Lecture: 41

So, we are entering into the 9th week of our ah course ah. Till now we discussed various circuit theorems ah with respect to ah DC source ah. These 2 weeks that is 9th and 10th week will devote more towards circuit analysis ah when the AC source is available. So, we will say that the AC circuit analysis with the help of what we discussed till now in case of DC source. So, let us start the discussion of ah the circuit analysis that is we will use the sinusoidal steady state analysis concepts to understand the various phenomena of AC circuits. Now, as we know that the concept of phasor ah was discussed in the first module, the concept of phasor we will use in the circuit analysis for the cases where we have the AC source available. So, the concept of phasor is very important in the case of AC circuit analysis. Now, we also know know that the previous discussions we had based on basic laws like Ohm's law, Kirchhoff's law ah the same can be applied to AC circuit also. So, what we will do in this module we will discuss particularly on how the theorems which we discussed in case of DC circuit can be used to analyze the AC circuits. So, whatever we techniques the techniques we used in previous lectures we can equally use those techniques for our AC circuit analysis. So, what we will do we will take the help of various examples to understand how the various theorems which we use in case of DC circuit can be utilized in AC circuit as well. Now, what are the ah steps to analyze the AC circuit? First what we will do we will transform the circuit to the phasor or frequency domain. That means, if the circuit is given in time domain we will first convert the circuit into phasor or frequency domain then we will solve the problem using ah various circuit theorems which we discussed previously in case of DC circuits and then transform the resulting phasor to the time domain back. Now, the step one is not necessary if the problem is already specified in frequency domain. So, that means, if you already have the problem given in phasor or frequency domain we need not to follow the first step we can straight away start from second step. The second step is performed similar manner to what we did in case of DC circuits. The only change is that here instead of the numbers real numbers what we used in case of DC circuit here we will use complex numbers for AC circuit analysis. Now, the complex numbers we can easily handle because we have already discussed how to deal with the complex numbers when we were discussing about the phasors and other fundamental theorems of the circuit analysis. So, let us try to understand the nodal and mesh analysis first in case of AC circuit. So, the basis of nodal analysis is called Kirchhoff current law and the basis for mesh analysis that is also called as loop analysis is based on Kirchhoff voltage law. Now, since KCL and KVL both are valid for complex number also the nodal and mesh analysis can be used in case of AC circuits. So, let us see few of the examples. So, that we can understand how we will utilize KCL and KVL when we are asked to find the various circuit parameters in case of AC circuit. So, let us see the example in this case you will see that we have one sinusoidal voltage source given by 20 cos 4 t and we have the inductor of 1 Henry we also have the dependent current source and the 0.1 Farad capacitor and 0.5 Henry inductor. So, the first task what we have to do is that we have to convert circuit into frequency domain. That means, we will convert the sources into phasor and the elements which we see in the circuit will be converted into the frequency domain. So, now let us see this voltage source voltage source is given as 20 cos 4 t. So, what we can write since we do not have any phase angle given in case of this voltage source the reference would be 20 angle 0 degree. So, we take it as a reference and now what is omega? Omega is given as 4 radian per second. So, if you compare with say V m cos omega t omega will be equal to 4. So, now, this omega will help us to transform the various circuit elements into frequency domain. So, what will happen the 1 Henry inductor which we see in the figure will be converted into the frequency domain. So, it means that the inductor will be converted as j omega L. So, if you put the values of omega and L you will get the value of inductor as j 4. Similarly in case of 0.5 Henry inductor you will get the value as j 2 for the capacitor it is a 0.1 farad. So, the value of capacitor is 1 upon j omega c, c is given as 0.1 farad. So, the value which we get in case of capacitor is minus j 2.5. So, what we will do we will convert this circuit from time domain into frequency domain. So, your voltage source will be 20 angle 0, there will be no change in resistance and then we will put the value of the inductors and capacitors in frequency domain. Now, we will apply the KCL at node 1. So, if you apply KCL at node 1 what you can write the current direction is in this side which is going inside the node and other 2 currents are going outside the node. So, we can write that 20 minus that is voltage V 1 at node 1 divided by 10 is the value of current going inside the node equal to V 1 upon minus of j 2.5 that is the value of I x plus V 1 minus V 2 upon j 4. Now, if you simplify this equation you can write V 1 1 plus j 1.5 plus j 2.5 V 2 equal to 20. So, this is what you will get when you apply KCL at node 1. Similarly, if you apply KCL at node 2. So, that is if you apply KCL at node 2 what you can write. So, these are going inside the node and one is coming out. So, V 1 minus V 2 upon j 4 plus 2 I x equal to the current which is flowing in this particular ah j 2 ohm inductance. So, what we can write. So, the value of this is V 2 upon j 2 equal to V 1 minus V 2 upon j 4 plus 2 I x. Now, we also know that I x is nothing, but V 1 upon minus j 2.5 this is what we get from here. So, I x is nothing, but V 1 upon minus j 2.5. So, we replace the value of I x with V 1 upon minus j 2.5. So, we can write V 2 upon j 2 equal to 2 in place of I x we write V 1 upon minus j 2.5 plus V 1 minus V 2 upon j 4. So, when you simplify it you get the equation as 11 V 1 plus 15 V 2 equal to 0. So, first equation from node 1 you get this one and second you get this one. So, these two nodal equations we can express in the form of matrix. So, if you write in terms of matrix you will write the vector containing elements 20 and 0 and then you have the matrix having the elements corresponding to V 1 V 2. So, 1 plus j 1.5 j 2.5 will be coming here 11 and 15 will be coming here. So, now, you have compiled the these two equation in matrix form ah why we are compiling in matrix form because solving equation in matrix form is easier how? Let us first take the determinant value of this matrix. So, what we can write 1 plus j 1.5 j 2.5 11 and 15 we take the determinant value of this we get 15 minus j 5. Now, next is we need to find out the value of V 1. So, when you ask to find out the value of V 1 we will compile the corresponding determinant. So, the updated determinant will be 20 0 j 2.5 and 15 and the value of this determinant is 300 for V 2 we get 1 plus j 1.5 then here you will get 20 11 and 0. So, you get delta 2 that is determinant corresponding to voltage V 2 you get the value of that determinant as minus 220. Now, the value of V 1 you can simply write as delta 1 by delta. So, the value of delta which we calculated. So, V 1 value you will get as 18.97 angle 18.43 degree. So, the quantity will be in volt similarly in case of V 2 you will find out the value of V 2 as delta 2 divided by delta and when you simplify you will get the value as 13.91 angle 198.3 degree volts. Now, the value of current I x I x we know that is equal to V 1 upon minus j 2.5. So, what we will do we will put the value of V 1 in the expression and we simplify we get the value of I x as 7.59 angle 108.4 degree. Let us take another example where we are asked to find out the value of V 1 and V 2 in the circuit. So, we will again use the nodal analysis Now, if you see this particular figure the voltage node V 1 and V 2 are connected through one voltage source. So, what we can do we can create one super node which includes V 1 and V 2 because these V 1 V 2 are connected through some voltage source. So, what we can do we create one super node which includes V 1 and V 2 and we will apply KCL for this super node. So, if you apply KCL at the super node. So, what you get you get 3 ampere as a current going inside and other 3 currents are going out. So, what we can write 3 equal to V 1 upon minus j 3 V 2 upon j 6 plus V 2 upon 12 when you apply that is again V 2 upon 12. So, now, what we will do we will simplify this and when we simplify we get the equation as 36 is equal to j 4 V 1 plus 1 minus j 2 V 2. Now, we also know that the voltage source is connected between nodes 1 and 2. So, what we can say that V 1 is nothing, but V 2 plus the value of tan angle 45 degree. So, we will write V 1 is equal to V 2 angle tan 45 degree. Now what we can do we can place the value of V 1 as this in this equation. So, what we can write we can write 36 angle. So, 40 angle 135 degree will be when you put the value of V 1 in this expression this will become V 2 plus tan angle 45 degree multiplied by j 4. So, when you simplify you will get the expression as V 2 equal to 31.41 angle 81 87.18 degree volt. So, now, using these 2 equation you have simplified the equation and found the value of V 2 the same value of V 2 you can put here and you can get the value of V 1. So, using nodal analysis you can easily find out the value of V 1 and V 2. Next we see how we will utilize the concept of mesh analysis in the previous 2 examples we used nodal analysis in this example we will try to understand how we will apply the mesh analysis. So, now, if you see this figure in this figure we see there are 3 loops. So, we give the value of current as I 1 for this loop I 2 for this loop and I 3 for the top loop. Now what we have to do we have to apply the KVL. So, if you apply KVL to mesh 1 what we can write we can write 8 plus j 10 minus j 2 multiplied by current I 1 minus of minus j 2 because I 2 is in opposite direction of I 1. So, this will become minus of minus j 2 I 2 similarly I 3 is also opposite to I 1. So, this will become minus j 10 I 3 equal to 0. So, from mesh 1 we get this equation similarly for mesh 2 you can write the equation as 4 minus j 2 minus j 2 into I 2 minus of minus j 2 into I 1 and then minus of minus j 2 into I 3 for this element plus the voltage source because we are going from positive to negative. So, voltage source will become plus 20 angle 90 degree and for loop 3 if you see I 3 is simple you can say I 3 equal to 5 because angle is 0. So, you will simply write I 3 is equal to 5. So, now what you can do you can put the value of I 3 in these 2 equations and simplify. So, when you simplify what you will get you will get 8 plus j 8 into I 1 plus j 2 I 2 is equal to j 50 this is what you got from first equation and from the second equation you will get j 2 I 1 plus 4 minus j 4 I 2 equal to minus j 30. So, this we you will get by simply putting the value of I 3 in these 2. Now again what we can do we can compile these 2 equation in matrix form. So, we can write j 50 and minus j 30 would be the elements of the left side vector, then you will have matrix containing elements 8 plus j 8, then j 2, then j 2 for I 1 and 4 minus j 4 for I 2 and the vector for I 1 and I 2. Now as we did in the previous example in this case also we will first determine the value of determinant of this matrix. So, if we determine the value of determinant it comes out to be 68. Now next is we need to find out the value of delta 2 because we need to find out the value of I 2. So, for that what we have to do we will just replace this column with this column. So, we get the determinant delta 2 as 8 plus j 8 j 50 j 2 and minus j 30 and we when we find the value of determinant we get the value as 416 dot 7 angle minus 35.22 degree. When you calculate the value of delta and delta 2 you get the value of delta as 68 delta 2 as 416.7 angle 35.22 degree. Now let us try to find out the value of current I 2. So, I 2 will be equal to delta 2 divided by delta. Now put the value of delta 2 as 416.7 angle minus 35.22 degree divided by delta which is 68 you get the value of current I 2 as 6.12 angle minus 35.22 degree MPa. Now in the question we need to find out the value of current I naught. So, what is I naught? I naught is nothing but minus of I 2 and we have calculated the value of I 2. So, the I naught value will be equal to minus of I 2. So, what we can write? We can write minus of I 2 as minus of 6.12 angle minus 35.22 MPa. Now since the negative sign which you see in the expression you can simply replace it with 1 angle 180 degree MPa. So, when you write minus sign as or you can say minus is nothing but minus 1 multiplied by this current expression. So, minus 1 you can simply write as 1 angle minus 1 angle 180 degree. So, when you simplify you can simply write 6.12 angle 180 degree minus 35.22 MPa that is equal to 6.12 angle 144.78 degree MPa. This is how we get in case of the value of current I naught. So, next let us try to find out the value of V naught in the circuit using mesh analysis. So, what we will do here? We see first that we have one current source coming between two meshes. So, that is between two loops. So, what we can do? We can create a super mesh because if you have current source between two meshes we can create super mesh and solve it. So, we will utilize that concept. So, meshes 3 and 4 will now form the super mesh due to current source between them. So, we will apply KVL first to mesh 1. So, when you apply KVL to mesh 1 it will become that is minus 10 plus 8 minus j 2 into I 1 minus j 2 into I 2 minus 8 I 3 equal to 0. So, this you get from mesh 1. Similarly, from mesh 2 if you see the value of I 2 is equal to minus of 3. So, for the super mesh the third loop what you will get? You will get 8 minus j 4 into I 3. So, 8 minus j 4 into I 3 minus 8 I 1 plus 6 plus j 5 into I 4 6 plus j 5 into I 4 minus j 5 into I 2 equal to 0. So, this what you will get from the super mesh. And now if you apply KCL at this node what you will get? You will get I 4 is nothing, but I 3 plus 4. So, now, we have 4 equations using I 2 value and the value of I 4 we can put these 2 values in these 2 equation and simplify. So, when we simplify we are reduced to only 2 equations. So, what are those 2 equation? First we will have 8 minus j 2 I 1 minus 8 I 3 is equal to 10 plus j 6. And similarly for second one we will get minus 8 I 1 plus 14 plus j I 3 equal to minus 24 minus j 35. So, we have now these 2 equations which we can compile in the form of matrix. So, we will have first left side we will have vector containing elements as 10 plus j 6 and minus 24 j minus j 35, then the matrix having element 8 minus j 2 and minus 8 and minus 8 and 14 plus j and then the currents. So, that is I 1 and I 3. Now if you solve for I 1 what you will do? You will first find the value of determinant that is 8 minus j 2 and minus 8 minus 8 and 14 plus j the value of this determinant will be 15 minus j 20. To find out the value of delta 1 delta 1 is nothing, but what you will do you will replace this particular column with this. So, this will become 10 plus j 6 and minus 8 then minus 24 minus j 35 and 14 plus j. So, when you find the value of determinant you will get minus 58 angle j 186. So, now, to find out the value of I 1 you will find the value of delta 1 by delta delta 1 and delta if you put the values you will get the current I 1 as 3.618 angle 274.5 degree ampere. Now what is the value of V naught? V naught you can get if you go back to the equation. So, the value of V naught is the value across j 2 ohm. So, what are the currents flowing in this? The current will be I 1 and I 2. So, I 2 we already know that is minus 3 I 1 we have just found. So, what we will write V naught is equal to minus j 2 I 1 minus I 2. So, we solve and we get the value of voltage V naught which is across the capacitor. So, using these techniques which we discussed in case of DC circuit analysis same we applied in case of AC circuit analysis and solve the circuit. So, with this we close our today's session. Thank you.