00:01
19 .29.
00:03
So we have this circuit here and we want to find the voltage between points a and b.
00:13
We also want to find the terminal voltage of each battery and i've sort of dotted boxed in the emf and internal resistance for each of these just to make it clearer what what those points are that we need.
00:27
So one voltage bd for one and g e for the other.
00:36
And so also this circuit is the same as one used in an example problem, 19 -8.
00:47
So if you want after this, you can go back and look at that.
00:50
If something's not clear, it may be getting the explanation for the chapter will be helpful as well.
01:01
But in any event, so we want to find the voltage between a and b, which means if we find current three here, that will give us this voltage because the current times the 47 oms here will be the voltage across this resistor, which since its terminals are at a and b will give us the voltage between a and b.
01:28
So from kirkoff's junction rule, we know that current three here has to be equal to the sum of currents 1 and 2.
01:52
So that's good.
01:54
Next, we have, we go around.
02:00
In a loop, in a couple of loops.
02:06
We have one around the entire thing here, one around the sort of top part, and one around the bottom part, and the sum of all the voltage changes around each of those loops is zero.
02:23
So for our first one, we have negative 34 oms times i, one, plus the 45 volts, from this battery and then we have a total of 48 oms of resistance here times i3 that has to be zero next we want to go around the sort of outer loop so we have again negative 34 i1 and just to save us writing all the units out that's just it's assumed that this is in oms plus 19 i2 because we're going in the opposite direction of i2 here, so that's why there's a positive sign, minus the 85 volts from the other battery is equal to 0.
04:09
So now we can find i2, from this one is 85 volts plus 34, 34 i -1, divided by 19 oms, which is 4 .474 plus, plus 1 .789 times i1.
05:45
Then we have i3.
05:53
We do the same thing, except it's with this one.
06:00
And so we find that that is 0 .938 minus 0 .708 times i1.
06:29
So from this, we know we have an expression for i3, i1 and i2 that relate them.
06:50
And we have expressions for i2 and i3 in terms of i1 so we can put them into here and solve for i1 and then find the other two currents and so doing that we find that i1 is negative 1 .011 amps just underline that so we don't lose track of it so it turns out we got the arrow for you know i1 here wrong but that's okay because when you do that you just get a negative sign and so then putting that back into our other two we get that i2 is 2 .665 amps and then i3 which is what we want for finding the potential difference in part part a they find that that is 1 .6 5, 4 amps.
08:30
So now, for part a, we said that the potential difference between points a and d, a and d.
08:53
This is going to be the current times, times the resistance between those points.
09:01
So v, d minus va...