00:01
So i have the diagram drawn here already, and then let me write out the equation that will guide the principle we're going to use to solve this.
00:07
So recall that we have that l, the i, the t, plus ri plus 1 over cq equal to e of t.
00:22
Now, this is the equation we're going to be using.
00:24
And then also recall that l as units of h.
00:32
Right and then r as units of oms sorry that's that's a mistake and then we also have that so this as units of oms and then we have that c as units of farad so these are the guiding principles that will guide us now so this is the diagram i've not changed anything so we have that this is going this way this is going this way.
01:02
So this tells me that this arrow would just also be going this way.
01:06
So it's going anti -clockwise.
01:09
Similarly in this place, this goes this way and then this will be going this way.
01:14
I should have this arrow going this way and then i have this arrow going this way.
01:18
So everything is going towards this.
01:22
Now, i'm going to do something and i'm going to name this.
01:25
Let's name this loop one, loop one, because it's going to counter.
01:32
Clockwise let's name this loop two and now let's name the outer one the one that goes this way let's name this loop three so i'm trying to get that so let's name that loop three now let's see how we can go about solving this now let's start solving this so for loop one i have that using this equation so i have that my l so l will be 10 because it has the same units of um eris as this so i would have 10 multiplied by what current is flowing through 10 that is i 1 so this matches up with this so that'll be 10 the i 1 the t plus um i have a i also have a 10 times i 3 because this is oms so this matches with r so i have 10 times i 3 i actually don't have a cue because there's nothing that looks like a farad in this place so this would be equal to recall that i forgot to put this that e of t as unit of voltage so this would just be 20 because everything is going towards 20 so that's loop 1 now for loop 2 for loop 2 i'm going to add that's do i have an l yes i have an l which is 30 so i have a 30 and what current takes goes there that's i2 because this is parallel today so 10 30 the i2 the t so problem just just note that so this is i1 and this is i2 that's what flows in this and so that's 30 the i2 do i have an r yes, in short, i have two r's in this loop 2.
03:46
So i'll have plus 40 times the current flowing here is also i2.
03:56
Right.
03:56
So i'll be 40 i2 because this goes this way, i2.
04:02
And then i have also a 10 multiplied by i3.
04:06
But let's look at the direction.
04:07
So this is going this direction, going this direction, going this direction.
04:10
But when you come here, this i -3 is going in an opposite direction.
04:15
So instead of saying plus 10 i3, it would be a minus 10 i3.
04:23
Now, in this loop 2, do i have a voltage? no, i don't have a voltage in this closed loop.
04:30
So this would be equal to zero.
04:33
Good.
04:34
Now, let's look at for loop 3, for loop 3, what do we have? we have a um so we have do we have something like an like h yes we have two because looking at the outside loop i have 30 h i have 10 h now 30 is the i 2 d t 10 is the i 1d t and then in short i even have a 40 r so this is 40 times i 2 then equal to 20 voltage so that will be that would be a long equation.
05:12
So that will be 10, the i1, the t, plus 30, the i2, the t, then plus 40 is my r.
05:28
So that's 40 i2 equal to 20.
05:33
So let me just re -explain that.
05:35
So i have 10 ampers.
05:38
So this is an h, this is an h.
05:40
And recall that l always multiplies the derivative of the cost.
05:45
So this will be 10 the i1 the t plus 30 the i2 the t then i also have a 40 times i 2 because this is ohms which is r so that is 40 times i 2 and then i have everything is going towards 20 volts so that's my 20 et so now let's um so that's why i have this equation so now let's also look at um so let me name this equation, equation three.
06:21
Let me name this equation equation one.
06:25
And let me name this equation, equation two.
06:29
Now, let me, i want to do that.
06:31
I want to explain that thing again.
06:33
Let me name this point, point a, and then let's name this point b.
06:39
Now, if you look at point a, what is going into point a? i3 is going into point a, i3 is going into point a, whereas i, 1 is living point a.
06:50
So at point a, we'll have that i2 plus i3 because that is going into point a, then minus i1 because i1 is living.
06:59
So that's, so that implies that at point a, i have that i2 is positive, i3 is positive, then i1 is living and that will be equal to zero.
07:17
At point b let's see what is happening in point b at point b we have i2 is living i1 is going inside i3 is living so only i1 is positive i2 and i3 will be negative so i have i1 minus i2 minus i3 will be equal to zero now if you look at this together this is still saying just that if you look at this equation 1, it's still the same thing as this is still equivalent to move everything to the right -hand side.
07:55
You have i1 minus i2, minus i3, equal to 0.
08:01
So you see that at point a and point b, they still give the same expression.
08:06
So in both cases, at both points, or in both points, in both cases, we have that i1, minus i2 minus i3 equal to zero now let me name this equation star now let's assemble let's assemble equations 1 2 3 into a system and let's take uh when we have let's take d to be d d t so we'll replace it by d now, that implies that if you go back to equation one, equation one is, that'll be 10 d i1 plus 10 i3 equals to 20.
09:01
So i'm just going to write that down.
09:03
So that will be 10 d of i1 plus 10 i3 equal to 20.
09:14
And then we have that 30 d i2 plus 40 i2, minus 40.
09:24
Minus 10 i3 equal to 0.
09:30
And then lastly, that equation three gives us that 10 the i1 plus 30 the i2 plus 40 i2 equal to 20.
09:47
Now let's rewrite it as, we write as, let's go back to equation one.
09:55
For this, i can divide side by 10 also divide both sides by 10 and then so what we are doing is write as uh write by let me use it by dividing by 10 dividing by 10 so this would give us that the i 1 plus i 3 equal to 2 and then 3d plus 4 multiply by i2 minus i3 equal to zero and then the i1 plus 3d plus 4 i2 equal to 2 all i did was just divide both sides of those three previous equations by 10 now um let me name this equation equation um let's say 1a let's say 1b let's say 1c so that implies that now let's go back to equation star so apply from you call that from star let's go back to star which is this so i have i1 minus i 2 minus i 3 equals to 0 apply d to both sides of star right so that implies that we have that the i 1 minus d i minus the i3 equal to if you differentiate zero you have zero recall that d is equal to d t so this implies that i can write this as the i1 equal to the i2 plus the i3 now substitute this into equation okay let me let me name this equations as yeah into equation 1c that doesn't matter so sub this into 1c for the i for the i for the i one so what i'm trying to say is that this di i1 replace it yeah so that implies that this c equation 1c becomes my di i 1 is not the i 2 plus the i 3 plus 3d plus 4 i2 equal to 2 so if you expand this so this would be 3d i2 plus the i2 so that will be 4 d i2 plus the i 3 plus 4 i2 equal to 2 so i can pull together the i 3 plus 4 i 2 equal to 2 so i 2 so i can pull together the i so that'll be 4 d plus 4 into bracket i 2 plus d i 3 equal to 2 now let me name this equation equation 4 don't mind um sorry let me name it equation 5 just to follow numbers let me name this equation 5 so this implies so now let's add apply d let's apply d to to 1b.
13:57
Now what is 1b? 1b is d.
13:59
So let's apply d to both of this side.
14:02
So you're going to have that we're going to have apply d.
14:06
So this is d.
14:08
You're going to have 3d squared plus 4d of i2 minus d i3 equal to 0.
14:23
Let me name this equation 6.
14:25
So all we're just trying to do is just a so now i have i2 and i3 together and i can add equation 5 and equation 6.
14:35
So add 5 and 6.
14:40
Right.
14:41
So when i add 5 and 6, what i'm going to have is 4d plus 3d squared plus 4 d into brackets of i2 will be equal to zero because what i will have is that the i3 plus minus the i think that will be equal to zero and then two plus zero that will be equals to two and then when i add this together i'm going to have the left hand side so actually this should actually be equal to two there's a mid a mistake here so this should be equal to two now let's see if you can add this together so we'll add that this would be 3d squared plus 8d plus 4 of i2 equal to so now the goal of all i did so far let me just explain that to you starting from equation 1a 1b 1c is to have an equation in just one of the is maybe i 1 i 1 2 or i 3 so what i what i what i systematically did in all these was to make sure that i have the equation just in terms of i 2 and and that's why when i started here, i applied d to both sides of star, i substituted that into these so that i can have i2 and i3 and then add up those equations, which is what i did in this place.
16:17
So i've been able to achieve my goal of having everything in terms of i2.
16:21
So what i do is now that the homogeneous equation, the homogeneous equation of the above, it has the auxiliary form.
16:33
So this is, to recall that d is, let me just write it down so that you don't forget, recall that d equals to d d d, t.
16:46
So this would be like three d squared c, d squared, i, dt squared...