Chapter Six series and parallel circuits. There are two basic ways in which to connect more than two circuit components series and parallel. Here in this slide we have three resistors labeled r one r two and r three connected in a long chain from one terminal of the battery to the other. It should be noted that this subscript labeling those little numbers to the lower right of the letter R are unrelated to the resistor values in ohms, they are only to identify one resistor from the other. The defining characteristic of a series circuit is that there is only one path for current to flow. In this circuit current flows in a clockwise direction from point one, to two, to three to four, and back to one We say that the three resistors are connected in series.

Again we have three resistors, but this time they form more than one continuous path for electrons to flow. Each individual path through r one r two and r three is called a branch. The defining characteristic of a parallel circuit is that all components are connected between the same set of electrical electrically common points. Looking at the schematic diagram here we see that points 123 and four are all electrically common. So, our points 876 and five. Note that all resistors as well as the battery are connected between these two sets of points.

We say that these two resistors are connected in parallel. In this circuit we have two loops for electrons to flow through. Notice how both current paths growth go through resistor number one. In this configuration we say that r two and r three are in parallel with with each other, while r1 is in series with the parallel combination of r two and r three. The basic idea of a series connection is that components are connected end to end, in a line to form a single path for the current to flow, in this case from left to right, or it could be the currents going to flow from right to left. The basic idea of a parallel connection, on the other hand is that all components are connected.

Cross each other's leads in a purely parallel circuit. There are never more than two sets of electrically common points no matter how many components are connected. There are many paths for the electrons to flow, but only one voltage across the components. The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path or the current to flow in a series circuit, the rate of flow at any point in this circuit at any specific point in time must be equal. This brings us to the second principle of series circuits.

The total resistance of any series circuit is equal to the sum of the individual resistances This should make intuitive sense, the more resistors in series that the electrons must flow through, the more difficult it will be for those electrons to flow. In the example problem, we have a three k 10 k, and a five K resistor in series. In other words, it's 3000, ohms, 10,000, ohms, and 5000 ohms, all in series, giving us a total resistance of 18 K or 18,000 ohms. In essence, we've calculated the equivalent resistance of r one r two and r three combined. Knowing this, we could draw the circuit with a single equivalent resistor representing the series combination of r1 r two and r three. Now, we have all the necessary information to calculate the circuit current because we have the voltage between the two points nine volts and the resistance between the same two points 18 k ohms current is given by ohms law which is the voltage divided by the resistance or in this case nine divided by 18,000 which would give 0.0005 amps or point five milliamps since 1000 milliamps is equal to one app because the current is 500 milliamps, then the 500 milliamps will flow through each resistor and a voltage drop will apply across each resistor, the voltage drops across each resistor will be given then by ohms law for r1.

It's point five milliamps times three K, which is equal to 1.5 volts. For our two, it's point five times 10 k point five milliamps times 10 K is five volts and four r three is point five milliamps times five K, which is equal to 2.5 volts. Notice that the sum of the voltage drops 1.5 plus five plus 2.5 is equal to the battery supply voltage nine volts. This is a principle of series circus The supply voltage is equal to the sum of the individual drops around the circuit. This is known as kerchief Voltage Law, or sometimes referred to as kerchief loop or mesh rule, which states that the direct sum of electrical potential differences voltage drops around any closed network is zero. State in another way, the sum of the emfs in a closed loop is equal to the sum of the potential drops in that loop.

Notice the polarities of the resistors. They go from plus to minus, which is voltage drops 1.5 plus five plus 2.5 equals nine volts voltage drops supply voltage goes from negative to positive, which is a voltage rise, negative voltage drops, which is a negative voltage drop if you want to say it, minus nine. So if you add up the voltage drops, you get 1.5 plus five plus 2.5 minus nine is equal to zero. The first principle to understand about parallel circuits is that the voltage is equal across all components in the circuit. This is because there are only two sets of electrical common points in a parallel circuit and the voltage measure between sets of common points must always be the same at any given time. Therefore, in the circuit we have here, the voltage across R one is equal to the voltage across R two which is equal to the voltage across our three, which of course is equal to the voltage of the battery.

We can in immediately apply ohms lock of each resistor to find the current in each resistor, because we know the voltage across the resistor nine volts, and we know the resistance of each resistor, therefore the current in resistor r1 is point nine milliamps. The current in r two is 4.5 milliamps, and the current in r three is nine milliamps. At this point, we still don't know what the total current or the total resistance for this parallel circuit is. So we can't simply apply ohms law either to the total resistance. However, if we think carefully about what is happening, it should become apparent that the total current must equal the sum of the individual currents in each resistor or an each branch. If it wasn't the case, then we would have a buildup of current or charge in one part of the circuit.

And that isn't happening, we have a continuous flow of current at any one time, so that the total current must equal the sum of the branch currents. Okay, let's let the voltage drop across all the resistors equal the T, which is equal to the supply voltage. And we're going to let our T equal the equivalent parallel resistance of all three resistors. We know from ohms law, that the voltage the total voltage is going to be equal to the individual voltage drops across each resistor which is I one times r one, which is equal to i two times r two which is equal to r Three times r three. We also know that the from ohms law that the supply voltage V t is equal to the total current times what would be the equivalent resistance of the parallel resistors. Or we can rewrite the equation where the total current is equal to the voltage of the supply voltage over the equivalent parallel resistance R T. We know that the sum of the currents flowing into the branch circuits, i t is equal to the sum of the branch circuits I one plus i two plus three or we can rewrite that that equation substituting a voltage over the resistance for the current terms.

So, I t becomes vt over RT which is equal to V T all over r1 plus vt all over r two, plus vt over r three. Now, if we divide both sides of the equations by Vt, then we're left with one over RT is equal to one over r1 plus one over r two plus one over r three. So we now have a method of calculating the equivalent parallel resistance. With this equation. As I said in the previous slide, this gives us a rule for calculating equivalent resistors in parallel. The, if we let the equivalent resistance of parallel resistors be r t, then one over r t is equal to one over r one plus one over r two plus one over r three This also leads us to kerchief current law, which states, the current entering any junction or node is equal to the current leaving that junction or node, a node being defined as an electrically connected point in the circuit.

Another way of stating purchase current law is all currents entering a node must sum to zero currents entering our positive currents leaving our negative. This is another way of looking at the same electrical circuit. In this case, we've brought the node to one common point. And it's easier to see that current shops current law holds true that the current entering the node is equal to the current world. Leaving the node or the sum of the currents entering a node is zero. So, now with the knowledge learned we can reduce any mesh network of resistors to the equivalent of one resistor as seen by the power supply.

For example, in this circuit r one r two r three and r four can be replaced by one resistor R EQ, that will draw the same current as r one r two r three and r four. We start by calculating the single resistor equivalent for r one and r two in parallel and r three and r four in parallel, which results in r one r two in parallel giving 71.429 ohms and r three and r four in parallel gives 127.27 ohms We can then add these two resistances because they are in series and they will give us 198.70 ohms, which is the required single resistor, we wanted to calculate, which will draw the same current as r one, r two, r three and r four all together. As I said before, any mesh network of resistors can be reduced to one single resistor as in this case, and this one looks a little bit more complex, which it is because there's more resistors involved.

And it's drawn in such a way that it's not intuitively obvious how we would reduce this. So sometimes it's easier to redraw the circuit so that the parallel series combinations will jump out At you a little bit easier, then you can start to reduce this in steps. For example, r three and r two are in parallel, they can be reduced to one resistor of r two in parallel with r three, it is now a read one resistor in series with our four. So that can be reduced by the parallel combination plus our for each series, we've now replaced those resistors by one resistor, and we can see that that equivalent resistance is now in parallel with R five. So we can reduce that to one resistor, which is r five in parallel with a combination r two in parallel with r three in series with r four, which gives us one resistor.

And that resistor is now in series with our seven so that can be replaced by one resistor, which is the equation have our seven in series with our five in parallel with a combination are two in parallel with r three plus r four, that is now in parallel with R six and can be replaced with one resistor and I'm not going to go through repeating it all again, we now have that resistance as a single resistor. Now in series with r1, which can be replaced now by one resistor, and that resistor is now what we're trying to achieve in the first place and we can calculate the current that the power supply supplies to that single resistor. This ends chapter six