Example- Connection of Resistances Series and Parallel
All right. Let’s do an example related to the connections of resistances. Let’s assume that we have several resistances which are connected in this form. R1, R2 is connected here, and R3 is connected here. R4 is here and then we have resistance R6 here, which is eventually connected to resistance R7 here. Then, let’s say we apply a potential difference across the whole combination by connecting the ends of these connections to the terminals of an electromotive force which generates ε volts.
Let’s see, the question is, what is the power dissipated through each resistance? Now, to be able to determine the power, if you recall that it was equal to power dissipated — let’s use a subscript of dissipated — was equal to i2 times the resistance R. Square of the current passing through the resistance times the resistance value of that resistor will give us how much energy is converted into heat in every second.
Well, obviously let’s say that the numerical values of our resistances are given. Let’s say R1 is equal to 2 ohms, R2 is equal to 6 ohms and R3 is equal to 12 ohms. R4 is equal to 4 ohms, and R6 is equal to 10 ohms and finally, R7 is equal to 3 ohms. So, knowing the numerical values of these resistances, therefore if we can determine the amount of current flowing through each resistor, then we will be able to determine the power dissipated through each resistor.
Okay. To be able to do that, first we will try to determine the equivalent resistance of the circuit and then we will trace it back to be able to determine the current flowing through each resistance. If we recall Ohm’s law, which was simply equal to, from the definition of resistance, and that was equal to the ratio of the potential difference between two points to the amount of current passing through those points.
When we look at this law, we see that to be able to determine any one of these quantities, voltage, current, resistance, we need to know the other two of these quantities. So, we are going to therefore try to determine these quantities since the resistances are given, we will try to determine the current, and to be able to do that, we need to determine the potential difference across each resistor. So by applying the properties associated with parallel and series connections, we will determine those quantities, therefore eventually the current through each resistance and once we determine that, we will be able to calculate the power dissipated.
Okay. If we look at our circuit, we see that first, these three resistances R2, R3, and R4 are connected in parallel. First let’s mark the junctions points. These are the points in which we are ending up with different branches for the current to flow. When we look at here, the circuit, we see that we have a junction point here, another one here, another one there, and another one here. So we can easily see that these three resistances are connected in parallel.
In other words, I can redraw this circuit in this form. This is R1 and now from this junction point, I can redraw it as R2, R3, and R4, which are connected through these junction points, which I can also draw that this way. We have R6 and then we have this segment of the wire, just the wire, and connect it in parallel, which is eventually connected to R7 in this manner. Therefore the current can be redrawn in this form to be able to see the connections clearly and this is equivalent to our original circuit.
So we redraw the circuit to be able to see the connections clearly and we can easily see that these three resistances now are connected in parallel. First, therefore, we will replace them with their equivalent. As we do that, we will redraw the circuit, and here is the electromotive force, resistance R1. We are going to be replacing this with, let’s call this one as Req1 and let’s also just concentrate on this parallel connection over here. When we look at that segment of the circuit, following the current which is emerging from the power supply, which will be separated into three branches as it goes through R2, R3, and R4, and it will join and then it will make i again.
Now, whenever the current comes to this point, charges will see two available paths. One of them is this one and the other one is this one. So as they try to go through these two branches, they will see that along this branch, there is resistance R6, whereas along this branch, there is no resistance. Therefore, naturally the charges will choose the path that there is no resistance. In other words, they will go directly through this path and none of the current will go through resistance R6. Therefore, as equivalent of this segment here, we can just write down or replace it with a simple wire with no resistance at all and then we will end up with the next resistance, which is connected in series to this one that is R7. So the circuit will be induced to this simplified form.
Now let’s just go ahead and calculate the Req1 since that is the equivalent of the parallel connection of R2, R3, R4. 1 over Req1 will be equal to 1 over R2 plus 1 over R3 plus 1 over R4, from the properties of parallel connections of resistances. Moving on, 1 over Req1 is going to be equal to, R2 is 6 ohms, 1 over 6, plus R3 is 12 ohms, 1 over 12, plus R4 is 4 ohms, so we’ll have 1 over 4 here. Now, if you have common denominators, then we have to multiply this ratio, both numerator and denominator by 2, this one just by 1, and this one by 3 because they will have a common denominator of 12.
Moving on, 1 over Req1 is going to be equal to 2 plus 1 plus 3 over 12, which will be equal to 6 over 12, and that is equal to 1 over Req1. To be able to get Req1, we will take the inverse of this, which is going to be equal to 12 over 6 and that will give us 2 ohms. Therefore, the value of Req1 will be equal to 2 ohms. In other words, if we just take this parallel unit out and place 2 ohms of resistance here, it will do the same job that these three are doing in parallel connection.
Now, we will go one more step and find the equivalent of these three resistances which are connected in series. Before we do that, let me show you also why we will have no current flowing through R6. Again, if we just think of this as the parallel connection of resistances and try to figure out the equivalent resistance of this unit — let’s call that one as 1 over Req2 — the equivalent resistance of this unit over here is going to be equal to 1 over R6 plus 1 over the resistance of this segment. We’re going to assume that the resistance of the wire is much, much smaller than the resistance of this resistor, so we’ll assume that it’s almost 0. In other words, we will have 1 over 0, or 1 over a very, very small number, therefore, if we divide 1 by an extremely small number that is going to go to infinity, so it’s going to be 1 over 6 plus infinity. Of course then, this whole term is going to go to infinity, because it doesn’t make any difference whether we add a number to a very large number or subtract it.
From there, if we solve for Req2, that’s going to be inverse of this. In other words, a number divided by infinity will go to 0, because if we divide 1 by a very large number, then we are going to end up with an extremely small number. Therefore, the equivalent resistance of all of this unit is going to be equal to 0. It means that no current is going to be flowing through this resistance. All of the current will prefer to go through this much less resistive path, in other words, almost 0 resistive path.
Okay. Now once we figured out that the whole current is going through this branch, then we can easily calculate the equivalent resistance of all these three resistances in series connection. That’s going to be equal to the equivalent resistance of the whole circuit. Let’s give a numerical value for this electromotive force also. Let’s assume that that quantity is also given and it is equal to 7 volts. Here then, we can calculate the equivalent resistance of the whole circuit and that’s going to be equal to the equivalent of all these three resistances, which is R1 plus Req1 plus R7. Substituting the numerical values, equivalent resistance will turn out to be, R1 is 2 ohms, plus Req1 was also 2 ohms, plus R7 is 3 ohms, and that will add up to 7 ohms. Therefore, the equivalent resistance of the whole circuit is 7 ohms.
Now, once we determine the equivalent resistance, then we can determine how much current is going to be flowing through this circuit, in other words, how much current is drawn from the 7 volt electromotive force. To do that, we will apply Ohm’s law since we know that Req is going to be equal to potential difference across the resistor, which is going to be equal to whatever the potential difference supplied by the power supply, and that is ε volts, divided by the amount of current flowing through this resistor, and that is i. Solving for i, we will have ε over Req. ε is 7 volts and Req is 7 ohms, so therefore this is going to be equal to 1 amperes. 1 ampere of current, therefore, will be flowing through this circuit.
Once we have determined that, will trace our circuits backwards, so we’ll go from here to this circuit. Now, when we go there, we know that since Req is doing the job that these three resistances are doing in the circuit and the power supply is the same, therefore the current i is going to be flowing through this circuit. Since all these three resistances are connected in series, then the current flowing through each one of them will be the same current, which is the same i, or in other words, 1 amperes.
Now, we know that 1 A of current is flowing through R1, through Req1, and through R7. Therefore we can easily calculate the power dissipated through each one of these resistances. Well, let’s call for the first one as P1. P1 is going to be equal to i12 times R1, but i12 is equal to i, so that’s going to be i2R1. i2 is 1 A squared times R1, and R1 was 2 ohms, therefore that is going to be equal to 2 watts of power dissipated through resistance R1.
Similarly, we can calculated the power dissipated through R7 since we know the current flowing through R7. That’s going to be equal to i72 times R7, but i7 is equal to i. i2 times R7, and that’s going to be equal to, again, 1 A squared times R7 and R7 was 3 ohms, which will give us 3 watts of power being dissipated as the current flows through resistance R7.
Okay. We have determined the power dissipated through two resistances. Now, we’re going to go furthermore, trace it back to the first circuit. We see that the Req1 is the equivalent of these three resistances and those are connected in parallel. We know that in parallel connection, the potential difference across the whole combination is equal to potential differences across each resistor, which are connected in parallel. Therefore, if we find the potential difference across Req1, let’s call that one Veq1, that will be the same potential difference across each resistor in parallel connection. So therefore all these three resistances will have the same potential difference of Veq1.
We can easily calculate that. Since we know the resistance, we also know the current. Then we know two of these quantities, resistance and current, therefore we can easily figure out the potential difference. In this case, Veq1 is going to be equal to i times Req1, which will be equal to 1 A times Req1, let me calculate it, is +2 ohms. Therefore 1 times 2 is 2 volts is the potential difference across Req1 resistance. Now, once we trace it back to the original circuit over here, then both for all R2, R3, and R4, the potential difference between their ends will be the same and they’re going to be equal to Veq1.
In this case now, we know the potential difference and the resistance. Therefore we can figure out how much current is flowing through each one of these resistances, because as you can see, once current i comes to this junction point, some fraction will go through R2. Let’s call that one i2 for example. Some fraction will go through R3 as i3, and then the remaining fraction will go through R4 as i4. So knowing the potential difference as well as the resistance value, we can calculate i2, i3, and i4.
Again, by recalling the Ohm’s law, the current will be equal to V over R, so i2 is going to be equal to V2 over R2, but V2 is equal to Veq1, and substituting the numerical values, Veq1 is 2 volts, divided by R2, and R2’s numerical value was 6 ohms and that is going to give us 1 over 3 amperes.
i3 will be equal to V3 over R3, which is going to be equal to Veq1 over R3, because the potential difference across R3 is equal to Veq1. Again, Veq1 is 2 volts and R3 was 12 ohms. This is going to be equal to 1 over 6 amperes.
Finally, i4 will be equal to V4 over R4, which is going to be equal to Veq1 over R4. That will give us 2 volts divided by R4, and R4 was equal to 4 ohms. This will give us 1 over 2 amperes. So these are the currents flowing through R2, R3, and R4.
Knowing these current values, we can figure out the power dissipated through these resistors and therefore P2 will be equal to i22 times R2 over 32 times R2 and R2 was 6 ohms, which is going to be equal to 1 over 9 times 6. Again, we can simplify this as 2 and 3 here, therefore the answer is going to be 2 over 3 watts. This is the amount of power dissipated through resistance R2.
Similarly, P3 will be equal to i3 squared times R3 and i3 is 1 over 6, so 1 over 6 times R3, and that is 12 ohms, which is going to be equal to 1 over 36 times 12. Again, we can make a simplification over here, which will give us just 1 over 3 watts. Finally, the power dissipated through R4 will be equal to i42 times R4 and i4 was 1 over 2 amperes, so we’ll take the square of that, times R4 is 4 ohms, which is going to be equal to 1 over 4 times 4 and that will just give us simply 1 watt.
Now, in doing these calculations, therefore, we have determined how much power is dissipated through each one of these resistances. Let’s calculate the total power dissipated. It’s going to be equal to the sum of the power dissipated through each resistance, so P1 plus P2 plus P3 plus P4 plus P5 plus P6 plus P7. That is equal to, through P1, we have found 2 watts, plus through P2, 2 over 3 watts, through P3, 1 over 3 watts, and through P4, we have found 1 watt, and through P5, I actually skipped P5 and denoted this one as R6, so let’s erase P5 from our equations. Turns out to be that I skipped one of these numerically. Okay.
Through P6, we have found that no current is flowing through R6, therefore no power dissipation, since i is 0 through P6 and that was 0. Through P7, we have found 3 watts of power is dissipated. So the total power dissipated is going to be equal to, here we will have 2 over 3 plus 1 over 3, will give us 3 over 3, so that’s going to make 1. 1 plus 1 is 2, plus 3 is 5, plus 2 is 7 watts. 7 watts of power is dissipated. In other words, this much energy is converted into heat in every second.
Let’s also calculate the power supplied. Power supplied, by definition, is equal to the amount of current drawn from the power supply, and that is i, times the potential difference between the terminals of the power supply, and that is ε volts. As you recall, we have found that the current drawn from the electromotive force was 1 amperes, and the potential difference between the terminals of the power supply was 7 volts. That will be equal to, therefore, 7 watts.
This verifies that our calculation is correct, and when we compare power dissipated with the power supplied, we see that these two quantities are equal to one another. That’s a good checkpoint, because we know from the conservation of energy principle, these two quantities have to be equal to one another. If they were not equal, then we were supposed to go back and check our numerical calculations and see where the mistake was. Therefore, this is a good checkpoint whenever you are dealing with these types of circuit problems.