noc19-ee36-lec45

So, in last session we discussed about the magnetically coupled circuits. So, we will continue our discussion today and we will talk about energy stored in magnetically coupled circuits and then also we will discuss about the ideal transformer. So, let us start the discussion of today's session. The energy stored in inductor as we know can be given by W is equal to half L i square. Now, to determine the energy stored in magnetically coupled coil, let us consider the following circuit. Here I 1 and I 2 are the currents on left and right coils, V 1 is the voltage across first coil and V 2 is voltage across second coil. They have inductances as L 1 and L 2 and M is the mutual inductance. Now, let us assume that first the current I 1 and I 2 are initially 0, that means that total energy stored in the circuit is equal to 0. Now, I 1 is now allowed to increase from 0 to capital I 1, that is one arbitrary value we are considering while maintaining current I 2 is equal to 0. In that case, the power in coil 1 can be given by P 1 at any time t is equal to V 1 I 1. So, we can say as I 1, V 1 can be replaced by L 1 d I 1 by d t and then the energy stored in the circuit can be given by say it is W 1. So, you integrate the instantaneous power from 0 to capital I 1 and solve it. So, when we integrate the above equation with respect to t, we get half L 1 I 1 square. Now, if current I 2 is allowed to increase from 0 to say capital I 2, that is again an arbitrary value we are considering while maintaining the current same in the first coil. So, in that case, the induced mutual voltage in first coil will be M 1 2 d I 2 by d t, while in coil 2 it will be 0 because there is no change in I 1. So, d I 1 by d t will be 0 when we are talking about the mutual induced voltage in coil 2. So, for coil 2, we will have the mutual voltage, mutual induced voltage as M 1 2 d I 2 by d t. Therefore, the power in coils due to increase in current I 2 can be given by say it is P 2. So, it will have now two components, one is because of mutual voltage and second is because of the self inductance. So, this will become I 1 M 1 2 d I 2 by d t and V 2 I 2, V 2 can be given as L 2 d I 2 by d t because of the inductance self inductance of the second coil. So, we can write I 1 M 1 2 d I 2 by d t plus I 2 L 2 d I 2 by d t. So, energy is stored in the circuit because of change in current in the second coil from 0 to I 2 will be you will integrate the instantaneous power from 0 to I 2. So, when you integrate you get the energy value as M 1 2 I 1 I 2 plus half L 2 I 2 square. Now, in that case the total energy stored in the circuit will be considered when both I 1 I 2 have reached the their constant value. So, total energy stored in the circuit will be now W 1 plus W 2. So, this will become L 1 I 1 square that is W 1 plus M 1 2 I 1 I 2 plus half L 2 I 2 square. Now, if we reverse the order by which the current reach their final values that means, first we will increase I 2 from 0 to capital I 2 and then we increase I 1 from 0 to capital I 1. In that case the total energy stored in the circuit can be given as half L 1 I 1 square plus half L 2 I 2 square plus M 2 1 I 1 I 2. So, in this case the mutual inductance is M 2 1 because the current which is increasing in the circuit first is I 2. So, mutual inductance will be because of I 1. So, you get the mutual inductance in the second coil as M 2 1 I 1 I 2. Now, since the total energy stored should be same regardless of how we reach the final conditions in that case if you compare these two equations you can simply say that M 1 2 is equal to M 2 1. So, mutual inductance is this mutual voltage induced because of either first current is increasing or first current I 2 is increasing or first current I 1 is increasing will remain same. So, you can simply say M is equal to M 1 2 or equal to M 2 1. So, M 1 2 or M 2 1 in these two equations can be simply replaced by M. So, total energy stored in the circuit we can write as half L 1 I 1 square plus M I 1 I 2 plus half L 2 I 2 square. Now, this equation we derived by assuming that the currents in both coils are entering the dotted terminal that means the mutual voltage which induced is in either of the coil will be positive. So, in case if one current enters the dotted terminal while the other leaves the other dotted terminal the mutual voltage now will become negative. In that case the mutual energy will also be negative. So, the total energy stored in the circuit will now become half L 1 I 1 square minus M I 1 I 2 plus half L 2 I 2 square. Now, we have considered I 1 I 2 as some arbitrary values. So, we can replace with any other value like we replace with generic I 1 I 2 equations the symbols. So, what we can write the instantaneous energy which will be stored in the circuit can be given by some general expression, and the general expression is W equal to half I 1 I 1 square into L 1 plus minus M I 1 I 2 plus half L 2 I 2 square. So, this plus minus will be depending upon how current is entering and leaving the dots. Now, the positive sign we will select when mutual term in both the mutual the positive sign we will select for the mutual term if both currents enter or leave the dotted terminals otherwise we will consider as negative. Now, let us try to stabilize the upper limit for the mutual inductance the as we know the energy stored in the circuit cannot be negative because we are talking about the circuit which is passive in nature. So, what we can conveniently say that half L 1 I 1 square minus M I 1 I 2 plus half L 2 I 2 square should always be greater than 0. Now, we have two square terms in the expression. So, what we will do we will first create the complete square. So, to do that what we will do we will add and subtract I 1 I 2 under root of L 1 I 2 basically in the expression. So, when you add as well as subtract I 1 I 2 root L 1 L 2 you can simplify the above expression as half I 1 into under root L 1 minus I 2 into under root L 2 whole square plus I 1 I 2 root of L 1 L 2 minus M greater or equal to 0. Now, since this is the square term this can never be negative at most it can be 0. So, we can have the updated constraint of the criteria that the root of L 1 L 2 minus M should be greater or equal to 0. So, we get updated criteria now. So, if you simplify you get M should always be less than or equal to L 1 L 2 whole under root. So, that means the mutual inductance cannot be greater than the geometric mean of the self inductances of the coil. Now, this condition we can convert it into equality constraint. So, when you convert it into equality let us multiply with some constant K. So, we can write M is nothing but K into root of L 1 L 2 this K is called as coefficient of coupling. So, coefficient of coupling is the measure of magnetic coupling between two coils. So, the value of K will be ranging between 0 to 1. So, if K is 1 we will say that the circuit is perfectly coupled if K is less than 0.5 we say that circuit is loosely coupled and if K is greater than 0.5 we say it is strongly coupled. So, now let us take one example to understand what we discussed related to energy stored in the coupled circuit. We need to find out the energy stored in the coupled inductors at time t is equal to 1 second, where voltage V is given as 60 cos 4 t plus 30 volt. Now, if you see this particular circuit the L 1 L 2 are given as H 4 and H 4 5 and 4 Henry. So, K will become M is 2.5. So, K will become 0.56. So, since K is 0.56 we will say that the circuit is tightly coupled. If K is less than 0.5 we will say that it is loosely coupled and in case K is equal to 1 we will say circuit is perfectly coupled. Now, the next task is that to find the energy stored we have to calculate the value of current. So, for that what we need to do we need to first obtain the frequency domain equivalent of the circuit. So, when you convert it into frequency domain equivalent your voltage will become 60 angle 30 degree and the omega value is 4 radian per second. The 4 Henry inductor will be now J 20 as a reactance 2.5 Henry mutual inductance will become J 10, 4 Henry inductor will become J 16 as its reactance and then the capacitor will have the capacitive reactance of minus J 4. So, if you put all those values in the circuit you will get the circuit in frequency domain. Let us assume that the current I 1 is flowing in the left part of the circuit and I 2 is flowing in the right part of the circuit. So, if you see these two currents both are going inside the dot that means the mutual voltage will be positive. So, for mesh 1 what we can write for mesh 1 we can say the value of impedance that is 10 plus J 20 multiplied by I 1 plus mutual voltage that is J 10 I 2 equal to the applied voltage 60 angle 30. Now for mesh 2 what we can write we can write minus J 4 plus J 16 into I 2 plus J 10 I 1 because of mutual voltage equal to 0. So, from this we get I 1 is equal to 1.2 I 2 now this value if you put here the updated equation will be minus 12 plus minus J 14 into I 2 is equal to 60 angle 30. So, when you simplify you get the value of I 2 as 3.254 angle 160.6 degree. Now you can get the value of I 1 as 3.905 angle minus 19.4 degree ampere. Next task is that we will convert now the values of current we got into time domain. So, I 1 will become 3.905 cos 40 and minus 19.4 I 2 will become 2.354 cos 40 plus 160 degree 160.6 degree. Now we need to find the value of energy at time t is equal to 1 second. So, the in the voltage if you see the voltage value it is 60 cos 40 plus 30. So, that means that the value 40 will be the omega t value in the expression. So, in that you just put the value of t as 1. So, you get the angle as 4 radian this 4 radian we need to convert it into degree. So, if you convert 4 radian into degree you have to just multiply by 180 degree divided by pi. So, when you simplify you get angle that is instantaneous angle at t is equal to 1 as 229.2 degree. So, you put this value here. So, I 1 at time t is equal to 1 second will become now if you simplify this will become minus 3.389 ampere. Similarly I 2 will be 2.284 ampere. So, the total energy stored in the coupled inductors can be given as W equal to half L 1 I 1 square plus m I 1 I 2 plus half L 2 I 2 square. You put the values of I 1 and I 2 and L 1 L 2 and m and when you simplify you get the value of total energy stored as 20.73 joule. Now let us start the discussion about the another topic called ideal transformer. Now what is ideal transformer? The ideal transformer is the one which has perfect coupling. So, it means that if you have the magnetically coupled circuit and you have the coupling coefficient equal to 1 that means the circuit which has perfect coupling will be the ideal transformer. So, it consists of two or more coils with a large number of turns wound on a common core of high permeability. Since we have high permeability in the transformer core the flux which will be generated links to all turns of both of the coils thereby resulting in perfect coupling. So, what we can say that ideal transformer is the limiting case of two coupled inductors where the inductances approach infinity and the coupling is perfect. So, to understand this more clearly let us consider one circuit of the ideal transformer. So, in this case if you see V 1 is applied on the left side of the coil V 2 is at the right side of the coil m is the mutual inductance L 1 and L 2 are the self inductances of both coils. Now I 1 and I 2 both are entering the dot means the mutual voltage would be positive. So, what we will do we will write the equations for V 1 and V 2 in frequency domain. So, if you write equations in frequency domain you can write V 1 is equal to j omega L 1 I 1 plus j omega m into I 2 because of the mutual inductance. So, from this you can simply say I 1 is equal to V 1 minus j omega m I 2 whole divided by j omega L 1. Now for the second coil V 2 you can write j omega m I 1 plus j omega L 2 I 2. So, what you will do now you will put the value of I 1 in this equation and when you simplify you will get V 2 is nothing but equal to j omega L 2 I 2 plus m V 1 by L 1 minus j omega m square I 2 divided by L 1. Now you know that m which we have just now derived is equal to root of L 1 L 2 in case of perfect coupling. So, when you consider k is equal to 1 means perfectly coupled coils m will be equal to root of L 1 L 2. Now this value we will put in the previous equation and when we simplify we get V 2 is nothing but j omega L 2 I 2 plus root of L 1 L 2 in place of m into V 1 divided by L 1 minus j omega L 1 L 2 I 2 by L 1. So, when you simplify you get the value as V 2 equal to root of L 2 by L 1 into V 1. So, now if you just say root of L 2 by L 1 as n we can simply write V 2 is equal to n V 1. Now what is n? n is popularly known as the turns ratio of the transformer. So, if you see the expression if L 1 L 2 or m tends to infinity in such a manner that n remains the same the coupled coils will become the ideal transformer. So, to summarize we can say the transformer is said to be ideal if it has the following properties. Coils will have very large inductances means L 1 L 2 and m tends to infinity coupling coefficient is equal to unity that is k is equal to 1 and primary and secondary coils are lossless means the resistances of the primary and secondary coils are almost equal to 0 in case of ideal transformer. Now an ideal transformer is unity coupled lossless transformer in which primary and secondary coils have infinite self inductances. The circuit symbol of transformer we generally show like this where we have the coil 1 and 2 and have the turns ratio n 1 n turns number of turns in primary coil as n 1 number of turns in secondary coil as n 2 and the double lines in between two coils shows that the core is present in the transformer. Now, primary winding is having n 1 turns and secondary is having n 2 turns. So, what we will do now we will apply the sinusoidal voltage in the transformer circuit which we just saw and the magnetic flux in that case generated will goes through the both of the windings. So, you can simply apply the faraday's law to find out the voltages V 1 and V 2 when you apply voltage in the primary coil. So, what will happen you can say V 1 that is induced voltage equal to n 1 d phi by d t because this flux phi is linking both of the windings. So, V 1 will become now n 1 into d phi by d t and induced voltage in secondary coil V 2 will be n 2 into d phi by d t. Now, if you divide both of the equations we get V 2 by V 1 is equal to n 2 by n 1 this will be equal to n which we just derived which was the turns ratio of or the transformation ratio of the transformer. So, if in case of instantaneous voltages that is V 1 V 2 if you write phasor voltages that is capital V 1 and V 2. So, we can simply say that V 2 by V 1 is equal to n 2 by n 1 which is nothing but the turns ratio of the transformer. Now, the if for the reason of power conversion power conservation the energy supplied to the primary should be equal to energy absorbed by the secondary. So, when you follow the law of energy conservation the value that is V 1 I 1 should always be equal to V 2 I 2 in case of ideal transformer. So, what we can write now. So, if you write like V 1 upon V 2 will be nothing but I 2 upon I 1. So, if you write it in the phasor form you can write V 1 upon V 2 is equal to I 2 upon I 1 V 2 upon V 1 we have just now found that is equal to n 2 by n 1 which is nothing but equal to 1 by n. So, what we can write we can write that I 2 by I 1 is nothing but equal to n 1 by n 2 or equal to 1 by n. So, when n is equal to 1 we say transformer is called as a isolation transformer when n greater than 1 it means that the we have the step up transformer and when n is less than 1 it is called step down transformer. What is a step down transformer? A step down transformer is the one whose secondary voltage is less than the primary voltage whereas in case of a step up transformer the secondary voltage is greater than its primary voltage. So, now it is important to get the proper polarity of the voltages and directions of the currents for the transformer. So, if the polarity of V 1 or V 2 or the direction of I 1 or I 2 is changed the value of n which we have just saw may need to be replaced by minus n. So, how to find out whether we should use plus n or minus n we will simply use the following rules. What are the rules? If V 1 and V 2 are both positive or both negative at the dotted terminals then we will use plus n in the transformer voltage equation otherwise we will use minus n. If I 1 and I 2 both enter into or both leave the dotted terminals we will use minus n in the transformer current equations otherwise we will use plus n. So, if we follow these two rules the possible combinations for voltage and current are four. So, we can have four possible combinations of circuits for the ideal transformer. Let us see what are those four circuits. So, in first if you see V 1 and V 2 both are positive at the dotted terminal. So, we can say V 2 by V 1 is equal to plus n 2 by n 1. Now, since current I 1 is going inside and current I 2 is going outside the dotted terminal that is why I 2 by I 1 is also plus n 1 by n 2. Now, in this case the voltage terminals are positive at the dotted side of the terminal. So, we write V 2 is equal to n 2 by n 1 V 2 by V 1 is equal to n 2 by n 1, but now the direction of current is opposite n 1 and I 1 and I 2 both are going inside the dotted terminal. So, I 2 by I 1 will now become minus of n 1 by n 2. Now, in the third case the value of the voltages in left side the dotted terminal has negative polarity of the voltage and at the right side you have positive polarity of the voltage. So, we will write V 2 by V 1 is equal to minus of n 2 by n 1. Now, in case of current in both of the coils if you see in this case current I 1 is going outside the dotted terminal and I 2 is going inside the dotted terminal. So, I 2 by I 1 will become plus of n 1 by n 2. Now, in fourth case your voltages are having the same polarity as we saw in this case where V 1 is positive at the dotted terminal while V 2 is negative at the dotted terminal. So, we can write V 2 by V 1 is equal to minus of n 2 by n 1 while in case of current if you see current is going I 1 is going inside the dotted terminal at the same time I 2 is also going inside the dotted terminal because if you trace the path of the current I 2 will go inside the dotted terminal from here and come out from the upper side. So, in that case what will happen you will write I 2 by I 1 equal to minus n 2. So, with this we can close our today's discussion in which we discussed about the energy stored in magnetically coupled circuits and also discuss about the ideal transformer and their various connections and the polarity. We will continue our discussion related to the ideal transformer also. Thank you.