 This is the 17th lecture on phasors and their applications in AC circuit analysis. As you recall in the last lecture, the 16th one, we introduced an artificial concept of phasors which basically transforms a voltage or a current which is of the form root 2 v cosine of omega t plus theta into a phasor which is represented as V e to the power j theta and then the cosine omega t and root 2 is taken care of by root 2 e to the power j omega t and we take the real part of this. This is obviously equal to this, real part of this and here since all currents and voltages in this circuit shall contain this factor, we decided to ignore this and we also decided to ignore the real part. In other words if these 2 are cut out, this will be implied for all voltages and currents, then the quantity that is left is represented by a phasor V with a bar above it and we call this capital V e to the power j theta and in the complex plane we represent this by means of a vector whose magnitude is capital V and whose angle from the real axis is theta. If we do this then obviously this is a transformation from the time domain into the complex number domain, complex number domain and it has been shown that with the help of phasors, the circuit analysis, steady state circuit analysis or forced response of a circuit to a sinusoidal excitation becomes extremely simple. Let us consider the 3 basic elements, namely a resistance and inductance and the capacitance. A resistance suppose I excite this by means of a current I, I which is root 2 I let us say cosine of omega t then obviously the corresponding phasor is I bar which is equal to I e to the power j 0 or angle, I also write this sometimes as I angle 0 alright. Now if if this passes through a resistance R, the corresponding voltage phasor V R shall be simply I times R alright. In other words the voltage, the actual voltage we must be able to go back and forth between the time domain and the complex number domain. The actual voltage shall be equal to root 2 I R cosine of omega t. This is equal to obviously you see the phase is 0. So there is no phase change between the current and the voltage it is true in the time domain, it is also true in the transform domain and the ratio of the voltage phasor to the current phasor is simply equal to R which we can write as R 0 degrees alright. Now this is a trivial case if we take if we take an inductance now and cause the same current I to flow that is I equal to root 2 I cosine of omega t which corresponds to the phasor I equal to I angle 0 degree. Then you know the voltage across this is given by L D I D t the actual voltage this will be minus omega L root 2 I sine omega t and sine omega t can be written as omega L root 2 times omega L I can be written as cosine of omega t plus pi by 2 agreed. So if I write this as a phasor obviously the phasor shall be omega L I V L phasor shall be omega L I this would be the magnitude and the angle would be pi by 2 angle would be pi by 2 this is the angle cosine omega t plus theta alright. I can write this as V L its magnitude angle pi by 2 so in the case of the resistor in the case of the resistor if we go back if we wish to draw a phasor diagram that is the currents and voltages then you see the current is in this direction let us say this represents the current phasor I then the voltage phasor is in phase with the current phasor that is the voltage phasor also has an angle 0 so the voltage phasor may be is up to this on different scales current and voltage cannot be represented on the same scale so this would be V R phasor. On the other hand if I represent the current voltage relationship in an inductor in an inductor the current phasor is this and the voltage phasor is this and therefore if the current phasor is along the real axis I then the voltage phasor shall be along the imaginary axis shall be along the perpendicular axis because it leads by an angle of pi by 2. So the voltage across an inductor the voltage phasor across an inductor leads the current by an angle of pi by 2 this may be represented as the voltage phasor where this angle is pi by 2 alright with respect to I as the reference the current as the reference the voltage phasor across an inductance leads it by an angle of pi by 2 you also notice that the ratio of V L phasor to I phasor the ratio of the 2 is simply given by omega L times omega L an angle pi by 2 now is that a phasor no is this a phasor it is not a phasor a phasor is used to transform a time domain quantity into the complex number domain now corresponding to omega L there is no time domain quantity. So this is simply a complex quantity whose magnitude is omega L and angle is pi by 2 what is such a complex quantity obviously it is J times omega L the magnitude is omega L and the angle is 90 degrees that is why it is purely imaginary J times omega L alright and you notice that for an inductor the voltage and the current if one of them is sinusoidal if the current is sinusoidal I am sorry if the current is exponential e to the S t then the voltage is S L times e to the S t and S L was defined as the impedance of the inductor and you see that this quantity of voltage phasor to current phasor when the excitation is sinusoidal is simply given by S L with S replaced by J omega and therefore this can be called as the impedance Z L of the inductor to a sinusoidal excitation of frequency omega that is instead of e to the S t what we have is e to the J omega t and therefore S is replaced by J omega and J omega L is the impedance of the inductor to make this explicit we say Z L of J omega that is impedance of the inductor at a sinusoidal frequency of omega that is S is replaced by J omega alright in that sense what is the impedance of the resistor irrespective of the value of S whether it is DC or an exponential truly exponential quantity or a sinusoidal quantity the impedance is always equal to R and the angle is zero degree however I must caution you J omega L is not a phasor it is a complex quantity J omega L alright it is not a phasor a phasor is a current or voltage which is a representation of the corresponding quantity in the time domain it is a transform of a current or voltage. If I now take a capacitor and allow the voltage across the capacitor C to be let us say root to V cosine omega t alright then the corresponding current is C dV dt and therefore root to omega C minus V sin omega t alright the current is C dV dt is equal to this which I can write as root to omega C V cosine of omega t plus pi by 2 why is that this is minus minus sin omega t can be written as cosine omega t plus pi by 2 yes is that okay I do not want cosine 90 minus theta that is sin theta there is no sin change there is a sin change here cosine for cosine 2 you go to sin and therefore cosine omega t plus pi by 2 and therefore if the corresponding phasor quantities are denoted by V then V phasor is simply V angle 0 degree and I phasor I sub C the current in the capacitor is simply omega C V angle pi by 2 which indicates that is V sub no this is I sub C magnitude angle pi by 2 and therefore in the phasor diagram or the real imaginary or the complex plane diagram if this is V if this is the phasor V alright then the current leads the voltage phasor by an angle of 90 degrees and therefore the current I am sorry A S the current would be like this I sub C phasor and this angle this angle is pi by 2 alright so in a capacitor the current leads the voltage by an angle of 90 degrees we always take rotation in the anticlockwise direction as positive so you say the current leads the voltage by an angle of pi by 2 correspondingly in the case of an inductor the voltage leads the current by an angle of pi by 2 if I take the ratio of voltage phasor to the current phasor you see that the ratio is simply 1 over omega C angle pi by 2 now a complex quantity which is omega C magnitude omega C and the angle is pi by 2 obviously is a purely imaginary quantity and therefore its actual representation would be J omega C and since it is the ratio of a voltage phasor to a current phasor this is the impedance of the capacitor Z C alright this is the impedance of the capacitor Z C is that okay impedance Z C is not a phasor alright it is simply a complex quantity and you notice that it is quite in tune with the definition of an impedance for e to the S T excitation e to the S T excitation impedance was 1 by S C and here S is replaced by J omega and therefore you say Z C J omega alright impedance of the capacitor if I summarize if I summarize this relationships for a resistance inductance and a capacitance L C R then the relationships between the current and voltage phasors it is implied that all excitations all currents and voltages are of the form cosine omega T plus some constant theta root 2 times a constant all currents and voltages are of this form if that is so then the current in the resistance while we cannot represent only in terms of phasors let us do that this is I R phasor and the voltage phasor is V R this is I L phasor and the voltage phasor is V L I C phasor and the voltage phasor is V C with this polarity then the relationships are that V R is simply equal to R times I R phasor or the other way round I R is equal to V R divided by R which I can write as G times V R where G is the conductance alright G is the conductance of the resistance similarly for a for a inductor we have V L equal to J omega L I L J omega L is the impedance of the inductor or I L is equal to V L divided by J omega L alright which indicates that the admittance of an inductor Y L is simply equal to J omega 1 by J omega L that is correct and it is impedance Z L equals to J omega L alright for a capacitor for a capacitor V sub C well it is okay V sub C would be equal to I sub C multiplied by its impedance that is by J omega C is that okay or I sub C bar is equal to J omega C V C bar the impedance of the capacitor is 1 over J omega C and the admittance of the capacitor is J omega C whenever a quantity impedance or admittance is purely imaginary the real part of that impedance or admittance goes by a special name and we shall introduce the name here itself the name is if there is an impedance which is purely imaginary let us say let us consider the inductance for example if Z L is equal to J omega L then the real part of this is known as reactance that is what we do is we write this as J X L where X L X L is equal to omega L and is called the reactance of inductance L similarly when you write Y sub C as equal to J omega C I write this as J B C okay J times real quantity where B sub C is equal to omega C and this is called the susceptance of C susceptance alright it is not that we could not write the impedance also J the Z of a capacitor impedance of a capacitor is 1 over J omega C which I can write as J times minus 1 by omega C can I then we write this as J times X C and X C is called the reactance of the capacitor C the reactance of the capacitor is negative the reactance of an inductor is positive as you see the reactance of the capacitor is 1 minus 1 by omega C we can after we define the impedances for sinusoidal excitations we can combine them exactly in the same manner that we combined impedances for e to the S T excitation or combined resistances for example yes A plus J B that is right we deserve the symbol B for susceptance for impedance we will use the symbol capital X alright and the real part of the impedance an impedance can be of this form where capital R is the real part corresponds to a resistance capital X is a reactance is a reactance which corresponds to either an inductor or a capacitor or both a combination of them that could also form a reactance we shall come across this in a minute okay now as I said once we define impedance for sinusoidal excitation we can combine impedances exactly in the same manner that we did for e to the S T excitation for example the impedance of this combination Z of J omega would be would be equal to R plus J omega L plus 1 over J omega C so you simply sum up the impedances sum of the impedances this you can write as J omega L minus 1 over omega C and then we shall call this as the resistive part and this as the reactive part and you see that the reactive part you now consist of the combination of an inductor and a capacitor and because we also notice that the reactance of an inductor is an opposite sign to that of a capacitor and therefore depending on which is greater the reactance could be positive if a reactance is positive we call it an inductive reactance on the other hand if 1 by omega C exceeds omega L then the reactance would be negative and then we shall call it a capacitive reactance alright so reactances impedances all of them can be combined in exactly the same manner that we combined resistances or impedances to e to the power S T excitation to take another example suppose we have an inductor L in parallel with a capacitor C then I can either find the impedance or I can find the admittance suppose you find the impedance then Z of J omega shall be equal to J omega L multiplied by 1 by J omega C exactly like resistances J omega L plus 1 over J omega C which is equal to J omega L divided by omega squared well 1 minus omega squared L C is that okay and you see that depending on omega when omega squared L C is less than 1 the impedance is inductive that is the reactance is positive when omega squared L C is less than 1 then Z J omega is equal to J times a positive quantity so it behaves like an inductance on the other hand for frequencies omega squared L C is greater than 1 when the reactance is negative and therefore the impedance behaves like that of a capacitor and when omega squared L C is equal to 1 the 2 reactances cancel each other and the impedance becomes infinite what does it mean it means that no current can flow at that particular frequency that is omega equal to 1 by square root L C omega not squared L C equal to 1 at this particular frequency this circuit behaves as an open circuit and this is called the condition of resonance as we shall see later but let us enough of words let us see from a picture as to what this lead and lag mean in terms of actual sine waves alright we recall that the current and voltage in a resistor they are in phase there is no phase difference the current in an inductor leads or lags the voltage lags the voltage the current lags the voltage across an inductor leads the current by an angle of 90 degree and in a capacitor the opposite happens that is the current the current leads the voltage or voltage lags the current let us see what it means in terms of a picture is this visible on the monitor this curve I will explain 1 by 1 this curve that is the blue colored curve this is the current I equal to root 2 I cosine omega t plus theta and the corresponding drop in a resistor V R is simply this current multiplied by R which I have called I times R I have called V R root 2 V R cosine omega t plus theta and it is this curve obviously this and this are in phase that is the maximum minimum occur together whenever the current is a maximum the voltage is a maximum whenever the current is a minimum the voltage is a minimum and so on alright on the other hand this curve the black one represents the voltage across an inductor when this current flows through this and you can see that the voltage leads the current by an angle 90 degrees which means that on the x axis here it is omega t that when omega t is lagging when well 90 degrees ahead of the current the voltage arrives at a maximum the voltage reaches a maximum half a period is it half or one quarter one quarter period one quarter period earlier which corresponds to a voltage phase advance or phase lead by 90 degrees so this is what it is on the other hand the voltage across the capacitor if the same current passes shall lag behind in other words it will reach its maximum quarter of a period later this is what it means in terms of the actual voltages and currents is this clear omega t is represented in terms of degrees you see because root 2 i is cosine omega t plus theta we did not take omega t we said omega t plus theta to be completely general which means that the maximum will be reached at omega t equal to minus theta and that is what happens the maximum of the current is reached at omega t equal to minus theta this difference is theta and then the maximum of the voltage across the inductor is reached 90 degrees earlier the maximum of the voltage is reached voltage across a capacitor is reached 90 degrees later is this picture clear okay or is there any questions I am going to do an example now all right to illustrate these points but before the example how do k v l and k c l translate in terms of phasors you know that k v l in general is summation v i equal to 0 around the loop and k c l is summation i j equal to 0 at a particular node now if each of these voltages is of the same frequency as it would in a linear circuit it cannot change frequency all right then each of these voltages can be represented in terms of its phasor notation that is each of these could be represented as real part I can take the real part outside root 2 v i let us say e to the power j theta e to the power j omega t now this quantity root 2 e to the j omega t would be common to each of these voltages and the real part would also be common to each of these voltages and therefore I can get rid of this and I can simply write v i phasor equal to 0 is this okay a very simple argument says that k v l should be valid if we replace the time varying quantities by their transforms by their transforms that is that means phasors in an exactly similar manner the k v l k c l translates into i j phasor summation i j phasor equal to 0 all right so k v l and k c l can now be translated in terms of phasors phasor quantities also obey k v l k v l and k c l let us take an example suppose we have a line voltage 230 root 2 cosine of let us say 100 pi t let this line voltage be applied to a parallel combination of let us say resistance and inductance and the capacitance then and let us say the values are r l and c the resistance the currents in the 3 branches if added together shall obviously give the current that is supplied by the source if I represent the currents and voltages all of them by phasors then the corresponding phasor of this would be 230 then e to the power j 0 and therefore the current i r phasor representation would be 230 by r angle 0 the current through the inductor would be 230 divided by yes omega l and the angle would be plus or minus minus pi by 2 so I can also write this as 230 by j omega l that angle is taken care of automatically in a similar manner the current i sub c phasor shall be equal to 230 times j omega c that means 230 omega c angle plus pi by 2 and therefore the total current phasor i total current phasor i shall be i r plus i l plus i sub c I can manipulate these quantities algebraically rather than writing out differential equations and solving differential equations if the excitation is sinusoidal I can work solely in terms of phasors and this affords this offers tremendous simplification in in a circuit analysis. Let us take another example I have a voltage now represent by phasor the frequency is unimportant I represent it by phasor a voltage source I have a resistance in the capacitance I will do this very slowly resistance in the capacitance and I have an inductor l the current phasor in this is i l the current phasor in this is i r obviously i r and i sub c shall be the same right the same current flows to resistance and capacitance so it suffices to find i r it is given that r equal to 2 ohms l equal to 0.2 Henry and it is also given that i r is the actual current through the resistance is given by 10 root 2 cosine of 10 t plus 45 this is what is given alright is that clear a voltage source is connected across parallel combination of 2 branches 1 consist of an inductor l value 0.2 Henry and the other consist of a resistance in the capacitor in series resistance 2 ohms the capacitor is not known alright the question is on a phasor diagram on a phasor diagram show i r and v r on a phasor diagram show i r and v r and draw an arrow to show the direction to show the direction of of the voltage across the capacitor that is v sub c alright this is first part of the question now that is obviously very simple i r i r as you can see the current phasor would be simply equal to 10 angle 45 and the voltage phasor v r shall be equal to 10 times 2 that is 20 angle 45 this is volts and this is amperes alright so the phasor diagram phasor diagram would be take a vector 45 degree at an angle of 45 degree and you represent let us say this length by for the vector i r you take different scales and you represent this a vector in the same direction as v r as the voltage across the resistance alright now this length is 20 and this length is 10 this is in volts that is in amperes we are showing the direction is the same but we have different scales okay the next question is show the direction of v c now obviously v c yeah they are different scales okay we choose to have a smaller scale for v r alright now the current the voltage phasor across the capacitor obviously this will be i r divided by j omega c which means that it is i r by omega c and angle is minus pi by 2 minus pi by 2 with reference to whom with reference to the i r vector and therefore it goes like this this is the direction of v c is that clear the capacitor voltage shall lag the current by 90 degrees and therefore this angle must be 45 is that okay this angle is minus 45 okay the absolute value of the angle is 45 the point is that v c lags i c by 90 degrees so this part of the question is fine the next part says next part is interesting if the voltage phasor if the voltage phasor that is v we recall v is the v is the exciting voltage let me draw this circuit again okay if v is v 0 degree sorry then show v m v c on the diagram and determine v of t alright let us suppose the question like this that you know our original vector diagram was like this i had a v r in this direction and the v c was perpendicular to it v c direction was this let me show the direction this is the direction of v c alright now it is said that v which is the sum of v r and v c is has an angle of 0 and therefore v is in this direction is that clear v is in this direction and then what does it specify about v c v c is equal to say it again the magnitude of the magnitude shall be equal and therefore if i measure this would be my v c such that when i complete the parallelogram the parallelogram will now be a rectangle is not that right or a square it will be a square because this is 45 this is 45 and therefore this is my voltage phasor you are now required to find out v t that how do you find v t obviously obviously this angle is 45 so cosine 45 cosine 45 so interesting question cosine 45 would be base divided by hypotenuse so v r divided by v alright therefore v r is equal to well v equal to v r divided by root 2 no it is the other way round v would be equal to root 2 v r v would be equal to root 2 v r that is right and therefore what is v t then v t is simply yes come on i r 40 cosine 10 t that is what is v t that is all because the angle is 0 angle is 0 alright you see why did this come because small v r was 20 root 2 cosine of 10 t plus 45 that is what it was alright the magnitude of the phasor v the magnitude of the phasor v is root 2 times v r so root 2 times this will be 20 root 2 multiplied by root 2 which is 40 and the angle is 0 so it is simply i am set cosine 10 t is that clear okay please what we have seen is that our original equations let us say 10 root 2 cosine of 10 t plus 45 so what is v r then v r is 20 root 2 because r is 2 ohms 20 root 2 multiplied by cosine 10 t plus 45 and therefore the phasor v r is equal to 20 log 45 we have seen that the magnitude of v i want to find the phasor v the magnitude of v is equal to root 2 times v r which means that magnitude of v is 20 root 2 and the angle of v is 0 degree therefore the corresponding v t must be root 2 times the magnitude rms value which is 20 root 2 multiplied by cosine of omega t plus whatever the angle is the angle is 0 is that clear and therefore this is what we get we recall that root 2 i cosine omega t plus theta corresponds to a phasor which is i angle theta it is the other way round now we have to find v angle 0 degree what is the corresponding v t so multiplied v by root 2 to get the amplitude and cosine of omega t plus the angle which is 0 and omega is 10 omega is 10 and therefore this is the result where did I write it this is 40 cosine of omega t all right we will carry 2 more examples to illustrate this take another example let us say we have a v we shall make a we shall make a change over from time domain quantities to complex domain quantities and vice versa again and again you must be noticing that the small case letters are reserved for time and capital letters for phasors okay so there is a current i this is a voltage v and the circuit consists of r equal to 2 m and then an inductance l equal to 1 Henry and the capacitance c equal to 0.05 pharad and the output voltage is taken here v sub c these are the time domain quantities and now we specify that v of t let us say is 5 root 2 cosine of cosine of 5 t all right you are required to find out v sub c you are required to find out v sub c and in the process we are also required to find out a complete vector diagram complete phasor diagram that is i want to know what the current current through this resistance is what is its phasor i want to know what is this current let us say i l i r and i l is equal to i c then i want to know what is the relationship between i l and v c in terms of phasors and finally i want to know what v sub c is all right now i also want to know what this current small i is what is the corresponding phasor and what relationship does it bear to the phasor small v let us proceed systematically first of all the phasor representation of this would be v equals to 5 angle 0 okay this is the phasor representation if i want to find out i if that is my only concern then what i do is i find out the impedance all right now instead of impedance obviously it is easier to find the admittance because things are in parallel so let us find out y j omega and we will get very interesting results y j omega obviously is half plus admittance of this which is 1 by r plus 1 by impedance of this plus this so it will be j omega l minus 1 by omega c is that is that okay j omega l is the impedance of the inductor in series of the impedance of the capacitor minus j by omega c all right let us substitute the values and see what happens if you omega is 5 all right and l and c are given and therefore y of j omega is equal to half plus 1 over j omega is 5 and m is 1 so 5 into 1 minus 1 over 5 into 0.05 so this is equal to half plus 1 over j 5 minus 4 so this is simply half minus j all right 5 times 0.05 is 0.25 1 by 0.25 is 4 okay so this is the admittance then what is this current current phasor i this would be the admittance y j omega multiplied by the voltage phasor v all right the current phasor is admittance multiplied by the voltage phasor v so this is equal to half minus j times what is v 5 angle 0 degree and therefore what is the magnitude of this i can write this as 5 by 2 or i can simply write this as 5 half minus j angle 0 degree does not contribute anything and therefore what is its magnitude it is 5 square root of half plus 1 is it okay 1 quarter plus 1 all right and the angle is minus tan inverse not 5 by 2 tan inverse imaginary by real so tan inverse 2 minus tan inverse 2 is it okay now if i simplify this if i simplify this the current phasor is equal to 5 square root 5 divided by 2 is it okay okay and then minus tan inverse 2 i do not know how much minus tan inverse 2 is minus tan inverse 1 is 45 well whatever it is whatever it is you look at the phasor diagram now this is my v 63.5 all right minus 63.5 and what is 5 root 5 by 2 2.5 it is about 31 no no it is less than this how much 2.5 about 5.6 okay all right so the current phasor therefore lags the voltage by an angle of it is not to scale 63.5 and the current phasor is here if that is what i want if that is the only thing that i want all right and the corresponding current quantity i of t the corresponding current would be given by yes can you tell me what will be the current root 2 times that is 5 square root of 5 by 2 is it okay root 2 times this all right then cosine of 5 t minus 63.5 this will be the current i did not manipulate anything i simply manipulated some complex quantities all right now suppose suppose instead of this i want to find out the current through the resistance let us get back to the circuit this is the circuit what is i r i r is simply v small v by r and therefore the i r phasor would be equal to 2.5 angle 0 degree all right suppose i want to find out i l that would be more interesting now look at the phasor i l what do you think its relationship shall be to the voltage phasor v i should be happy if you if you can say v by j x l plus x c which can be either positive or negative how does what does this mean it means that the current made mad by 90 degrees or by 270 degrees is it right 780 plus 90 we shall continue this next time.