's distribute to negative 10, so that's going to be negative 10i1 plus 10i2, and that's equal to 0. So now let's combine like terms. So we have 4i2 plus 6i2 which is 10i2 plus another 10i2. So far we have negative 12 plus 20i2 and then we have negative 6 minus 10. So that's negative 16i1, that's equal to 0. Now let's take the negative 12 and move it to the other side so it becomes positive. So on one side of the equation, we're going to have negative 16i1 plus 20i2.
It's on the left side of this equation. On the right side of this equation, we have positive 12, but I've reversed the equation again, so now it's on the left side again. So now let's focus on those two equations.
Let's cancel I2. So all I need to do is multiply the first equation by negative 5. So it becomes negative 40, which is equal to negative 25I1 minus 20I2. And then I'm going to rewrite the other equation. Right beneath it. So if we add these two equations, we can cancel those two terms.
Negative 40 plus 12 is negative 28. And then that's equal to negative 25 minus 16, which is negative 41, times I1. So the value for I1 is negative 28 divided by negative 41. So I1 is equal to 0.6829 amps. Now, let's calculate I2.
So, I'm going to use this formula before I multiply it by negative 5. So, 8 is equal to 5 times I1, and I1 is 0.6829, and then plus 4 times I2. So 5 times 0.6829, that's equal to 3.4145. So 8 minus 3.4145, that's 4.5855, and that's equal to 4I2. So to calculate I2, we need to divide both sides by 4. So I2 is 4.5855 divided by 4. And so I2 is 1.1464 amps.
Now let's analyze the circuit completely. So we said that this was 3 ohms, this was 2 ohms, and this is 6 ohms, and this is 4 ohms. Here we have a 20 volt battery, a 12 volt battery, an 18 volt battery, and this is 10 ohms, and this is a 6 volt battery. So at this point, what I want you to do is to calculate the potential everywhere.
So let's call this point A. And this is going to be point B, point C, point D, point E. And then let's make this F, G, and H. Calculate the potential at each of those points.
Now let's talk about the currents. Notice that I1 is positive, which means that there is a current flowing in this direction and not in that direction. And so that current is 0.6829 amps. Now, I2 is also positive, which we define it to be a current going in this direction. So, we don't have a current going in that direction.
Now, what about I1 minus I2? I1 minus I2 was going in this direction. Now, if this result is negative, that means there is a current going in this direction, which is I2 minus I1.
So, let's see which one is the correct direction. So, if we take I1 minus I2, 0.68. 2 9 minus 1.1464 that's going to give us a negative answer negative 0.4635 so therefore this direction is not correct so this is the correct direction if we do I 2 minus I 1 is going to give us positive 0.4635 so now we know the correct direction of the current so we have this current which is 0.4635 amps So now we can place the appropriate signs across each resistor.
So across the 3-ohm resistor, I1 is flowing in that direction. So this is going to be positive, and this is going to be negative. Now I1 is also flowing in this direction.
So this is going to be the negative part of the resistor, and this is the positive part. I2 is flowing in this direction, so this is going to be the positive terminal of the resistor, and this is the negative terminal. Now, I2 minus I1 flows in this direction, so this should be positive, and this should be negative.
And it's also flowing in this direction, which means this is negative. And this is positive. So now we have the appropriate signs.
Now let's calculate the potential everywhere. So let's go from A to B. As we travel from A to B, we're going from low potential to high potential.
So that's going to be a voltage lift of 20 volts. So the potential at... B is 20 volts which I'm going to highlight in green now let's go from B to C so I one flows from positive to negative so as we go from B to C this is going to be a voltage drop of three times I one so that's gonna be three times point six eight two nine which is two point zero four eight seven so 20 minus that number that's going to give us a potential at C which is seventeen point nine five volts So now let's calculate the potential at D. So we're going from C to D. So we're going towards a lower potential.
So that's going to be a voltage drop. And it's I1 times 2. So 2 times 0.6829. That's 1.3658. 17.95 minus that number.
That's going to give us a potential at point D of 16.58 volts. Now let's move from point D to E. So we're going from positive to negative, so that's going to be a voltage drop of 12 volts. So 16.58 minus 12, that's going to give us a potential at point E, which is 4.58 volts.
Now, we don't need to go to A because we already know the potential at A. So we have the potential difference between points E and A. So that's the voltage of 4.58 minus 0, which is 4.58 volts.
If we take the voltage across the 4-ohm resistor and divide it by 4, that will give us the current flowing through the resistor. And so that current is 1.145 amps. And so this tells us that we're on the right track because this matches I2. Now this is a slight difference because there was some rounding in this problem. I mean, this is not exactly 17.95, it's like 17.951.
But however, this is close enough. 1.146 and 1.145. You could round that and say it's about 1.15 if you want to.
Now, let's start back at D, and let's go towards F. So as we travel from D to F, is this a voltage lift or a voltage drop? Notice that we're going against the current, and we're going towards a higher voltage.
potential from low to high. So that's a voltage lift. So to calculate the voltage across the 6 ohm resistor, it's going to be 6 ohms times 0.4635. That's the current flowing through it.
And so the voltage across that resistor is 2.781 volts. So as we go from D to E, I mean D to F, that's a voltage lift of that value. So we're going to start with 16.58 and then add 2.781 to it.
So the potential at F is 19.36 volts. So now let's go from F to G. So we're going from positive to negative, and that's going to be a voltage drop of 6 volts. So 19.36 minus 6 is 13.36.
So that's the potential at G. Now let's go from A to H. As we travel in this direction, we're going from negative to positive.
So that's the voltage lift of 18 volts. So that's the potential at H. And current is going to flow from a high potential to a low potential.
18 is greater than 13.36. So the current is flowing in this direction, which is in agreement with this current, which goes in that direction. So now let's calculate the current across the 10 ohm resistor.
So first, we need to calculate the potential difference between points H and G. So 18... minus 13.36 that's a voltage of 4.64 volts and if we divide that by 10 that will give us a current of.464 amps notice that it's approximately equal to I2 minus I1 which is 0.4635 amps.
So that means that we're on the right track. So everything is correct in this problem. Now we do have some rounded answers, but it all adds up. Now this is going to be the final problem that we're going to work on. So let's say this is a 2-ohm resistor.
And here we have an 8-ohm resistor. And a 4-ohm resistor. This one's going to be a 3-ohm resistor. 5 and 6 on the left we have a 20 volt battery and this is going to be a 6 volt battery and there's 8 volt a 12 volt and the last one is a 16 volt battery so go ahead and calculate the current flowing through every branch of the circuit And let's define this as point A, B, C, D. This is going to be E, F, G, H. At point A, the potential will be 0 volts.
So relative to that potential, calculate the potential at every point. Now let's define the current that flows through I2 as I1. and the current that flows through the 8-ohm resistor, that's going to be I2.
Now the current flowing through the 3-ohm resistor, that's going to be I1 minus I2. And through the 5-ohm resistor, we're going to call that I3. Now the current that flows in this branch, which is basically the same as through the 6-ohm resistor, that's going to be I1 minus I2. Minus this current here, which is I3 So let's start with the first loop We'll call that loop 1 So as we go in this direction We have a voltage lift of 20 volts Now, based on the direction of the current, this is going to be positive, this is going to be negative.
And for the 8 ohm resistor, this is positive, and this is negative as it follows I2. And for the 4 ohm resistor, I1 is flowing in this direction. So this is going to be positive and negative.
So as we flow in this direction, that is in a direction of I1, that's going to be a voltage drop. So that's negative 2I1. And then as we follow the loop through the 8-ohm resistor in the direction of I2, that's going to be negative 8I2. That's another voltage drop.
And then as we go through the 6 volt battery, we're going from a high potential to a low potential. So that's going to be a voltage drop of 6 volts. And then following I1, which goes through the 4 ohm resistor, that's another voltage drop of...
negative 4i1. So let's combine like terms. We have 20 minus 6, which is 14, and then we can combine negative 2i1 minus 4i1. And that's going to be negative 6i1 and then minus 8i2. So I'm going to take these two terms and move it to that side.
And so that gives me 14 is equal to 6i1 plus 8i2. Now I'm going to divide everything by 2. So it becomes 7 is equal to 3i1 plus 4i2. Now let's move on to the next loop. So let's call this loop 2. So let's go in this direction. So we have a voltage lift of 6 volts, and then as we flow through the 8-ohm resistor, that's going to be a voltage lift.
going from negative to positive against I2 so that's going to be plus 8I2 and then we travel in this direction in the direction of I1 minus I2 so that's a voltage drop and so that's going to be minus 3 times I1 minus I2 Now, as we flow through the 5-ohm resistor in the direction of I3, that's going to be a voltage drop of 5I3. And then let's go through the 8-volt battery. So we're going towards the lower potential. So that's a voltage drop of 8 volts.
And so that should equal 0. So now let's simplify the expression. We have 6 minus 8, which is negative 2. And then we have plus 8i2, and let's distribute the negative 3. So that's negative 3i1 plus 3i2 minus... 5i3.
So we can combine these two. So right now we have negative 2 and then minus 3i1, 8i2 plus 3i2, that's 11i2, and then minus 5i3. So I'm going to move the negative 2 to the other side.
So this is the second equation that we have. Negative 3i1 plus 11i2 minus 5i3. That's equal to 2. So let's save that equation for later.
And now, let's move on to the third loop. So as we travel in this direction, we have a voltage lift that is 5 I3. It's going against I3 as we go up.
So that's going to be positive 5 I3. And then we're traveling in this direction towards the positive terminal of the battery. So that's a voltage lift of 16 volts. And then we're following the direction of this current.
So that's a voltage drop. That's going to be negative 6 times I1 minus I2 minus I3. And then we're traveling in this direction towards the positive terminal, so that's a voltage lift of 12 volts. Now let's combine like terms.
So 16 plus 12 is 28. And then we need to distribute the negative 6. So it's negative 6i1 plus 6i2. and then plus 6i3 and that is equal to 0 and so we have 28 minus 6i1 plus 6i2 now let's combine these two so 5 plus 6 that's 11 and so now we have our third equation now what i'm going to do is simply take everything other than the 28 and move it to the other side So we can write this as 28 is equal to 6i1 minus 6i2 minus 11i3. Now, let's line up the three equations that we have.
So, the first one is 3i1 plus 4i2, and that's equal to 7. The second one, negative 3i1. plus 11 I2 minus 5 I3 that's equal to 2 now for the last one it's 6 I1 minus 6i2 minus 11i3, and that's equal to 28. So we have three equations with three variables. So we need to use the system of equations process to get the answer. And if you're not sure how to do this, you can check out one of my videos entitled System of Equations with Three Variables.
Just type in Organic Chemistry Tutor. It should come up. I have a 26-minute video that... gives you some examples on how to solve these types of problems. So go ahead and try this if you want to.
Now the best way to approach this particular situation is that we need to combine equations 2 and 3, and our goal is to cancel I3, because that's going to give us an equation in terms of I1 and I2, which we can combine that with equation 1 to either solve for I1 or I2. So what I'm going to do is I'm going to multiply the second equation by a negative 11, and so that's going to give me positive 55I3. And I'm going to multiply the third equation by positive 5, which will give me negative 55I3.
So negative 3i1 times negative 11, that's positive 33i1. And then this times negative 11, that's going to be negative 121i2. And negative 5i3 times negative 11, that's positive 55i3.
And 2 times negative 11 is negative 22. Now let's multiply the third equation by 5. So 6i1 times 5, that's going to be 30i1. And then negative 6i2 times 5, that's negative 30i2. Negative 11i3 times 5 is negative 55i3.
And 28 times 5. We know 20 times 5 is 100. 8 times 5 is 40, so this is going to be 140. Now we can cancel these two terms 33 plus 30 is 63 and negative 121 plus negative 30 that's negative 151 times I2. and 140 minus 22. 140 minus 20 is 120, and then 120 minus 2, that's 118. So now we need to combine this equation with this one. And so let's focus on I want some gonna multiply the first equation by negative 21 because 3i 1 times negative 21 is negative 63 i 1 and then 4i 2 times negative 21 4 times 20 is 80 and then 1 times 4 is 4 so this is going to be negative 84 i 2 and then 7 times 21 7 times 2 is 14 so 7 times 20 is 140 and then 7 times 1 is seven so this is going to be negative 147 now negative 151 minus 84 that's negative 235 and 118 minus 147 that's negative 29 so I 2 is equal to negative 29 divided by negative 235 and so I 2 is 0.1234 amps so that's the first answer that we need Now let's use the first equation before we multiply it by 7 to calculate I1.
So it's going to be 3I1 plus 4I2, or 4 times 0.1234, and that's going to equal 7. So 4 times 0.1234, that's 0.4936. So 7 minus that number, that's equal to 6.5064, and that's equal to 3i. So to calculate i, I mean i1 that is, we need to divide it by 3. So I1 is equal to 6.5064 divided by 3, which comes out to be 2.1688 amps.
So now that we have the value for I1, we need to calculate I3. And I'm going to use the third equation before multiplying it by 5. So this is going to be 6i1 or 6 times 2.1688 minus 6i2. and then minus 11 I 3 and that's equal to 28 so 6 times 2.1688 that's 13.0128 and then 6 times 0.1234 that's 0.7404 Now let's combine those two numbers. So 13.0128 minus.7404. That's 12.2724.
So we need to subtract both sides by that number. So 28 minus 12.2724, that's 15.7276. And now let's divide both sides by negative 11, and this will give us I3.
So I3 is negative 1.4298 amps. Because it's negative, we need to reverse the direction of I3. Now let's redraw the circuit. So this was the 2 ohm resistor and this was 8 ohms and then this was 3 ohms, 5 ohms, 6 ohms and this was 4 ohms. And this was a 20 volt battery.
Here we had a 6 volt battery and this is an 8 volt battery. And that's a 12 volt battery and then over here we have a 16. volt battery. Now I1, the current that flows through this branch, that's in the right direction.
So this is 2.1688 amps. And so this is positive, and this side of the resistor has to be negative. Now, I2 is flowing through this resistor, and because I2 is positive, we have the right direction, and I2 is 0.1234 amps, which we can write that later. But we know this is the positive terminal of the resistor, and this is going to be the negative terminal of the resistor.
Now I1 also flows through the 4 ohm resistor in that direction. So therefore, this side is positive and this side is negative. Now, I1 minus I2, if we subtract these two currents, that's going to give us a positive value.
And I1 minus I2 flows through the 3-ohm resistor. And so that's 2.1688 minus 0.1234. And so this current is 2.0454 amps. So this is going to be the positive side of the resistor, and this is the negative side.
Now, I3 is a negative value. Before, we defined I3 flowing this way. But in actuality, it's flowing in that direction. And so, therefore, this should be positive, and this should be negative.
Now the current that flows in this direction is I1 minus I2 minus I3. So that's 2.1688 minus I2, which is 0.1234, minus negative 1.4298. So the current that should be flowing in this branch is 3.4752.
Now it's flowing through this resistor. So this is the positive side of this resistor, and this is the negative side. So now that we have all the appropriate signs, Let's calculate the potential at every point.
So let's calculate it at, well we're going to assume point A is 0. This is B, C, D, E, F, G, and H. So if the potential is 0 at A, what is it going to be at B? So going in this direction, we have a voltage lift of 20 volts, so the potential at B is 20. now going from B to C we have a voltage drop because we're going towards a lower potential and so it's gonna be 20 minus IR or minus 2.1688 times 2 so the potential at C is 15.66 volts now let's go this way So we're going towards a higher potential, and the current through the 4-ohm resistor is I1, which is this value. So it's going to be 0+, because we're going towards a higher potential, plus 4 times 2.1688. So the potential at H is 8.675 volts.
now let's go up to this point and that should be labeled a new point so let's call this point I so as we go from H to I we have a voltage lift we're going towards a positive terminal so voltage lift of 6 volts and so 8.675 plus 6 That's going to give us 14.675. So that's potential at I. So now we can confirm the value of I2. So first, let's calculate the potential difference, or the voltage, across the 8-ohm resistor. resistor.
So, it's the difference between the electric potentials at C and I. So it's going to be 15.66 minus 14.675. So that's a voltage of 0.985.
And then divide it by the 8-ohm resistor. This will give you a current of 0.1231 amps, which is very close to this current. So that means that the information that we have in this loop is correct. Now let's move on from C to D.
So we have a voltage drop. And since we're going towards a negative terminal, so to calculate the potential at D, it's going to be the potential at C, 15.66, minus the voltage drop, so minus 2.0454 times the resistance of 3. And so the potential at D is going to be 9.524 volts. Now, let's go to H. Starting from there, let's go to G.
So that's a voltage increase of 8 volts. So 8.675 plus 8, that's going to give us 16.675 for the electric potential at G. So now let's calculate the potential difference between D and G to confirm the value of I3.
So notice that we're going from a high potential of 16 volts to a low potential of like 9.5 volts. And so it doesn't make sense that the current is going in this direction. So 16.675 minus 9.524, that's a voltage of 7.151. If we divide that by 5, that will give us a current of I3, which is 1.4302 amps. And, keep in mind, I3 is negative 1.4 going in this direction.
But once we reverse it, it becomes positive 1.4. It's no longer negative once we reverse it. And so 1.4298 is approximately 1.43. So this answer is acceptable.
Now let's confirm the current flowing through this resistor, which should equal this number. Because if you think about it, I3 is 1.4298, and we have this current, which is 2.0454. So these two currents should add up to this one.
So we add 2.0454 plus 1.4298. that will give you 3.4752 because this these two currents are the currents going into junction and this is the current going out of the junction and so based on Kirchhoff's junction rule the current that flows into a junction must equal the current that flows out of a junction Now we have the potential at D, so let's calculate the potential at E. So going from D to E, we have a voltage lift of 16 volts.
So 9.524 plus 16, that's going to give us a potential of 25.524 volts. Now let's calculate the electric potential at F. So we have it from G.
So let's go from G to F. And so going from a high potential to a low potential, so that's going to be a voltage drop of 12 volts. So 16 minus 12 is 4. So the potential is going to be 4.675 volts at F. So the potential difference between E and F, that's 25.524 minus 4.675.
And so that's a potential difference of 20.849. So that's the voltage across the 6. resistor if we divide it by six that will give us a current flowing in this direction which is 3.475 or 4.8 if you want around it and so that's very close to this answer so all the values make sense in this problem so these answers are indeed correct So now you know how to use Kirchhoff's Junction Rule and Loop Rule to calculate all the currents in a circuit. And now you know how to calculate the electric potential at any point in a circuit.
But you need to be given a reference value first. So relative to that reference value, you know how to calculate the potential at any point. So that's it for this video. Thanks for watching, and have a good day.