Hey friends, welcome to the YouTube channel all about electronics. So in this video, we will see the transient response for the ILC circuits. Now, couple of videos back, we had seen the transient response for the RL as well as the RCA circuits.
So, now in this video, we will see the transient response for the series as well as the parallel RLC circuits. Now, the study of this transient response for the RLC circuit is very important. The reason is that these RLC circuits are used in many applications. Like these RLC circuits are widely used as a tuning circuit in the radio communications. Apart from that, these RLC circuits are also used as a voltage multiplier.
as well as as a passive filters. So first of all, let's study the transient response for the series RLC circuit. So as you can see here, we have a resistor, capacitor and inductor are connected in a series. And a DC voltage is applied to this circuit at time t is equal to zero. And before this time t is equal to zero, there is no energy stored across any element.
That means all initial conditions are zero. So now let's redraw the circuit at time t is equal to zero. when this switch is getting closed. So, at time t is equal to 0+, if you look the circuit, it will look like this.
And let's say the current I is flowing through this circuit. So first of all, let's apply Kirchhoff's voltage law in this circuit. So applying KVL, we can write V is equal to Vr plus VL plus Vc, where Vr, VL, and Vc are the voltage across these three elements respectively. Now, the same current I is also flowing through this capacitor.
So, we can say that I is equal to Ic that is a current that is flowing through this capacitor C. And we know that the capacitor current C can be given as C into dVc by dt. Now, we can write this Vr that is equal to I into R as R into C dVc by dt. And similarly, we can write across the inductor VL as LxDi by Dt. So, now if we put the value of I in this equation, then we will get LxCxDvc divided by Dt square.
Let us say this is the equation number 1. So, now if we put all these values inside this equation number 1 and rearrange it, then we will get. d square Vc divided by dt square plus R by L into dVc by dt plus 1 by Lc into Vc that is equal to V over Lc. So, as you can see here, it is a second order linear differential equation. Now, as we had seen earlier, the total solution of this linear differential equation contains two terms, the complementary function and the particular integral.
Now, this complementary function is nothing but a transient response of this circuit, while this particular integral is nothing but a steady state response of this circuit. So, now in this circuit, if you see, at equal to infinity, the voltage across the capacitor is nothing but a voltage VV. So, we can say that Vc at time t is equal to infinity is nothing but a VV.
So, this will be the steady state response of this circuit at time t is equal to infinity. So, now let's find out the transient response or the complementary function for this equation. So, for that we will consider this term as a zero. That means there is no excitation in the circuit. And if we rewrite the equation, then we can write it as d square Vc over dt square plus R by L into dVc by dt plus Vc tiered by Lc that is equal to zero.
Let's assume that here d by dt is nothing but a capital D. So, the characteristic equation can be written as d square plus R by L into D plus Vc over Lc that is equal to zero. Then we will get two roots.
Let's say D1 and D2 are the two roots of this equation. And these roots can be given as minus R over 2L plus or minus. R by 2L whole square minus 1 over LC. And these roots are came from this basic equation. That is, minus b plus or minus under root b square minus 4ac over 2a.
Which is the roots for the equation ax square plus bx plus c that is equal to 0. So, if we compare this equation with this equation, then we will get these two roots. Now, let us say that this α that is equal to r over 12. So, let us say this r over 12 is nothing but α and this term which is in the under root is let's say it is a β. So, β is nothing but a under root. R over 2L whole square minus 1 over LC. So, we can say that d1 is nothing but minus alpha plus beta and d2 is nothing but minus alpha minus beta.
Let's understand one more thing about this RLC circuit. So in the RLC circuit, this LNC, that means the inductor and capacitor induces the natural oscillation in the circuit. That means they have a tendency to induce the oscillation inside the circuit.
While this resistor have a tendency to damping out these oscillations. So this LNC will induce the oscillation inside the circuit and this resistor will try to suppress this oscillation. So, depending on the value of this Lc and R, the response of the circuit can be found out. So, let's say the oscillation that is created by this Lnc is ω.
And ω can be given as 1 over under root Lc. And the frequency of oscillation is known as the natural frequency. Now, earlier we had assumed that The value of R by 2L is nothing but alpha.
And this alpha is also known as the damping coefficient of the circuit. Because this coefficient decides how well the circuit is able to dampen the oscillations. Apart from that, the ratio of this alpha over omega that is known as the zeta. And this zeta can be given as R by 2 under root C over L.
And this zeta is known as the damping factor of the circuit. So, this is basically nothing but a normalized damping coefficient. So, the value of this zeta decides how the circuit responses to the different excitations. So, earlier we had found the two different roots which can be given by this equation.
Now, here there are four different possibilities. So, first of all, let's take the first case in which this R by 2L whole square is greater than 1 over LC. So, in this case, All the roots will be real and negative.
So, we will get d1 as minus alpha plus beta and we will get d2 as minus alpha minus beta. Where alpha is nothing but r over 2L and beta is nothing but this whole term. So, now if we see the solution of this differential equation, it can be given as Vc that is equal to C1 e to the power minus d1 into T.
plus C2e-d2t, where d1 is nothing but minus alpha plus beta and d2 is nothing but minus alpha minus beta. So, the first case where this R over 2L whole square is greater than 1 over LC, then the response is known as the overdamped response. And we will see more about it graphically once we discuss all the four different cases.
So, now let us take the second case in which The R by 12 whole square that is equal to 1 over LC. So, in this case, the whole term will get 0. And we will have two real and equal roots. That means, D1 is equal to D2 is equal to minus α. And in such case, the solution Vc can be given as C1 plus C2t into e to the power minus α into t, where C1 and C2 are the coefficient.
And this coefficient can be found out using these initial conditions. And we will see about it when we will solve some problems based on these RLC circuits. So now, in the second case, where this R over 2L whole square is equal to 1 over LC, then in such case, the response of the circuit is known as the critically damped response.
So now, let's see the third case. So, in the third case, let's say the value of this R over 2L whole square is less than 1 over LC. So, in that case, if you see. The term inside this square root will be negative.
So, we will have two complex conjugate roots. So, D1 can be given as –α Jβ and D2 can be given as –α Jβ. So, we will get these two roots. And the solution Vc can be given as e to the power –α T.
into C1 cos and once again we can find the value of this C1 and C2 by using the initial conditions. So, this third case where the value of R by 12 whole square is less than 1 over LC, then the response of the circuit can be known as the underdamped response. So, now let's consider the fourth case. The fourth case, let's assume that the value of this resistance is zero.
So, in that case, the roots will be purely imaginary. The value of D1 is nothing but J into underroot 1 over LC. And we had seen that this under root 1 over else is nothing but omega.
So, we can say that the value of D1 is nothing but j into omega. Similarly, the value of D2 is nothing but minus j omega. So, this will be the two roots when the r is equal to zero. And the solution Vc can be given as C1 into cos plus C2 into sin. So, as you can see here, When the value of R is zero, then circuit will not provide any kind of damping.
The response of the circuit will have pure oscillations. So now, let's graphically represent these over-damped, under-damped, and critically damped responses. So, let's assume that to the circuit, we have provided a unit step function at time t is equal to zero.
Or we can say that at time t is equal to zero, we have applied some voltage to this circuit. And let's see the response of the circuit for the three different cases. So, if you see the over-damped response of the circuit, it will look like this.
So, as you can see here, the response of the circuit is very sluggish. And it will take a long time to reach its final value. Similarly, if you see the critically damped response, then it will look like this.
So, if you see the critically damped response, the critically damped response will be faster compared to the over-damped response. And in this critically damped response also, you will not find any kind of oscillations. And in this critically damped response, the circuit is able to reach its final value faster compared to the over damped response. So now, let's see the third kind of response that is the over damped response. Now in case of over damped response, you will see some oscillations into the response.
And how fast this oscillation dies out, it depends upon the value of R, L, and C. So in case of this under damped response, the circuit is able to reach its final value very fast. But it will take some time for settle down to its final value. And this settling time depends upon the value of R, L, and C. So, these are the three typical responses for the over-damped, critically-damped, and under-damped responses.
So, as I said earlier, over-damped response takes a long time to reach its final value. While in case of critically-damped response, it is able to reach its final value faster compared to the over-damped response. While in case of under-damped response, it reaches fastest to its final value. But after reaching the final value, you will see some overshoots.
And these overshoots will get settled down after the settling time. And the settling time depends upon the value of this R, L, and C. So, these three are the typical responses of the RLC circuit. And depending on the value of R, L, and C, for the given excitation, we can have any of the three responses in our circuit. So, now as we have understood about a series RLC circuit, now let's see the transient response for the parallel RLC circuit.
So, in this circuit, at time t is equal to zero, this current source I is applied to this circuit. And before this time t is equal to zero, let's assume that all the initial conditions in our circuit is zero. That means there is no charge or no energy across any of the elements inside the circuit.
So, now for finding the transient response for this parallel RLC circuit, let's find redraw the circuit at time t is equal to 0. So, at time t is equal to 0+, the circuit will look like this. Let's assume that at this node, the voltage is VV. So, applying KCL, we can write I is equal to 2 over R plus 1 over L into Vdt plus C dVc by dt. Well, this second term is nothing but the current that is flowing through this inductor L and the third term is nothing but the current that is flowing through this capacitor C. So, now if we differentiate this equation, then we will get C into d2V by dt square plus 1 over L into V plus 1 over R into dV by dt is equal to Now, again if we rearrange this equation, then we will get d2v by dt square plus 1 over rc into dv by dt plus 1 over lc into v that is equal to 0. And if we Write the characteristic equation then we can write it as d square plus 1 over RC into d plus 1 over LC that is equal to 0. So, now again this equation will have two roots.
Let us say d1 and d2 are the roots of the equation. And the roots can be given as minus 1 over 2 RC plus or minus under root 1 over 2 RC whole square minus 1 over LC. Let's assume that this α is nothing but 1 over 2RC that is this term and this whole term is let's say β.
So, β can be given as under root 1 over 2 RC whole square minus 1 over LC. So, we can write d1 as minus alpha plus beta and d2 as minus alpha minus beta. So, now let's once again see the 4 different cases.
So, in the first case, let's assume that this 1 over 2 RC whole square is greater than 1 over LC. So, once again we will have a real as well as the negative roots. So, we will get d1 as minus alpha plus beta and d2 as minus alpha minus beta where alpha is 1 over 2 Rc and beta is this whole term. And if you see the solution, this Vc once again can be given as C1 into e to the power minus d1 t plus C2 into e to the power minus d2 t. So, for the first case, once again it is a case of an over-damped response.
Similarly, in the second case, if the two values are equal, then the roots will be real and equal. So, we will have d1 as minus alpha and d2 as also minus alpha. And the solution? can be given as V that is equal to C1 plus C2 into e to the power minus alpha into T.
Well, any of you, if you don't know how we have arrived to this equation, then I will provide a separate link in the description for the mathematics which is involved for finding this solution. So, now for the second case, the response is once again can be known as the critically damped response. Similarly, for the third case where this 1 over 2 Rc whole square is less than 1 over LC, then we will have a complex conjugate roots. And D1 can be given as –α Jβ and D2 can be given as –α Jβ.
So, this will be the two complex conjugate roots for the given equation. In the solution, V can be given as e to the power –α T into C1 cos β T. plus c2 sin. So, this will be the solution for the case when the roots are complex conjugate. And in this case, the response can be known as the underdump response.
So, now in the fourth case, when r is equal to 0, once again you will get imaginary roots. So, the value of d1 is nothing but jω and d2 is nothing but minus jω, where ω is nothing but 1 over under root So, in this case, the roots can be given as V that is equal to C1 cos plus C2 sin. So, in this way, we have seen the four different cases for the parallel RLC circuit as well which is quite similar to the series RLC circuit. So, now let us take a few examples based on the series and parallel RLC circuits.
So, in the first example, we have been asked to find the equation of this current IT. And also, we have been asked to find the value of a current at time t is equal to 0.1. And in this circuit, we have been given that all the initial conditions are zero. So, let's first of all redraw the circuit at time t is equal to zero.
So, at time t is equal to zero, the circuit will look like this. Let's say the current I is flowing inside this circuit. So, if we apply a KVL. Then we can write it as a V. is equal to Vr plus Vl plus Vc.
Now, if we put the value of Vr, it is nothing but I into R. This value of Vl is nothing but L into Di by Dt and value of Vc is nothing but 1 by C integration of Idt. D2i by Dt square plus R by L into Di by Dt plus I over Lc that is equal to 0. So, the characteristic equation can be given as d square plus r by L into d square plus d plus 1 over lc that is equal to 0. Now, if we put the value of r, l, and c inside this equation, then we will get d square plus 9 into D plus if we put a value of C as 0.05, then we will get 20. So, the D square plus 90 plus 20 that is equal to 0. And if we solve, then we will get value of D1 as minus 4 and value of D2 as minus 5. So, this will be the two roots of this equation. So, the solution can be given as C1 into e to the power minus 40 plus c2 into e to the power minus 5t.
So, now we need to find the value of this c1 and c2 by applying the initial conditions. So, first of all, let's apply initial condition at time t is equal to 0+. So, at time t is equal to 0+, this equation can be given as c1 into e to the power 0 plus c2 into e to the power 0. It is nothing but a c1 plus c2. So, now as we have already given that all the initial conditions inside the circuit is zero.
So, the value of I is also zero. That means C1 plus C2 is equal to zero. Or we can say that C1 is equal to minus C2.
So, this will be our first initial condition. Now to get the value of C1 and C2, let's apply a second initial condition. So, now at time t is equal to zero plus.
This inducts the open circuit because it will oppose the instantaneous change of current. So, the entire 20 V will appear across this inductor. So, as we know that the inductor voltage VL can be given as L x Di by Dt.
That is 20 is equal to 1 Henry into Di by Dt. So, first of all to get Di by Dt of 0 plus, let's differentiate this equation. So, if we differentiate it, then we will get Di of Dt 0 plus that is equal to minus 4C1 into e to the power minus 0 plus minus 5C2 into e to the power minus 0. So, we will get minus 4C1 minus 5C2 that is equal to minus 4C1 minus 5C2.
equal to 20. Where 20 is nothing but a value of this di by dt 0+. And earlier we had seen that c1 is equal to minus c2. So, we will get value of c2 as minus 20 and value of c1 as plus 20. So, if we put these values inside this equation, then we will get I is nothing but 20e-4t-20e-5t.
So, this will be the equation of this current I during the transient. Now, we have been asked to find the value of this current I at time 0.1 second. So, for that let's put a value of t inside this equation as 0.1. So, we will get 20e-4t-20e-5t.
0.4-20e⁻⁵. And if we calculate the value, then we will get this value roughly as 1.26A. So this will be the current that is flowing through the circuit at time t is equal to 0.1 second.
Now, for the given example, up to this point it is enough. But for our understanding, let's plot this equation. For plotting this equation, either you can use any cow plotting software or you can also use a MATLAB. Or also you can use any online cow plotting software which are available.
So, if you plot this equation, then it will look like this. So, I have plotted this equation using the online cow plotting website. I will also provide the link for the same in the description. And as we have seen earlier, all the roots are negative and real. So, the response of the system will be the over-damped response.
So, you will not see any oscillation in the response. And at time t is equal to 0.1 second, as we have already found out, if you see, the value of the current is nothing but roughly around 1.26 A. So, this is how this response will look like when you plot the response. So, now let's see the second example based on the parallel RLC circuit.
So, in this example, we have been asked to find the value of this voltage. vt. And we have been given that all the initial conditions inside the circuit is 0 before this time t is equal to 0. So, now let's redraw the circuit at time time t is equal to 0+. So, now in this circuit, let's apply a KCL.
So, we can write it as I is equal to nothing but V over R plus C dv by dt plus 1 by L integration V dt. And if we simplify it, then we will get dv by dt square plus 1 over RC dv by dt plus 1 over LC into V that is equal to 0. characteristic equation is dv by dt square plus 1 over LC into V that is equal to 0. d square plus 1 over RC into d plus 1 over LC that is equal to 0. So, now let's put the value of this R, L and C inside this equation. And if we put it, then we will get d square plus 2d plus 2 that is equal to 0. And if we see the solution, we will get d1 as minus 1 plus j and d2 as minus 1 minus j. So, as you can see here the roots are complex conjugate. Now, if we see the solution, the solution can be given as e to the power my alpha t into c1 cos of Now, here the value of α is minus 1 and value of β is also 1. So, we will get solution as e to the power minus t into c1 cos plus c1 cos.
So, now we need to find the value of this c1 and c2 by applying the initial condition. So, first of all, let us see the initial condition at time t is equal to 0+. So, at time t is equal to 0+, we can write this equation as e to the power minus 0 into c1 cos of 0 plus c2 sin of 0. So we will get. C1 is equal to V and as we know all the initial conditions in our circuit is 0. So, the value of C1 will be nothing but 0. So, now in this circuit if you see at time t is equal to 0 plus this capacitor will act as a short circuit.
So, the entire 2 ampere current will flow through this 1 Faraday capacitor. So, we can say that. I at time t is equal to 0 plus is nothing but 2A.
And as we know, I can be given as C dVc by dt. That is equal to nothing but 2A at time t is equal to 0 plus. So now, to find the value of dV by dt, let's differentiate this equation. So now, this V can be given as e to the power minus t into C2 sin T as value of C1 is 0. So, now let's differentiate this equation with respect to T. So, we will get dv by dt as minus e to the power minus T into C2 sin T.
plus e to the power minus t into C2 cos of t. And we know that value of dvc by dt at times t is equal to 0 is 2 ampere. So, if you put this value, then we will get e to the power minus 0 into c2 cos of 0 plus 0. That means the value of c2 is nothing but a 2. So, in this way, we found the value of c1 as 0 and value of c2 as a 2. So, now the solution can be given as 2e-t sin. So, this will be the equation for the voltage for the parallel RLC circuit.
So, now if plot this equation, then this equation will look like this. So, as you can see here, as this equation involves the sine term, so you will see some oscillations in the response. But these oscillations will die out after a few seconds. So, I encourage all of you to change the value of R, L and C. in the given example and just see that how the response of the system changes and you can plot the new response which you have found using the any cow plotting software so i hope in this video you understood the transient response for the series as well as the parallel rlc circuit so if you have any question or suggestion please let me know in the comment section below if you like this video hit the like button and subscribe to the channel for more such videos