00:01
In this question we are going to find the power delivered to each resistor in the circuit shown.
00:11
To do that we're going to need to know a couple of things.
00:15
We need to know, well we'll start with this.
00:20
So to calculate power delivered to a resistor we can do that with the current through it multiplied by its voltage or we can do that with the current through it squared multiplied by its resistance or we can do it with the voltage across the resistor squared divided by its resistance.
00:39
And given that we would need to apply kirchhoff's rules in order to find current to get voltage.
00:50
Well we're going to make this a little bit easier for ourselves and we are going to simply find the current and then we already have the resistance for each one and then we don't have to go through and calculate the voltage as well.
01:04
So we are going to form three loops and then there are going to be only two junctions that are critical pieces here.
01:17
So i like color coding things so enjoy that with me.
01:23
I'm going to take as my first loop this left side loop going clockwise and i'm going to call it i1 going through the 2 ohm resistor.
01:43
And then we are going to have a junction right here.
01:48
I'm going to call that point a.
01:50
I'm going to call its friend down here point b that's going to be the same current involved.
01:57
And what we're going to say is that we have i2 going down from that junction but then we're going to have i3 heading out from that junction into that middle section of the circuit.
02:20
And then you can see at b we will end up having i3 plus i2 coming in and then i1 going out.
02:28
So we just generated two really important things.
02:31
First of all we can see that i1 will be equal to i2 plus i3 for our junction at a which we may want to rewrite i1 minus i2 minus i3 equals zero.
02:51
And then we can also generate our loop rule.
02:55
So i am going to write this sum of potential rises equals sum of potential drops.
03:03
Because what i'm ultimately going to do is set up a couple of matrices to solve this system of equations.
03:13
So 50 volts is our potential rise.
03:15
It's our only potential rise.
03:16
And that's going to be equal to our potential drop of i2 multiplied by 2 ohms plus, sorry i1 multiplied by 2 ohms.
03:26
Got ahead of myself there.
03:31
Plus i2 multiplied by 4 ohms.
03:34
And then there's no actual current at, sorry, no resistor that i3 goes through so we won't have an i3 term.
03:48
And next we can take the loop that starts with the 50 and goes through the 2 and then we're going to look at continuing on i3 after, i drew that just a little bit too far, continuing on i3 after junction a and then we're going to come down through the other 4 ohms and then we're going to come back to the left and back to that 50 volt battery.
04:19
And so you can already see we have i3 but we're going to note, i'll use yellow, i don't get to use yellow very often.
04:29
So i'm going to call this junction c and then down here it's friend close to the 2 is junction d.
04:36
We're going to say we have i3 coming into that.
04:40
With i4 going down through the 4 ohms and then i5 being the one that completes that outside loop through the 20 volts and the 2 ohms.
04:52
And this gives us another junction rule that i3 equals i4 plus i5 rather.
05:04
And so we could also write that i3 minus i4 minus i5 equals 0.
05:11
And yes, we are ultimately going to combine those two junction rules into a single junction rule because we'll substitute for i3 into the i1 and i2 equation.
05:26
So let's come back to our loop.
05:28
So we know that we're going to go clockwise around this loop.
05:33
And again, potential rise is going to be the 50 volts and then i'm going to have a potential drop of i1 through the 2 ohms.
05:48
And since there's no i3 term, i'm just going to skip like there's no never a resistor for current 3 to pass through.
05:55
So we're just going to skip right to current i4 passing through the 4 ohms and that's it.
06:03
And the reason i'm spacing it like this is because that will help me visually align which terms i'm going to set up for this matrix.
06:11
There's no i4 in that first equation, so i'll have plus 0 i4 is something that we'll end up filling in.
06:22
Ok, and now at last, we're going to take our 50 volts across through the 2 ohms, across into i3, and then we across through junction c to become current i5, which we're going to take down through the 20 volts and then across through that 2 ohms and then we will complete our loop back through some plain wires to that 50 volt battery.
06:57
So we have no new junctions, but now we're going to have a 50 volt potential rise and we are going through the 20 volt battery in the correct direction as well, like from negative to positive.
07:13
And so our, actually i want to keep my equal signs lined up so that all of my current terms lined up.
07:23
So i'm going to say 50 plus 20 equals, we still have that i1 times current 2.
07:29
We don't have any current 2 involved here.
07:33
We don't have, there's never a current 3, we don't have current 4, but we do have current 5 passing through our 2 ohms...