 This is the 12th lecture on Natural Response of Electrical Circuits. We have already defined the various kinds of responses that an electric circuit, electrical circuit can have and we have said that natural response is the one which arises due to internal stored energies and the circuit being left to itself to settle. There are no external forces or external excitations. If there are external excitations then the response that we get is called the forced response and in general, in general a circuit may have excitations, voltage sources or current sources are both and also internally stored energies like electrostatic energy in capacitors and electromagnetic energy in inductors and therefore in general the response of an electrical circuit shall be a combination of natural response and forced response and then two other terms which we introduced where the transient response that is the response that occurs at the beginning of excitation and after a long time has passed the response that you get from an electrical circuit is called the steady state response. So there are 4 kinds of responses, natural, forced, transient and steady state. Sometimes natural response and transient response are wrongly identified with each other. They are not necessarily the same. Similarly the forced response and the steady state response are also not necessarily the same. The previous day I had also introduced to you terms like order of a circuit and we said that a first order circuit is one which has effectively one energy storage element that is either one inductor or one capacitor alright. There may be more than one capacitors or there may be more than one inductors trivially connected to each other which behave as a single energy storage element and one of the examples that I gave was something like this. If you have let us say 3 capacitors like this and there is a resistance here then you know these 3 capacitors behave like a single capacitor. It would be C1 plus C2 C3 divided by C2 plus C3 and therefore this is the effective capacitance connected across the resistance R and this behaves as a first order circuit with an effective time constant of this total effective capacitance multiplied by the effective resistance across it. Similarly there may be more than one resistor here but effectively they behave like one resistor. Similarly we can have a resistance inductance circuit which will also be a first order circuit or a multiplicity of inductors connected in a trivial manner so that they could be reduced to one effective inductance then that is also a first order circuit and as an example I had taken this circuit. Let us say we have a battery V and a resistance R which is connected to let us say an inductor L in series with a resistance R alright. The switch is left in this position for a very long time so that the current that passes to the circuit would be given by V by R1 plus R. The inductance is not effective in dropping a voltage when the current is steady. So in the steady state the current in the circuit shall be given by V by R1 plus R and let us call this as I0. At t equal to 0 this switch as a connection here at t equal to 0 this switch is moved in the downward direction alright this is how we indicate this. At t equal to 0 this switch is moved to the downward position so that the initial current in the inductor, the inductor was carrying a current of I0 now the inductor and the resistor in series they are connected across a short circuit and therefore therefore the circuit will have a natural response. It is an example of a first order circuit having a natural response that is we are interested in the behaviour of the circuit at t greater than 0. The initial condition is that if I consider this current as I then I of 0 minus that is just before the switch is closed is equal to I0, I of 0 minus just before the switch is thrown in the other position I of 0 minus this is the indication this is the significance of the minus sign here is I0 and you know the energy of an inductor cannot change instantaneously and therefore this will be I of 0 plus also alright and you can see you can see that as time passes the inductor shall lose its energy to the resistor. The resistor will dissipate that energy and gradually this current should diminish with time and at infinity at time t equal to infinity the current should diminish to 0. The differential equation that one can write is simply application of KVL that the drop across the inductance plus the drop across the resistance should be equal to 0 because there is a short circuit and therefore the drop across the inductance is LDI DT the drop across the resistance is RI and this should be equal to 0. This is a first order differential equation and this is what defines the order of a circuit. If a circuit natural response can be described by a first order differential equation then it is a first order circuit. This is not the only one. We could have a capacitor resistor circuit for example let us say we have a battery V0 connected like this connected across a capacitor C and then there is a resistor waiting to receive energy from the capacitor C. This is the capacitor C and then we know that if the switch if the switch is left in this position for a long time in this position for a long time then the capacitor voltage V sub C if we call this V sub C with this polarity the capacitor voltage V sub C at 0 minus shall be equal to V0 it will assume the voltage of the battery and if the switch is thrown to the other position at T equal to 0 then the capacitor which has stored energy will now try to send a current through R. Let this current be I alright. It will try to send a current through R and in the process the capacitor will discharge or it will lose its energy. Gradually the this energy will be lost in the resistance as heat dissipation and gradually as time increases the current will decrease the voltage across the capacitor will decrease and it will take infinite amount of time for the capacitor to completely relax that is completely lose the energy that it has contained and once again since a capacitor cannot change its energy instantaneously the voltage across the capacitor at 0 plus shall also be equal to V0 and if you see the differential equation if you see the equation describing this once again once again it is an application of KVL or common sense alright it says that if the current IR the drop across the resistance IR should be equal to the drop across the capacitor that is V sub C. Now what is V sub C? V sub C had started with V0 at T equal to 0 and as time proceeds this current I it is important to understand this. This current I has to flow in a closed circuit and therefore this current I charges the capacitor in the opposite direction which is equivalent to saying that the capacitor is losing charge. What is this charge? It is equal to 1 by C times the charge integral 0 to T I dt. This is the equation is this is this equation clear? The charge the voltage across the capacitor at any instant of time would be the initial voltage minus the voltage that it would have acquired in the opposite direction because of the current I flowing in a closed loop. Now this is not a differential equation it is an integral equation in general in general we may get an integral differential equation. In the previous case in the LR case it was an differential equation pure differential equation in the RC case it is a pure integral equation nevertheless one can make it into a differential equation by one more differentiation RDI dt plus I by C if you differentiate this it simply I over C that would be equal to DV0 dt is 0. So this is also this is also as you can see a first order differential equation and therefore this circuit is also a first order circuit. Let us see what kind of solution you get you are acquainted with the solution to such differential equations but in the context of electrical circuits we wish to bring in some more physical concepts and I want you to pay attention to this. Let us take the second circuit first where our differential equation was RDI dt plus I by C equal to 0. Now the right hand side of this equation is 0 which means that this is a homogeneous differential equation homogeneous differential there is no forcing term well there cannot be a forcing term because we are investigating natural response it is quite logical that our right hand side is 0 that this is a homogeneous differential equation of the first order first order homogeneous differential equation. Now in order to solve this equation we know mechanically how we do it you put an integrating factor and things like that or you take the this quantity to the right hand side and integrate change of variables and so on. You know all this we shall do it in a slightly different manner which is more physical in character which throws more light into the physical happening in the circuit. We argue that any solution to this differential equation has to satisfy this D I dt equal to minus I by C any solution has to satisfy this which means that D I dt shall be equal to minus 1 by R C times I all right. So what you usually do is bring I here D I by I minus 1 by R C dt that you integrate both sides and so on but no we are not going to do that. What we are going to argue is now comes the physical picture into the into the into the situation what we argue is that my solution must be of a form such that D I dt cancels I by C or D I dt and minus I by R C must be of must be identical and a function which amplifies eminently qualifies for this kind of a job is the exponential function. You know that the exponential function the differential coefficient is again an exponential function and therefore you argue that it must be of the form e to the power some constant s small s times t. Time has to be there time is the independent variable it is an exponential function of time it could be a constant multiplied by time and then we argue that the in general the solution must be of this form some arbitrary constant multiplied by e to the s t. So is this is this physical reasoning clear that the solution must be of this form a e to the s t a is an arbitrary constant which will be determined by the initial conditions or the boundary conditions of the problem. Exponential function we took because D I dt D dt of this must be of the same form as I same form of the function and therefore it must be an exponential function. Now if I substitute this in the original equation then what I get is our original equation is R D I dt plus I by C equal to 0 and I substitute I equal to a e to the s t this is called a trial solution we try a solution of this form then I get R A s e to the s t plus a by C e to the s t equal to 0 where a e to the s t a e to the s t cannot be equal to 0 because this is the form of the solution if this is 0 then of course the current is identically equal to 0 current identically equal to 0 obviously satisfies the equation that cannot be a solution because we know that if the capacitor has an initial energy it shall send a non-zero current and therefore I cannot be 0 therefore a e to the s t cannot be 0 the only other alternative is that R times s plus 1 by C should be equal to 0 which means that s should be equal to minus 1 over R C alright s should be equal to minus 1 over R C in other words the solution to the equation shall be of the form a e to the power minus t divided by R C and you notice that R C the product R C must have the dimensions of time and therefore we denote this as a time constant and we say a e to the power minus t by capital T the only problem that remains is to find out capital A alright that is found out from the initial condition you know that if we if we draw the circuit again it is a C in parallel with R this is I and you know that V C 0 plus is the same as V C 0 minus we recall that this is equal to V 0 and therefore what is I of 0 plus it must be V naught by R our solution is a e to the power minus t by t so I of 0 plus is obtained by putting t equal to 0 which means that a must be equal to V 0 by R and therefore the total solution is V 0 by R e to the minus t by t alright now let us call this as I 0 e to the minus t by t okay now the the if you plot this current versus time if you plot this current versus time then it starts at I 0 it starts at I 0 and falls exponentially and you know how to define the time constant that it is the time after which the current has fallen by how much 1 by e of its original value and so on and so forth the point is the current is decaying exponentially it is only at infinite time small t equal to infinity that the current shall be minus to 0 what is the charge that flows after all the capacitor started with a certain amount of charge which was equal to C times V 0 alright so how much charge does the capacitor lose in the process alright that obviously will be obtained by integrating this over from 0 to infinity now so the charge that is passed shall be equal to I 0 integral 0 to infinity e to the minus t by t dt and it is very easy to see that this is simply equal to I 0 times capital T okay you integrate this and put the limits I 0 times capital T I 0 is V 0 by R and capital T is RC and you can see that this is precisely the total charge that has that the capacitor loses in the process of sending a current to the resistance is precisely the charge that it started it so it loses its whole charge is that clear this is the physical reasoning that I wanted to bring into the solution of the differential equation you should not do it blindly must have the physical picture in mind in an exactly similar manner we can take care of the inductance resistance circuit L R which is short circuited with the current I here I of 0 minus equal to I of 0 plus equal to let us say I 0 and the differential equation satisfied by the circuit is L di dt plus R i equal to 0 if you proceed exactly similarly that is you put small i equal to a e to the power s t incidentally what should be the dimension of s 1 by t second to the minus 1 and 1 by t is a frequency and therefore small s is called a frequency will come to this a little later the physical interpretation of small s and you will see how nicely it tunes with physical concepts of frequency but if you try a solution of this form and obtain and substitute this initial condition that is I 0 equal to I 0 you shall be able to show that I t is exactly of the same form that is I 0 e to the power minus t by capital t where capital t is now equal to R upon L instead of RC it is R upon L all right and the same kind of argument about the total charge that passes in the circuit if you integrate this it would be I 0 times capital t this also incidentally brings in another definition of capital t is that right what is the what is the definition another definition of capital t one definition of time constant is the time required for the current to drop by 1 by e th of its value at t equal to 0 another definition could be in terms of charge that it is effectively effectively the initial current if the initial current had passed for a time equal to capital t then the charge that is lost would be equal to the actual charge lost by the capacitor or actual charge flowing in this circuit in this circuit for infinite amount of time is that clear let us go back here time constant you see can be written as t equal to V 0 C divided by I 0 if I go back to the original problem why did it take so long thank you this is the correct story all right now an interpretation of the time constant it is more it is more evident in the capacitor resistor circuit the capacitor started with a total charge of V 0 V 0 C and you see if it if it had sent a current at the rate if it had sent a constant current I 0 then it requires only and a time equal to capital T to lose the whole charge so you can define capital T as that amount of time is the is the is the time required for the capacitor to lose its charge if it maintains a constant current at the initial value this is another interpretation or definition of the time constant now enough about the first order circuit let us look at the second order circuit let us consider a capacitor C which is charged to which is charged to a voltage V 0 all right and then and then at let us say at let us say t equal to 0 this capacitor is first charged to a voltage V 0 maybe by connecting a battery across this a battery of voltage V 0 it has in the circuit in the circuit and inductance and resistance and a switch if the switch is open then the capacitor is sits pretty it does not have to lose charge but suppose the switch is closed at t equal to 0 the switch is thrown into the lower position at t equal to 0 then obviously the capacitor cannot sit pretty it shall have to lose charge a current shall flow like this a current shall flow like this capacitor is charged to this plate positive so this charge will effectively behave like a battery you remember the Thevenin equivalent circuit which will tend to send a current in this direction and as time passes as time passes the current this current I has to pass through the capacitor in the opposite direction which means that the capacitor gets charged in the opposite direction which is equivalent to saying that the capacitor loses charge and as time proceeds it loses more and more charge and therefore the current gradually diminishes the charge diminishes and after infinite amount of time the circuit will again be relaxed that means it shall have no initial energy. Let us look at the equation what kind of equation does this circuit obey? Obviously once again it is a simple application of KVL and you notice that the inductor voltage L di dt plus R i this should be equal to V sub C but V sub C starts with V 0 then this charges that is the current charges in the opposite direction 1 by C integral 0 to T i dt this is the equation that is obeyed by the circuit and you notice that this is an integral differential equation with the initial conditions that what is the initial condition i at 0 minus is equal to 0 and since this is a current through the inductance which cannot change instantaneously i of 0 plus shall also be equal to 0. One initial condition and the other is that V C 0 minus is equal to V 0 is equal to V C 0 plus alright this is the other initial condition. Now an integral differential equation as I have told you is always converted it can be converted into a differential equation by one more differentiation we want to get rid of the integral sign so what I get is L d 2 i dt 2 plus R di dt and in the process d dt of V 0 becomes equal to 0 and therefore we get simply i by C which taken to the left hand side becomes 1 by C times i equal to 0. If you notice this equation is also a homogeneous differential equation of the second order and therefore this circuit is called a second order circuit and we are indeed talking of natural response. The initial energy in this circuit comes from the capacitor voltage it could also instead of the capacitor voltage it could also have come from an initial electromagnetic energy in the inductor for example the inductor could have a current of i 0 and then t equal to 0 t equal to 0 it is connected across the capacitor and the resistor exactly the same way and we would have the same kind of differential equation a second order differential equation thus defining a second order circuit. Let us write this equation again L d 2 i dt 2 plus R di dt plus i by C equal to 0 I want to make a point regarding the initial conditions can you go back to the previous slide before I pass to the to the solution of the equation this you see what am I going to do is to solve this solve this differential equation for i all right. So i of t and i of t I shall obtain in terms of unknown constants which I shall try to evaluate from the initial conditions one of the initial conditions is that i of 0 is equal to 0 i of 0 minus and 0 plus are the same. The other initial condition I say it is not on i it is on the capacitor voltage can we convert it to a condition on i this is very illuminating by looking at this equation this equation all right. Let us consider t equal to 0 plus at t equal to 0 plus what is the value of i it is 0 di dt is not 0 not necessarily i can be 0 but it can be rising so and what about this integral 0 and therefore do not you see that my initial condition is L di dt at t equal to 0 should be equal to v0 and therefore di dt at t equal to 0 should be equal to v0 by L this is the other initial condition. So there are 2 initial conditions it is a second order circuit it stands to reason that you have to solve the second order differential equation you require 2 initial conditions all right and the 2 initial conditions are that the initial current is 0 and that the initial rate of change of current is simply the initial voltage across the capacitor divided by the inductor the resistance does not come into play anywhere in this why not because i of 0 is 0 so there is no drop in the resistance at this at this 2 initial conditions clear can we go ahead to the solution of the circuit once again once again we shall proceed from a physical reasoning rather than integrating directly physical or using an integrating factor as we do in mathematics we do not want to lose sight of the physical situation and so we say all right we notice that i di dt and d2 di dt2 must be of the same functional form otherwise the sum cannot be 0 if one is a sign the other is a cosine and the third is an exponential obviously it cannot it cannot cancel and therefore we argue once again that i must be of the form of A e to the st A e to the st now if i substitute this here and clear a e to the st a e to the st cannot be 0 clear a to the st then what we will get is s squared l plus s r plus 1 over c shall be equal to 0 do you see this i have omitted a couple of algebraic steps i simply substitute this take 2 differentiation i shall get s times r here and s squared time cell a e to the st i cancel because it cannot be equal to 0 and this equation yes that is a t equal to 0 a 0 at that is a t equal to 0 okay all right we shall apply that boundary condition later all right this is a first start this is not a first order circuit so this cannot be the complete solution we are trying the form of the solution we say is this does this qualify as a form of the cell in the let me clarify this in the first order circuit we are trying a complete solution we know the solution has to be of the form a e to the st here obviously since it is a second order circuit the solution must be more complicated than a e to the st because at t equal to 0 as you rightly pointed out since i of 0 plus is equal to 0 a has to be identically equal to 0 so this cannot qualify as the total solution but does this qualify as a form of the solution this is what we wish to find out and the physical situation will immediately bring in the total solution as you see here as you see here this equation s r by l plus 1 by l c equal to 0 this is a quadratic equation so there are 2 values of s which satisfies the equation there are 2 values of s originally it was only in the first order circuit it was only 1 value of s here there are 2 values and these 2 values are s 1 2 you know you know what this means subscript 1, 2 that we write both of them simultaneously you can see that this is minus r by 2 l plus minus square root of r by 2 l whole square minus 1 over l c there are 2 possible values of s and therefore there are 2 possible solutions of the original differential equation and these solutions are my equation is l d 2 i d t 2 plus r d i d t plus i by c equal to 0 and the 2 solutions are minus r by 2 l plus minus square root of r by 2 l whole square minus 1 over l c and therefore the solutions are e to the s 1 t and e to the s 2 t both of them shall satisfy the original differential equation and multiplying by any arbitrary constant will not affect the solution and therefore we say these constants are let us say a 1 and a 2 both of them individually satisfy the original differential equation so the sum of them should also satisfy the original differential equation and this qualifies as the general solution to the original differential equation. We shall not apply the initial conditions till we reach this point till we get the general solution to the equation a e to the s t was a trial solution we see that indeed it is of the correct form but it does not satisfy the initial conditions therefore why is it so because it is a second order circuit and therefore there must be 2 linearly independent solutions is this phrase known to you what does linear independence mean that by multiplying this by a constant I cannot get the other solution you see e to the s 2 t cannot be obtained from e to the s 1 t by simply multiplying by a linear constant all right these are 2 independent solutions and now we can apply the initial conditions the first initial condition is that I 0 is equal to 0 that is a 1 plus a 2 should be equal to 0 which tells me that a 1 should be equal to minus a 2 all right in other words the general solution therefore shall be of the form a 1 e to the s 1 t minus e to the s 2 t a 2 is simply minus a 1 and the other initial condition is that d i d t at 0 should be equal to v 0 by L which means that a 1 now make a d d t and put t equal to 0 d d t of e to the s 1 t is s 1 e to the s 1 t and if you put t equal to 0 then it is simply s 1 minus similarly s 2 is that clear I have omitted the differentiation step and therefore a 1 is equal to v 0 by L s 1 minus s 2 is that okay all right and therefore therefore my total solution i of t is equal to v 0 by L s 1 minus s 2 times e to the s 1 t minus e to the s 2 t this is the total solution total solution to the equation now so far i have not said anything about the nature of these 2 quantities s 1 and s 2 if you recall s 1 and s 2 are given by minus r by 2 L plus minus square root of r by 2 L whole square minus 1 over L C the nature of these 2 solutions obviously these 2 constants obviously shall depend on the discriminant that is d which is r by 2 L whole square minus 1 over L C if d is greater than 0 then obviously the 2 roots shall be real distinct anything else that you can say negative is it not right minus r by 2 L plus a quantity which is less than r by 2 L obviously it will be negative quantity when the sign is negative minus r by 2 L minus obviously the whole thing is negative so all are distinct these are real distinct and negative if d is equal to 0 obviously that is possible by a combination of r L and C you could make them equal to 0 they are real coincident we do not say equal they coincide they coalesce upon each other and they are still negative okay if d is less than 0 these roots are complex why complex because there is a real part and d is less than 0 so this will be j times okay square root of minus 5 is j times square root of 5 where j is equal to square root of minus 1 alright so you will have minus a real quantity plus minus j times another real quantity so it will be complex distinct we can say positive or negative conjugate here suppose r is equal to 0 then the roots are not only complex they are purely imaginary they are purely imaginary because this is 0 this is 0 square root of minus 1 over L C so purely imaginary to emphasize you say purely okay purely imaginary distinct and conjugate we shall talk about this pardon me there of course conjugate in the case of repeated roots in the case of repeated roots as you know in the differential equation we have to have a 1 plus a 2 t multiplied by e to the multiplied by the exponential but we will not bother about that we will go by this and see what exactly comes from the physical picture alright we will look at it from more a physical point of view because it is introduction to electronic circuits that we are learning not solution to differential equation this is incidental we have to solve the equation to be able to find a solution let us consider this case d greater than 0 that is roots are S 1, S 2 are real distinct and negative alright and you already know that the current solution is equal to V 0 by L S 1 minus S 2 e to the S 1 t minus e to the S 2 t, d greater than 0 means that R L and C are such that this quantity R by 2 L whole square let us look at this this is also interesting d greater than 0 means what that R by 2 L whole square should be greater than 1 by L C which means that R should be greater than R square should be greater than 4 L by C is that right that is R should be greater than 2 L by C square root of L by C there are 2 things I want you to notice that it depends on the total resistance in the circuit okay if the total resistance in the circuit is large compared to is greater than twice square root of L by C well the roots will be real and distinct if you if you take a resistance at a restate for example in the circuit and gradually reduce it then you can see all the phenomena that is you can see real effect of real distinct negative roots effect of real coincident negative roots effect of complex roots and effect of purely imaginary roots if you simply go on adjusting this resistance when it is 0 you will see completely imaginary roots and so on alright the other thing that I want you to notice is that square root of L by C obviously has the dimension of resistance is that right so if you wish to express the dimension of L in terms of the dimension of C and R square then obviously it would be this right L would be equal to C R square now let us take a numerical example for this case let us suppose about what how does the circuit behave this is what I am going to show you with numerical example we have an example in which let us say L equals to 1 Henry C equal to half Farad and R equal to 4 ohm alright if this is the combination of the elements then you see S 1 2 is minus R by 2 L so it would be minus 2 is that okay 4 divided by 2 into 1 minus 2 plus minus square root 4 minus 1 over L C 1 over L C is a bit mistake let us take with you pardon me let us take C as 1 third Farad I want to keep life simple alright let us take C as 1 third Farad then it will be 4 minus 3 which means that it is minus 2 plus minus 1 the roots are real distinct and negative so the roots are minus 1 and minus 3 and therefore I of t is equal to V 0 by L is 1 S 1 minus S 2 is obviously 2 is that right minus 1 plus 3 minus 2 and then e to the minus t S 1 is minus 1 and S 2 is minus 3 so e to the minus 3 t okay let me write it down again I of t is equal to V 0 by 2 e to the minus t minus e to the minus 3 t how do you think the current will behave with time how shall it vary with time obviously at 0 minus the current was 0 at 0 plus the current will be 0 put t equal to 0 it is 0 at infinity at infinity when t goes to infinity this term goes to 0 this term goes to 0 0 minus 0 is 0 so at infinity it must again be 0 let me show infinity on a finite plane okay by a break let us say this is the point at infinity so infinity is also 0 in between the current is positive it does not change direction it cannot be 0 and therefore it must have a maximum it must have at least one maximum in a higher order circuit it can have more maximum minimum alright so it must have at least one maximum so it must go to a maximum and then goes to 0 we show a break because we are showing infinity on a finite plane alright so the current attains a maximum value somewhere and one can by differentiating this find out at what value of time at what value of time the current attains a maximum and then what happens one can also find out the total charge that is delivered to the circuit what do you think will be the total charge it will again be C times V naught that is one third V naught if you integrate this from 0 to infinity you shall get the same expression on Monday we will consider the case of a complex route that is D less than 0 and we will have fun more fun there.