Video Lectures created by Tim Feiegenbaum at North Seattle Community College.
We're continuing in section 7-3, and we're considering series and parallel connections. In this section, we will be doing some calculations. As with other components, capacitors may be connected in series or in parallel. Here, we have series capacitances. When connected in a series the total capacitance is less than the value of any of the individual capacitances. Here, we have CT for total capacitance and we're going to see that it's one over the quantity of one over C1 plus one over C2 plus one over C3 and on and on. Now, this looks very similar to formulas we have used previously. When we looked at resistance in parallel this was one over R. When we looked at inductances it was one over L. This is capacitance and this is not parallel. When we looked at resistance and inductance it was in parallel. This is in series.
Let's look a circuit. Here, we have a circuit with three one microfarad capacitors and the plate separation accounts for the difference in our calculations. Now, remember that when we said from our earlier simulations we saw that as plate separation increased, capacitance actually decreased. Now when a series circuit we have the plate separation of this particular capacitor, but the circuit as current attempts to move in this circuit it doesn't only see that one it see this separation and it sees this separation. It sees all three plate separations together, so that is going to result in a capacitance that's smaller than any of the three.
If we did the calculation, let's bring up our calculator and let's do the calculations. We would say one exponent minus six and that is effectively one microfarad, and we're going to say one over that value. According to our formula, we're going to add all of them together and since they're all the same value we could just say times three, and there we have the combined reciprocals of the three capacitances, and then to complete the formula we needed to take the reciprocal of that value, so one over that. Here, we come up with a value 333 but notice it is to the minus ninth which is nano. Let's note that value, 333.3 to the exponent minus nine. That is nanofarads.
Now that is going to be the equivalent of 0.333 exponents to the minus sixth which is one-third of this value. Using our formula we see that the capacitance that is experienced in this circuit is smaller than the smallest value.
When capacitors are connected in parallel the total capacitance is equal to the sum of the individual capacitances. Here, we have a circuit connected in parallel, and this will be really simple since I have got them all the same size. If you had them in different sizes, but you'd just simply add them up. The reason that we're going to do this is that the plate area is effectively increased. Remember, in a simulation as the plate area increased the capacitance also increased. For purposes for current flowing in this circuit, it is seeing the plate area not of one capacitor but of three capacitors. The calculation, in this case, would just be one plus one plus one and the capacitance total would be three microfarads.
Series capacitive reactance is the opposition to AC offered by a capacitor. We talked about that in a previous discussion. Remember this is measured in ohms. Capacitive reactance in a series circuit is found by summing the individual reactances. Remember, this is resistance, and so in a series circuit if we have resistances we just add them up. We'll go back to this circuit again.
Let's calculate the reactance of the individual first. If we went in and we said here we have this particular component and what is this reactance? We'd say one divided by, and let's use our bracket here, divided by the quantity of two times 3.14 times the frequency which is 60 Hz, and then take that times the capacitance which is one exponent minus six, then we would end parentheses, equals, and we have about 2.65K. We would have 2.65K, and remember that this is X of C capacitive reactance and this is measured in ohms. We would have the same here, 2.65K and 2.65K, and the total, what's the total if we said this value times three equals about 7.962. Total 7.962 would be the total X of C and this would be in ohms.
If you wanted to go into Workbench you could do this and set this simulation up. If you do that, you'll get, let's see, if you open the circuit, let's say pretend we opened the circuit here. We insert a voltmeter. This would be one way to confirm this and measured current. If you do that in Workbench the value you'll get is 1.257 milliamps. The way you would confirm this is if you took 10 volts and you divided by the current of 1.257 milliamps, remember voltage divided by current equals what resistance, and this is the value that you would get. That could be a check just to verify that this is in fact true.
Series and parallel connections. Parallel capacitive reactances are calculated according to the following, and again, this would be the same thing that you would do in a resistive circuit. Just remember that this was an R, R, R, and if kept calculating resistance would be the same thing, and this would give you R, but we're calculating capacitive reactance. Here, we have capacitive reactants in parallel and we had previously calculated the reactance of each of these components. We said that it was 2.65K, and so 2.65K, 2.65K. Again, if we brought up our calculator we could say 2.65 and exponent three. That's 2.65K and we wanted to do one over that, and then since all of these are the same value we could take that times three, and then again do the one over function and our capacitive reactance would be 883 ohms. That would be X of C, and that would be the resistance that circuit would experience. Again, you could do the same thing we did. You could insert a voltmeter in here in Workbench and find the current, and then you could voltage over this resistance and that's the current that you should get.
Once all of the capacitive reactance values are calculated for all capacitors in a given circuit, voltages and currents can be solved using the strategies presented earlier. We did quite a bit of calculating with resistors in series and parallel. Once the reactances are calculated, then you can just use all those same formulas.
This concludes section 7-3C.
Video Lectures created by Tim Fiegenbaum at North Seattle Community College.
by Jake Hertz
by Jake Hertz