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# KVL in the frequency domain

## Video transcript

as we do AC analysis and we do operations in the in the frequency domain we need to bring along Kirchhoff's laws so that we can we can make sense of circuits so in this video I'm gonna basically show that Kirchhoff's voltage law works in the frequency domain and what I have here is a circuit that has some voltage source an AC voltage source let's put AC on it like that and it has three impedances connected inside each of these boxes is an R and L or a C and I we're not going to show which because we're going to carry these along just as general impedances so in AC analysis the voltages are all cosine waves so V N equals some voltage amplitude times coasts of Omega T plus we'll call it we'll call it V zero some starting phase shift so this is our input signal now let me label the voltages on everything else we'll call this v1 and well I'm going to label it this way here and this will be v2 and I'll put it upside down like that Plus and this will be v3 minus plus v3 and OH now that I have v1 here let's let's change the name of this to v-0 just so I don't get I and one mixed up so we'll change that to zero so the input source is v-0 and that's that voltage now when we apply KVL to this Kirk causes voltage law what it says is if we start in a corner if we start somewhere in the circuit let's start right here and go around the loop it should add up to zero volts that's KVL for normal DC Circuit's and we're going to see how that applies to AC circuits here so in time domain we say that v-0 plus v1 plus v2 plus v3 equals zero so let's talk about how this is going to turn out well what do we know right now well we know that v-0 is a cosine wave at some some phase angle now we know about the other voltages in this in AC analysis what we're doing is we're looking for a forced response so we've let the natural responses to die out there's no switch in this circuit and we just assumed this circuit has been in this state forever so the natural response the natural response has died out and that means we're looking for the forced response so what we know is we have three voltages we have three voltages we know that all these voltages are going to resemble the input voltage so they're all going to be sinusoidal all the voltages here are going to be AC sinusoids because the forcing function is a sinusoidal are gonna all have the same Omega the frequency of this voltage and this voltage in this voltage is going to be identical to Omega here I'm going to put a big bang here that's really important in an AC circuit when you're driving it from a frequency every other frequency in the system is the same frequency this is a linear system and linear components if we all the analysis we've done linear components don't create new frequencies they're all Omega now some other things we know there's going to be phase shifts involved here remember when we do impedance we are multiplying in by J and rotating things by 90 degrees so we're gonna have different different fee for each one and the other thing we're going to have is we're going to have different the amplitudes of our sinusoids are going to be different the amplitude of v1 it could be different than the amplitude of v2 so this is what an AC solution is going to look like let's let's move on a little further here what I'm going to do now is we're gonna take this input voltage plus these things that we know here and we're gonna see how Kirchhoff's voltage law works in the frequency domain when we work with these transformed Z's these impedances okay let's go ahead and do that okay let's do a little more in the time domain and we'll write out our KVL equation again so the KVL equation was V naught cosine Omega T plus V 0 plus V 1 that's the amplitude of V 1 cosine Omega T plus some different phase angle we don't know what that is yet plus V to my a platoon of V 2 cosine Omega T plus V 3 plus v3 amplitude of V 3 cosine Omega T plus V 3 all equals 0 and Omega all these omegas are the same exact number the same Radian frequency all the fees are different and all the v 2s and v 3s are different ok now I'm going to switch to complex exponential notation we're just changing notation here we could represent this number here as this is the real part of V naught V 0 e to the J times Omega T plus V 0 that's exactly the same as this this cosine can be represented as the real part of a complex exponential with this frequency and I can write out the rest of these V 1 e to the J Omega T plus V 1 plus the real part of V 2 e to the J Omega T plus V 2 parentheses plus V 3 oops real part of V 3 e to the J Omega T plus V 3 equals 0 all right now one thing I can do next we can start to factor this we can start to take this apart a little bit so I know that if I have the expression if I have the expression e to the J Omega T plus fee just in general I can change that but just by exponent properties to e to the J fee let's cut the parentheses in there like that e to the J fee times e to the J Omega T so I'm going to do this transformation on all four of these terms here let's keep going so we're still working on this let's go the real part now I'm going to take apart Omega T and V zero here and I get V naught e to the J V 0 e to the J Omega T plus real part V 1 e to the J V 1 e to the J Omega T plus real part V 2 e to the J V 2 times e to the J Omega T plus real part V 3 e to the J fee 3 e to the J Omega T all equals 0 and here's a nice simplification we take out this common term we're going to factor out this common term across the entire equation and what do we come up with we come up with the result is the real part of V 0 e to the J V 0 plus V 1 e to the J V 1 plus V 2 e to the J P to see the pattern all that times e to the J Omega T and we closed that and that equals zero and we're getting close we're getting close all right how do we make this equation zero does e to the J Omega T ever become zero well e to the J Omega T e to the J Omega T is a rotating vector it's never zero so that's not going to do it so how do we get it well that means that this other term here has to be equal to zero so how am I going to do that let's I'm going to make one more notational change this kind of a number here is is called a phasor it's some amplitude times e to a complex one angle and there's no time up here there's no time the time is only over here this is the only place that time appears in the equation and this is the only place that Omega appears in the equation and these are just phase angles these are starting phase angles so my notation for a phasor is going to be this is going to be called I'm going to call it v-0 and I'm going to put a line over it to indicate that it's a complex vector and that equals V naught e to the J V naught so when you see the vector symbol and the aught that's that right there and we can right now finally the real part of V naught plus v1 phaser + v2 phaser + v3 phaser equals zero so this is KVL in the frequency domain and fortunately it looks like it looks exactly like KVL that we remember from our DC analysis the sum of the voltages going around the loop is equal to zero and in this case it's the sum of the phasers going around the loop is equal to zero