Example- Kirchoff’s Rules
Lecturer: All right. Let’s do an example related to the Kirchoff Rules.
Let’s assume that we have a simple circuit with three electromotive forces, Epsilon1 connected to a resistance r1, Epsilon2 connected to a resistor r2, and finally, Epsilon 3, connected to a resistor of let’s say r3. And you can easily in this circuit we can not go ahead and directly apply Ohm’s law because of the existence of these three electromotive forces, or three batteries.
Let’s assume that we are interested with the energy dissipated per unit time through each one of these resistances. So, the question is find the power dissipated through each resistor. Now, if you recall that the power dissipated through a resistor was equal to i squared times r, in other words, the current flowing through that resistor squared times the resistance of that resistor.
To be able to determine this quantity therefore, assuming that we are given the numerical values of Epsilon1, Epsilon2, Epsilon3 and as well as r1, r2, and r3, and we need to determine the current flowing through each one of these resistances.
Let’s say let Epsilon1 is equal to 2 volts, Epsilon2 is equal to 4 volts, and Epsilon3 is equal to let’s say 10 volts, and r1 is equal to 1 ohm, r2 is equal to 2 ohms, and finally r3 is equal to 4 ohms for example.
So, knowing the values of these electromotive forces and as well as the resistances, to be able to determine the power dissipated through each resistor, we need to find out how much current is flowing through each resistor, and to do that, we will apply the Kirchoff’s rules.
Well, in doing that, we can follow two different methods. Let’s say method one, which we call this method as branch current method. Okay, first, we’re going to mark the direction of the emf arrow, and since they point from negative to positive, for Epsilon1 it’s going to be pointing up, for Epsilon2 again it’s going to be pointing up, and as well as for Epsilon3 because this is the negative end and this is the positive end for these electromotive forces.
As second step, we’re going to assign a current for each branch of this circuit. What we mean by the branch is the paths between the junction points. Well, as you can see, this is one junction point and this is the, another junction point. These are the only points that we have the wires are connecting to each other or separating two different paths.
Therefore, this one is one branch, this one is the other branch, and this one is the other branch. So, as a first step, we will assign a current for each branch, and as we do this, the direction of flow of current is irrelevant. In other words, we can choose any direction that we’d like to.
For this one for example, let’s choose current i1 flowing through the first branch, i1. Therefore i1 is moving along this branch. Again, I chose the i1 in clockwise direction for that branch. I could have easily chosen just the opposite direction, and that will be fine.
So, if we look at the middle branch, in this case I’m going to choose, which is this path, current flowing in downward direction, and I will call this one as i2. And finally I’m going to choose for the third branch that the current is flowing again in clockwise direction, so I will call that one as i3.
Again, the direction of flow of these currents completely arbitrary. We can choose it any direction that we’d like to. In other words, we could have easily chosen just the opposite directions of these flows, but we choose it arbitrarily and then we mark it for each branch as i1, i2, i3, and so on an so forth.
After assigning the branches, the next step is we’re going to choose a direction to traverse the possible loops. As a first glance, we can easily see that we have one loop over here and another loop over here. Let’s mark those loops as loop one and loop two.
For the loop one, we will just choose a direction to traverse this loop. Well, I’m going to choose clockwise direction, and I will show that direction in dashed lines with an arrow in my circuit diagram. And for the other loop, again, I’m going to choose clockwise direction and show it in the diagram.
As we are choosing these directions to traverse the circuit or these loops, that direction is again completely arbitrary, we can choose them in counterclockwise direction, we can choose one of them clockwise, the other one in counterclockwise direction.
So, it’s all up to our choice, and we have complete freedom in choosing those directions. Once we choose the direction that we will traverse each one of these loops, then we will go ahead and write down the loop equations.
Let’s say loop one. We’re going to start at an arbitrary point. We can start just for example right before the emf Epsilon1, and starting from that point, we will trace this loop in clockwise direction.
Now, since we’re going in clockwise direction, as we cross Epsilon1, we will cross this in the same direction with emf arrow, and from the rules we know that if we do that the potential will increase by +Epsilon1 volts. And then we will continue. Now we’re going to cross Epsilon2. In this case we will cross it in opposite direction to the emf arrow. Therefore, the potential will decrease by -Epsilon2 volts.
Now we will cross r2 in the direction of flow of current, and as you recall, if you cross a resistor, a resistance in the direction of flow of current, then the potential will decrease. So, we’re going to have minus sign over here by the current, which is i2, times the resistance, r2.
We will continue in clockwise direction. Now we’re going to cross r1 in the direction of flow of current. Again, the potential will decrease by current times, in this case it is i1, times the resistance, which is r1.
The loop is completed. Algebraic sum of the changes in potential will add up to zero, and that is our equation number one associated with loop one. We will do the same thing for loop two.
We’re going to start again at a certain point. Let’s start before Epsilon2. Again, we will be tracing the loop doing clockwise direction. In doing so, we will cross Epsilon2 in the same direction with the emf arrow. Therefore, the potential will increase by Epsilon2 volts. And then moving on in clockwise direction, we will cross Epsilon3 in opposite direction to the emf arrow, so the potential will decrease by -Epsilon3 volts.
Now we will be crossing r3 in the same direction with the current, so again, we will experience a decrease in potential, and that will be equal times, which is i3, times the resistance, and that is r3.
And moving on in clockwise direction, now we will cross r2 in opposite direction to the flow of current. Therefore, the potential will increase by current times the resistance, and that is i2 times r2.
Loop is completed. Algebraic sum of these changes in potential will add up to zero, and that is equation number two. So, when we look at this equation, we’re given the values of Epsilon1, Epsilon2, r2, r1, Epsilon3, r3, and r2. So, our unknowns are the currents, which are namely i1, i2, and i3. And here we have three unknowns but two equations. To be able to solve for the unknowns, we need one more equation, and that is going to come from the junction rule, and then the junction rule simply says that sum of the currents into a junction should be equal to sum of the currents coming out of that junction.
And when we look at the junction points, which are these two points, we can pick any one of these. If we look at this one over here, we see that i1 is going into that junction while i2 is coming out and as well as i3 is coming out. So, the junction rule says that i1, which is going in, should be equal to i2 plus i3, and those are the ones that they are coming out from that junction.
If we use the other junction point over here, that will give us the same equation. In that case, i2 and i3 are going in and i1 is coming out, so i2 plus i3 will be equal to i1, which is the same equation that we have over here.
Now, as you can see, we have one equation over here, the other equation over there, and the third equation here. We have three equations and three unknowns, therefore we can easily solve these equations simultaneously to be able to obtain i1, i2, and i3.
Once those currents are obtained, then the power dissipated from the first resistor will be i1 squared times r1, through the second resistor will be i2 squared times r2, and through the third one will be i3 squared times r3. And in a similar way, power generated by the emf one, Epsilon1, will be equal to, let’s call it this one p-sub-Epsilon1. That’ll be equal to i1 times Epsilon1.
Epsilon2, power generated by Epsilon2 will be i2 times Epsilon2, and similarly for Epsilon3 we will have i3 times Epsilon3 because those are the currents which are pumped or generated by these electromotive forces.
Now, I mentioned that as we were choosing the direction of these currents through each one of these branches, we said that we’re completely free in a direction that we’d like to choose. At the end of our algebraic calculation, it turns out to be — if it turns out to be that the i1 is negative, it simply implies that the actual flow direction of the current is just the opposite that we have chosen.
So, important thing over there is the information of the magnitude of the quantity, the value that we will find once we solve these three equations simultaneously.
Again, when we look at the power generated by these power supplies depending upon the sign of these currents that we will end up, if we end up with positive current, it means that that battery is supplying the energy to the circuit and if we end up with -current value, therefore the power associated with that power supply will be negative, indicating that that power supply is getting charged, that battery is getting charged in that circuit.
And of course, power supplied should always be equal to power dissipated. In other words, sum of the powers, energy per unit time supplied, should be equal to sum of the energy per unit time dissipated through each resistor and as well as if there are some batteries which are getting charged in the circuit, then the power consumed by those batteries.
I’m not going to go to the details of solving these three equations simultaneously. I will leave that to you as an homework, and of course you will just replace these symbols with their numerical values over here, and to be able to solve them simultaneously, you can just solve one of the unknowns in terms of the other two and substitute that into the following equation. Therefore —
For example, here I want it to equal to i2 plus i3, and if you just take this i1 and replace it or substitute it into the first equation, then you’re going to end up with two equations in terms of i2 and i3, the first equation and the second one. You can solve one of them for one of the unknowns, say i2 in terms of i3, and then substitute it into the first equation. Then you’re going to end up with one equation and one unknown, which you can easily solve. Once you obtain one of the currents, then you can trace it backwards to be able to get the other two equations.
Now I will show you another method of solving the same circuit, and in order to do that I will redraw the circuit over here. Epsilon1 with resistance r1 here, Epsilon2 with resistance r2 here, and emf Epsilon3 with resistance r3. So, this is the original circuit. Let’s call it over here method two, and this method is also called as loop current method.
In this case, again let’s first draw the emf arrows directions for each one of these power supplies pointing from negative to positive end. And so in this case instead of assigning current for each branch, we will assign loop current for each loop in the circuit, and we will see that net current i1 flowing through the first loop in clockwise direction and for the second loop let’s choose a current i2 flowing in counterclockwise direction. So this is clockwise direction and that one is counterclockwise direction.
Therefore instead of assigning branch currents we’re assigning loop currents. Again, we’re choosing the direction of flow of these currents completely arbitrary. I just chose one of them as clockwise direction, the other one as counterclockwise direction.
Again, I’m going to trace, I’m going to choose a direction to trace each one of these loops, so I will choose this one in clockwise direction, and I will choose the other one — I can choose it in clockwise direction or counterclockwise direction. Let’s choose it in this case in counterclockwise direction.
Again, all these directions are arbitrary. You can choose any direction that you’d like to choose. And this is our loop two and we will write down our loop equations starting from the loop one. Let’s again start just right before the electromotive force one, Epsilon1. We’re going to be crossing this in clockwise direction, therefore from negative to positive terminal, in other words in the direction of emf arrow, therefore the potential will increase by Epsilon1 volts.
Moving in clockwise direction now, we’re going to cross Epsilon2 in opposite direction to the emf arrow, therefore the potential will decrease by Epsilon2 volts. Now we’re going to be crossing r2 in the direction of flow of current. Why? When we look at this middle branch, i1 is going, or flowing, in downward direction and as well as i2 is flowing in downward direction.
So, the net current flowing through this middle branch is sum of these two currents. As we cross this in clockwise direction, we’re going to be crossing r2 in the same direction with the flow of current, therefore the potential will decrease by the net current, which is i1 plus i2 times the resistance r2.
Now moving on, we will come back to resistance r1. We will be crossing this again in the same direction with the flow of current, therefore the potential will decrease by current times, and that is i1 times r1. The loop is completed. Algebraic sum of the changes in potential will add up to zero.
Now let’s look at the second loop. In the second loop, starting just right before Epsilon2 and going in counterclockwise direction, so we’ll start over here, because in this case we chose the direction — I chose the direction to traverse in counterclockwise direction, so I’m going to be crossing Epsilon2 in opposite direction to the emf arrow. Therefore, the potential will decrease by -Epsilon2 volts.
Going in counterclockwise direction, I will cross r2 in the same direction with the net current flow. Therefore, again we will experience a decrease in potential, minus. Net current is along the middle branch i1 plus i2 and times resistance r2. Moving on in counterclockwise direction, we will cross r3. Current is also flowing in the same direction, in, therefore in the direction of the flow of current. We will experience a decrease in potential, and that’ll be i3 times r3.
Now we will cross Epsilon3 as we go in counterclockwise direction, in the same direction with emf arrow, therefore the potential will increase by Epsilon3 volts.
We have come back to this point that we started with. Loop is completed. Again, the algebraic sum of these changes in potential will add up to zero.
As you see now, in this case we have only two unknowns, i1 and i2 loop current, and we have two equations. So, when we apply loop current method, the junction rule is irrelevant, and we will have equal number of equations to our unknowns once we express the loop equations, and we can easily solve that algebraically.
If you prefer to use this method, you should be careful while you’re assigning the currents. Here I chose the first loop’s current in clockwise direction, the other one as counterclockwise direction. In doing so, I ended up the net current flowing through the middle branch as the sum of these two currents because they flow in the same direction along this branch.
Well, if I had chosen i2, which I can do that. There’s no restriction for that. The i2 also in clockwise direction. If I do that, then the i2 will be flowing up through the middle branch while i1 was going to be flowing down through the middle branch. Then the net current was going to be the difference between these two currents.
And since before we solve the problem we will not know which one is greater than the other one, then the only way by determining the net flow of — net flow direction of the current, is by assuming one of them as greater than the other one. So, in that case, we will make an assumption and we’ll say that net i1 is greater than i2 for example.
In that case, since i1 is going downward direction whereas i2 is in upward direction, and i1 is greater than i2, the net flow will be in downward direction. And then we keep that assumption throughout our loop equations as we write down those loop equations.
Again, at the end of the calculations, if it turns out to be that i2 is greater than i1, which we’re going to end up in that case with a negative value for the net current through this middle branch, then it means that the actual flow direction is just the opposite that we have chosen. But the magnitude of the current will be the same in any direction that we choose.
Again, once the current values are obtained, one can determine the power dissipated by r1, which is going to be i1 squared times r2, but for the resistance r2, p2, power dissipated, is going to be equal to net current flowing through r2 squared, which is i1 plus i2 squared, times r2, and r3, again, let me just write down. p3 is going to be equal to current through r3, and that is i2 squared times r3, and p1 will be i1 squared times r1.
So, using these expressions, then we can easily determine how much energy is dissipated or converted into heat in every second.