Lecture: 42

[Music] [Music] namaskar so in last class we were discussing about the nodal and mesh analysis with the help of kirchhoff current law and voltage law today we will discuss about two more theorems which with respect to ac circuit analysis we try to understand how we will implement those theorems in ac circuits so let us understand the superposition and the source transformation theorem with respect to ac circuit analysis now as ac circuit is also linear you can utilize the superposition theorem in the same way as you did in case of dc circuits now here the important thing which you have to keep in mind is that the processing of the superposition theorem is little bit different as we did in case of ac circuit because your source operating frequencies can be different so if you have multiple sources connected in the circuit and these those sources are having a different frequencies then the problem will also become interesting and we will see how we will solve those kind of problems with the help of superposition theorem whenever we have different voltage or current sources operating at different frequencies impedance offered by the inductance and capacitance will be different so when will analyze the circuit with the help of superposition theorem the total response will be obtained by adding the individual responses which we get from the superposition theorem and we will convert the response in time domain for the final result now it is important to note here that the responses which we get with the help of superposition theorem cannot be cannot be mentioned in the form of phasor because the frequencies of different sources may be different so in that case the phasor form cannot be used for representing the answer so rather we will use the time domain as a method to give the answer which we get finally after the circuit analysis so now let us understand the superposition theorem with the help of a few examples under ac circuit now in this particular example we need to find the value of i naught with the help of superposition theorem so i naught is the current which is flowing from voltage source the voltage source is having magnitude as 20 angle 90 degree and we also have one current source that is five angle zero so let's assume that the current total current is composed of two components i naught dash and i naught double dash where i naught dash and i naught double dash are the currents because of voltage and current sources separately so when you take one source at a time suppose if you take the voltage source at one time you will get i naught dash then you take current source only then you get i naught double dash the total response will be the summation of i naught dash and i naught double dash so let's first take the voltage source so when you take the voltage source your current source will be open circuited so in that case the value which we need to find out is i naught dash and the modified circuit will be as shown in the figure now the voltage source is 20 angle 90 degree so you can conveniently write it as j20 in complex form now you have the circuit as shown in the figure in which the capacitor is connected in parallel with this series combination of 8 ohm and 8 ohm resistance and j 10 ohm impedance now what we will do we will first try to find out the total impedance of this particular block so let z is the parallel combination of the j 2 minus j 2 ohm that is the inductance ah the impedance offered by capacitance and then the in series with atom and j 10 which is connected in parallel with this capacitance so when you solve it the z value is minus j 2 into 8 plus j 10 divided by minus j 2 plus 8 plus j 10 so when you simplify you will get the value of impedance offered by this particular section is 0.25 minus j 2.5 now the value of i naught dash you can calculate as j 20 is the voltage source divided by 4 minus j2 plus the impedance that is this particular section will have impedance equal to z which we just now calculated so we can say the i naught dash is nothing but j 20 divided by 4 minus j 2 plus in place of z you can write 0.25 minus j 2.5 and when you simplify i naught dash you will get as minus 2.353 plus j 2.353 now next is you consider the current source and short circuit the voltage source so your modified circuit will now be as shown in the figure here you have 5 ampere current source back in the circuit and the voltage source is short circuited now in this case we see there are three loops so what we will do we will utilize the kirchoff voltage law to find out the loop currents so for say loop 1 or you can say mesh 1 the equation you can write as 8 plus j 10 minus j 2 into i 1 minus of minus j 2 into i 2 because i 1 i 2 are in opposite direction here and then minus j 10 i 3 because i 3 and i 1 are also in opposite direction so from mesh 1 we get this equation for mesh 2 what you get you get 4 minus j 2 minus j 2 into i 2 minus of minus j 2 into i 1 because i 1 and i 2 are in opposite direction similarly minus of minus j 2 into i 3 because i 2 and i 3 are opposite in the this particular capacitor so we get another equation for mass two if you see mesh three you can directly say that i three is nothing but equal to five ampere which is the current source so what next you have to do you put the value of i 3 in these two equations so it will be simplified as 8 plus 8 j into i 1 plus j 2 into i 2 is equal to j 50 and j 2 i 1 plus 4 minus j 4 i 2 equal to minus j 10 so now you get the two equations and 2 unknowns i 1 and i 2 you can represent this equation in matrix form so matrix form will contain one left side vector which contains j 50 and minus j 10 as element and then we compile the matrix containing the coefficients of i1 and i2 that is 8 plus j8 and j2 and j2 and 4 minus j4 and then the unknown quantities that is i 1 and i 2 now to solve it the objective for finding out the solution for this particular equation is to find the value of i naught double dash so if you see i know double dash it is nothing but minus of i 2 so it means that we if we find the value of i 2 that is sufficient for the calculation so what will be the value of i2 first we will take the determinant of the matrix so determinant of the matrix will come out to be 68 and then the determinant of sub matrix which corresponds to the variable i2 so what you will do you will simply replace the second column with the vector which is given on the left so when you replace you get the modified determinant as 8 plus j 8 j 50 j 2 n minus j 10 so when you find the value of determinant you get 180 minus j 80 now what is the value of i 2 i 2 will be nothing but delta 2 divided by delta so when you simplify you get again the value of i 2 as 2.647 minus j 1.176 ampere now as you know that i 2 i naught double dash is nothing but minus of i 2 so this will be now minus 2.647 plus j 1.176 ampere now the i 2 value will be i naught dash plus i naught double dash i know this we found here that is minus 2.353 plus j 2.353 and you also got the value of i naught double dash then when you sum up you get the value of final current i naught that is 6.12 angle 144.78 degree now let's take example this example will be really interesting because here we need to find out the value of v naught using superposition theorem and if you see this particular circuit you will have three sources and all three are operating at a different frequency so this is a dc voltage source so it means that your frequency is zero here you have current source omega is 5 here you have voltage source where omega is 2 so when you have different sources operating at different frequencies we will utilize the superposition theorem by taking one source at a time and solve the circuit and finally we sum up the output so that we get the final output of the circuit so in this case we need to find out the value of v naught so let us assume that v 1 v 2 and v 3 are the individual voltage responses when you take d c and then the 10 cost 2t volt and then finally v3 is corresponding to the current source that is 2 sin 5 t taken into the consideration so the final voltage will be v naught is equal to v one plus v two plus v three now let's take one voltage at a time first let's take the dc voltage five volt so this will be open circuited this will be short circuited now you have to keep in mind since we are talking about the dc voltage source it means that inductor will also be short circuited and capacitor will be open circuited so in that case what will happen your modified circuit will look like as shown in the figure you will left with only the resistances that is 1 ohm resistance and 4 ohm resistance so 1 ohm resistance you got from this particular section and 4 ohm is from this section where you have capacitor in between but capacitor will be open circuited and this is short circuit this is short circuit so the complete left to one ohm resistance will also be short circuited so you will left with only one ohm and four ohm resistances and five volt voltage source now what we have to do we need to find out the voltage across 1 ohm resistance so what we can do we can utilize the voltage division so here the direction of voltage is opposite so what we will say we will say this is minus v 1 equal to 1 upon 1 plus 4 into 5 volt so minus v 1 you get as 1 volt that is v 1 is equal to minus 1 volt now the second case is that in case of v 2 will set 5 volt dc source and the two sine five t current source zero so that means if you see the figure this is zero and this is again zero so now this will be short circuited this will be open circuited so in that case what will be the modified circuit you will have 4 ohm resistance in parallel with now the capacitor and the inductor and resistance will be in series that is 1 ohm resistance in series with the inductance so the next task what we have to do if you remember in case of the nodal and mesh analysis we discussed that first the elements need to be converted into frequency domain so to analyze the this particular circuit we will first transform the circuit into frequency domain so 10 cos 2t we can write simply as 10 angle 0 where omega is 2 radian so the 2 henry inductance will be converted into frequency domain so you can write it as j4 and then the 0.1 farad capacitor can be written as 1 upon j omega c so it will come out to be minus 5 now your circuit is ready for the analysis we need to find out the voltage v 2 across 1 ohm resistance now since these two elements are in parallel so you can take the parallel combination of these two elements and you can find out the impedance ah so what is the impedance jet so in this case minus j5 in parallel with four so when you solve you get the impedance as 2.439 minus j 1.951 now you have found this value we need to find out the voltage across 1 ohm resistance so you can simply utilize the voltage division method and find the value of voltage v2 so what will be the voltage v2 that is the voltage source value that is 10 angle 0 divided by 1 is the 1 ohm resistance we need to find so 1 will be in numerator divided by the 1 ohm plus j 4 plus the parallel combination of the resistance and capacitor so we found the value of z so we place the value of z here and simplify then we get the value of v 2 as 2.498 angle minus 30.79 volt now this you have got in phasor form next task is that you need to convert it back into time domain so if you convert it into time domain what you will write v2 is nothing but 2.498 cos 2t minus 30.79 degree we get 2t because the original source was having frequency of omega is equal to 2 so the voltage across the 1 ohm resistance that is v 2 will be 2.498 cos 2t minus 30.79 degree volt now the third one is the current source so we need to find out the value of v3 when the 5 volt dc and 10 cos 2t voltage sources are set to 0 it means that both of them will be short circuited so your modified circuit will look like shown in the figure where the current source will now be in parallel with the inductance then we have 1 ohm resistance and in series with parallel combination of the minus j 2 impedance offered by capacitance and 4 ohm resistance now first task is that again we have to transform this into frequency domain so we will convert it into frequency domain the given signal that is the given current source is 2 sine 5t now if you convert into phasor what you can write 2 angle minus 90 degree because sine you have to first convert into cos to convert it into phasor form so what you can write 2 sine 5 t you can write 2 sine cos 5 t minus 90 degree so when you convert it into phasor it will become 2 angle minus 90 degree here omega is now 5 radian per second so you have to convert again all the inductances capacitances into frequency domain so 2 henry inductance will now become j 10 ohm as an impedance 0.1 farad capacitance will offer minus j 2 ohm as an impedance so we get now the modified circuit where you have j 10 ohm as impedance offered by inductance and minus j 2 ohm as the impedance offered by capacitance then what we have to find is the voltage across 1 ohm resistance so that is v 3 which we have to find so first what we have to do we have to first take these two the parallel combinations and simplify it so when you take these two we let us say the effective impedance is z1 where minus j2 is in parallel with 4 ohm resistance so you get effective impedance as 0.8 minus j 1.6 so this you have got now we utilize the current division to find the value of current flowing through 1 ohm resistance so here if you apply current division then you can say that the current is nothing but 2 angle minus 90 degree into 10 divided by 1 plus z 1 plus 10 so the current i 1 which is flowing through 1 ohm resistance that is i 1 equal to j 10 divided by 1 plus j 10 plus the value of z 1 that is 0.8 minus j 1.6 and then multiplied by current source that is 2 angle minus 90 degree so when you simplify you get the value of current i 1 now what is the value of v 3 v 3 is nothing but i 1 into r r is 1 so you get the value of v 3 as 2.328 angle minus eighty degree so now this is again in phasor we will convert it into time domain so when you convert it into time domain it will become two point three two eight cos five t minus eighty or you can again convert it back into sine because sine was the initial value when you saw the current source this the current was written in terms of sine 5t so we will convert it back into sine so it will become 2.328 into sine 5 t plus 10 degree because cos 5 t minus 80 degree can be written as sine 90 plus 5 t minus 80 degree so this will be simplified to 2.328 sine 5 t plus 10 degree now the total response v naught you can write v1 that is minus 1 plus v2 that is 2.498 cos 2t minus 30.79 degree plus v3 that is 2.328 sine 5t plus 10 degree so now you can see that superposition theorem works well when you have voltage or current sources connected to the circuit and having the different frequencies now let's talk about the uh source transformation now in case of source transformation the in frequency domain the source transformation involves the transforming a voltage source in series with impedance to a current source in parallel with impedance or vice versa that means if you have current source in parallel with impedance can be converted into voltage source in series with an impedance how we will express them we will express it as vs so this is voltage source in series with zs that is impedance so vs is nothing but zs is equally it can be represented as is where i s is nothing but vs upon zs and then in parallel with impedance zs so you can transform the voltage source in series with impedance to a current source in parallel with impedance as shown in the figure so we'll utilize this concept to solve the circuit so let us take one example we need to find out the value of v x using source transformation so in this case how will carry out the source transformation you see that you have voltage source 20 angle minus 90 degree connected in series with 5 ohm so this particular segment you can utilize to convert it to equivalent current source in parallel with resistance so how you can write so this can be converted into one is that is current source so what will be the value of current source current source will become 20 angle minus 90 degree divided by 5 that is the resistance so it will become 4 angle minus 90 degree or in complex form you can write i s is equal to minus j 4 ampere and this current source will be in parallel with 5 ohm resistance now if you see the circuit these two legs are in parallel so you can further simplify then the parallel combination of 5 and 3 plus j4 will be simplified to say z1 so the value of z1 is 2.5 plus j 1.25 now again this is converted into equivalent impedance so if you see that the equivalent impedance is z1 connected in parallel so you can say again the current source is now in parallel with one impedance so you can convert it back into a current a voltage source in series with impedance so what will be the value of voltage source in that case the value of voltage source will be is into z1 so now the voltage source you can write vs is nothing but is into the value of z1 which you have calculated that is 2.5 plus j 1.25 multiplied by is that is minus j4 so finally you get the value of vs that is 5 minus j10 and now this will be in series with the impedance that is z1 now if you see this circuit it is much simplified and you can easily find out the value of vx so what is the value of vx in this case you will simply utilize the voltage division you can say vx is nothing but 10 divided by the [Music] elements which are connected in series now 10 plus 2.5 plus j 1.25 plus 4 minus j 13 multiplied by the v s that is voltage source 5 minus j 10 when you simplify you get the value of v x now you can say that by successive implementation of source transformation in the circuit you can simplify the network and find the value of vx very easily so with this we can close our today's session in which we discussed about the source transformation as well as superposition theorem in next session we will discuss about ah two more important theorems which we discuss in case of dc circuit that are your thevenin's and norton's theorem ah we will discuss when we consider the ac circuit for the analysis thank you [Music] [Music] [Music] you