00:01
Okay now the question here is asking to do using kvl and kcl how much power 20 volt source extracts from the circuit now for the first question let me redraw the circuit we have current source we have 16 ohm resistance this one is 5 amperes here we have one ohm connected to 20 volts and a 3 ohm right here now when we are doing kcl or kvl it's always best to look at the circuit and find out which one is the best case now just by looking at it i can already see that here well i'm going to choose one of the we are always free to choose where our ground is everything is relative of course so for that i'm going to choose this node right here and ground it this is our ground this is 0 volts right here that means that i know this node right here which is 20 volts and i have two more nodes that i don't know so i can use kcl and calculate those two nodes one is this one and one is this one right here let's name them va and vb so we can go on and basically use the kvl so when you're writing kvl there kcl there are a bunch of different um basically you can always do the currents flowing into a node or out of a node so it's just a convention whichever you prefer but what we are going to do we are are going to assume all of the currents are flowing away from the node you have this one this one and this one and we can write the equation for each of these currents basically for example the current here i1 just for demonstration i1 is equal to minus 5 amperes because we have the current flowing from here right so that using that we are going to write our kcl equations.
02:27
Alright, so for the first question, for node va, we have three general currents.
02:37
One is flowing in, as we saw.
02:39
That one, the second one from the 16 ohm right here, we can write va minus vb divided by 16 ohms.
02:53
Basically, va right here minus vb divided by 16 16 ohms and then the last one is going to be a minus 20 volts across the one ohm resistance right here divide by one equal to zero this is our first question you can always of course simplify it and basically have multiplying both of the sides by um for example 16 let's say here we will have 80 minus 80 multiplied by 16 plus va minus vb plus 16 va, 320.
03:52
And then we can simplify it further.
03:56
We will have va plus 15 vb is basically equal to 320 and 8400.
04:07
400 we move it to the other side to have it positive.
04:10
This is our equation 1 so for vb we are going to do the same scenario we know we can write in order this current 1 2 and 3 and basically sum them up so for the first one we have vb minus 0 divided by 3 ohms plus vb minus va divided by 16 ohms and plus 5 is equal to 0.
04:48
Now we can do the same thing.
04:50
We can multiply both sides of these equations by 48 basically and then we are going to have this is just equal to 16 times 3 i just don't want to get rid of all of the fractions to make life easy now we're going to have 16 times vb minus 0 of course plus 3 times vb minus 3 times va vba plus 5 times 48, which is 240, equal to 0, right? if just you're wondering how we got that, 48 divided by 3 times vb minus 0 is that number, 16.
05:45
This is 16.
05:47
All right.
05:48
Further simplification.
05:49
You have 19 vb minus 3 va plus 240 equal to 0 right now this is our second equation we can use basically the first equation right here we have va equal to minus 15 vb plus 400.
06:30
You can use, i name it equation 3.
06:34
Inserting equation 3 into equation 2, we're going to have 19 vb minus 3 times minus 15 vb plus 400 plus 240 equals 0.
06:51
Then for your simplifications, oops plus 45 vb that's 1200 plus 240 equals zero now um this means that 64 vb is equal to simplifying it simply 960 then we have our vb right here which is 15 volts we can calculate va2 using equation 3 we know what vb is we have va equal to to minus 15 times 15 plus 400, which is 175 volts.
08:13
This is our va.
08:15
Now, we have these two numbers, and we can calculate basically any current or any power.
08:21
Well, we only need this current right here.
08:25
Let me use another color so that it is obvious.
08:28
This is i and this is the current that is flowing through our 20 volt source and the question asked what is the power which is the power of 20 volt source is going to be because the current is flowing in this direction that i showed here and we are assuming that it doesn't have to it can be just a negative or positive power but the way the problem suggests is that extracts power which which means that it loses power, and as a result it's going to be 20 times i.
08:58
So we need to calculate i, and we are done for this part.
09:02
To calculate i, i should be equal to va minus 20 divided by 1 over i across this one.
09:15
It's just a simple kcl kind of a thing to calculate that.
09:18
And we know what our va is.
09:20
I is going to be 175 volts minus 20 divided by 1 and this gives 155 peers now the power is simply v 20 volt times i just 20 times 155 and this is equal to to 3100 watts this is basically the power that um is extracted basically from the circuit now let's move on to the next part part two just doing a sanity check everything is in order now we have um a circuit of this form we have the voltage source 256 volts volts then we have 20 ohm connected like that it's going to be very similar to what we did i think based on judging by how the circuit looks 16 ohms 240 and then we have the last part and here 128 volts and done and we have a bunch of currents um it is told what they are it doesn't matter when you solve it you can assume all of these currents in any direction but based on how it is done means that the currents will flow most probably as positive numbers in this direction so okay now we are asked to using kvl and kcl just basically find what are these currents now we are going to do the same thing here i'm going to choose this node because it's a large node all of them as ground v equal to zero i know what that is i know this node right here is basically if it is 128 volts less than ground which i can just use one negative 128 volts it's just a number it doesn't matter voltage is just difference matters and we have two nodes that we don't know what the voltages are this guy and this guy and we are going to use in kcl again doing this repeating the same things we are going to find them so kcl for node va we are going to have i forgot to say well this is going to be of course 256 volts it is from ground from zero it's 256 more so back to kcl let's start with va across 20 ohm what is the current flowing out of va yeah i just do that sum them up and equate them to zero like the previous one 256 is the voltage drop across 20 ohm resistance plus now let's look at a branch that has ib right here it's gonna be va minus zero i'm not writing divided by 240 and then the last node va minus vb divided by 16 is equal to zero let's simplify this you can can multiply both sides by 240 then you have okay 240 divided by 20 is going to be 12 12 va and 12 times 256 is going to be 3072 that's that one plus va this one times the second term times 240 is that 240 divided by 16 is going to be 15 plus 15 va minus 15 vb equal to 0.
14:58
Now we can do some simplifications here you can see the va is right here we have a total of 16 plus 12 is going to be 28 ea when's 15 maybe equal to 3072 and of course you can choose any of them and replace in the other equation and solve for basically all of them now what i'm going to do maybe i can find va in terms of vb which is going to be va is equal to 1 over 28 times taking everything to the same side plus 15 times vb let's do one more step here just to make sure we have better numbers which is not in this question unfortunately numbers are rather gigantic 15 divided by 28 0 .5357 vb now this is our equation 1 let's move to vb again from the circuit i'm going to use first let's start with 40 ohm resistance right here vb minus minus 128 divided by 40 is gonna be my first sorry this is vb the second one across 320 oh right here along id i i of d is going to be vb minus 0 of course divided by 320 oh plus can write it right here if you want to and then the last one, vb minus va divided by 20, which is right across this 20 ohm.
17:01
Again, i'm writing the opposite of ic because it's just a matter of convention.
17:07
It doesn't matter really.
17:08
We are going to calculate all of them.
17:10
We just need these two voltages.
17:12
This is equal to 0.
17:14
Now i do some simplification here, multiplying both sides, all of the sides by 320.
17:20
320 divided by 40 is going to be 8.
17:25
8 times vb minus minus is plus.
17:30
8 times 128 is going to be 1024 plus vb.
17:39
Plus 320 divided by 20 is going to be 16.
17:44
16 vb minus 16 va equal to 0.
17:48
Further simplification.
17:49
We have the vb terms right here, right here and right here which is going to be a total of 25 vb plus minus, actually 16 times va.
18:05
Now i can insert va from equation 1 right here this guy, 109 .7143 plus 0 .5357 plus 1024 24 equal to 0 now i can further simplify this 16 times 109 point 7143 is going to be 1 ,755 .4 minus minus 16 minus 16 right here times this guy so minus 16 times 0 .5357 it's gonna be 8 .5712 vb plus 1024 equal to 0 again simplification 25 vb minus 8 .5712 is going to give us 16 .4288 vb.
19:19
And then for the rest, we have 1024, sorry, 1024 minus 1755 .4 is minus 731 .4 equal to 0.
19:44
Then from here we can just simply write 731 .4 divided by 16 .4288 is going to be our vb.
19:54
We can calculate vb as 44, roughly, 94 volts.
20:08
This is our vb.
20:10
And now using this one and inserting that to our equation 1 inserting 3 to equation 1 gives va is equal to 109 .7143 plus 0 .5357 times 44 .5194 here, va is 5357 plus 109 .7143.
20:56
It's going to be 1220.
21:01
Wait, i made a mistake, for sure.
21:05
Let's repeat this.
21:09
44 .5194 times 0 .5357 plus 109 .7143.
21:20
Okay, now that makes much more sense.
21:25
133 .5633 volts.
21:28
Now we are ready to calculate basically all of the currents.
21:31
Now let me bring back our circuit.
21:34
It we wrote quite a lot after that oh no here it goes i copy it all right let's calculate the currents we have the voltages we can write them right here va is calculated to be 133 .5633 volts volts, vpe is 44 .5194 volts.
22:19
Now, ia.
22:22
So, ia is going to be 200, this voltage right here, 256 minus va divided by 20 ohms.
22:31
Simple.
22:31
The voltage drop divided by the resistance across that is going to give us ia...