Peter Wheeler: Welcome back.
What we're going to look at now is, again, a continuation on with this concept of series
and parallel circuits.
I've got a circuit here where I've got two voltages now, two voltage supplies, and again,
a network of resistors.
The question starts to become, this is interesting.
First of all, is this unusual?
No.
There are times when people do hook up batteries in order to get more current out of the two
combined, but the voltages may not be the same.
Being the technologists that we are, we need to do some calculations to determine just
exactly how much current it is.
Another analogy is if you look at the electrical network of, say, Ontario, you could have Bruce
Generating Station and Darlington Generating Station, and maybe Pickering ... They're all
generating voltages.
When you start to do a grid analysis, then you have to see the contribution from each
one.
We're going to ask the question, what is the current in R3 equal to from these two sources?
What current are we seeing in there?
How do we go about doing that?
The first thing we want to do is, in this analysis, in super position, is we need to
look and see what is the contribution from each source by itself.
We're going to break our problem down and check one out.
The 10 volt source, for example.
Then check the other one out, the five volt, without interference of one with the other.
When we're doing the 10 volt, there won't be a five volt source, and when we're the
five volt, there won't be a 10 volt source.
What are we going to do with those sources if we're not going to consider them?
Well, as per Thevenin, in order to do an analysis of a circuit, we're going to short them.
In super position, we will short them.
In this case here, I'm going to draw our 10 volt contribution.
As you can see, R2 is in parallel now with R3 as a result of shorting the five volts.
Okay, so we've got, in this case here, 200 ohms, in parallel with 50 ohms, and we're
going to add the 100 ohms, which is in series.
The first rule for parallel is that our result has to be less than 50.
I'm going to use the R1 times R2, over R1 plus R2 formula, to get the total resistance
of the two, in parallel, of the two in parallel, which is basically two and three.
Okay?
That's going to be 200 times 50, over 200 plus 50.
25 into this is 40 ohms.
The two in parallel is 40 ohms, and the total resistance of one, two, and three, is equal
to 40 plus 100, 140 ohms.
Okay?
140 ohms.
If that is my total resistance, and Ohm's law said, that voltage equals current times
resistance, and I've got, in this case here, as far as voltage goes, let's say ... 10 volts
we said?
Yeah, 10 volts.
Then I have 10 volts over 140 ohms.
Trying to think, in my head, what would that be?
Actually, what I'm going to do is something different.
I'm going to say what is the voltage across here?
If that is the case, I'm going to use the resistor divider rule that we just had.
I'm going to have 40, over 140, times 10, to figure out what the voltage is, over here.
Once I've got the voltage over here, I can just divide 50 into it, and I'll have the
current.
In this case, here, I've got 14 into 40, what is that, 2.86 volts?
I think so.
I would have 2.86 volts, across 50 ohms, and that will give me my current.
Should I leave that with you?
Okay.
40 over 140 times 10, will give me my voltage, 2.86 volts.
Divided by 50 ohms, will give my IR3 due to the 10 volts, okay?
That's one.
Let's just stop there for a second, and I'm going to show you, now, the contribution from
five volts.
Okay, so we calculated two resistors in parallel, plus one in series, total current, and then
current, that being the case, we got a voltage drop.
We can do it that way, or we can take this as a percentage of the total, come up with
the voltage at this point.
If we've got a V and we've got an R, then basically I is equal to V over R.
The contribution, in this direction, since the current is going clockwise, is 0.057,
from the 10 volt source.
We have to do the same thing from this side, now.
We will end up doing a five volt battery, into a 200 ohm resistor, shorting out the
10 volt battery.
We've got 50, we've got 200, we've got 100, and we've got five volts.
Now, two in parallel.
Let's use R1, R2, over the sum.
50 times 100 over 150 basically is equal to 5,000 divided by 150, is equal to 33.3, if
you want, three ohms.
Make sense?
Two resistors in parallel result in a resistance lower than that.
We have to add to that 200.
Our total will be ... our total, as seen by the five volts, is going to be 233.33 ohms.
Two in parallel, one in series.
Okay, so now that we've got that, if we said, for example, we can do something like, I need
to know, what is the voltage across this resistance here?
This so called 33.33 ohms package, as opposed to the total voltage, so that I can do a calculation.
Once I've got a voltage, here, I can calculate the current through there.
We can do that, if we do, for example, a voltage divider.
We had 33.3 ohms ... ohms.
We divided it by the total, which is 233.33, okay?
If we multiply that by five volts, we'll get basically ... what are we going to get here?
0.714 volts, okay?
0.714 volts is sitting here, and if we divide 0.714 volts by 50, we're going to see what
the current contribution is from that.
It's going to be on the order of 0.014, and you need to check my math, because I'm kind
of trying to do some of this in my head, three amps.
This is the contribution from the five volts, and this is the contribution from the 10 volts.
We've got ... let's just do it in milli-amps.
14.3 milli-amps.
We've got 57 milli-amps.
So our total current is 71.3 milli-amps, as a result of the two sources.
Again, what we were doing was shorting voltage sources, to do them independently, and then
doing a series parallel combinations, and also checking voltage drops across various
resistors, in order to see, what's the resulting voltage across the resistor of interest, okay?
Thank you.