# Source Free Series Rlc Circuit || Example 8.4 || Practice Problem 8.4 || Lca 8.3(2)(New)

We'll have another him. This is the second video on Thor 3 series of the circuit. And here we'll be solving two problems. So lets device learn that.

S 1 can be written as this as to the 2 roots of the characteristics equations. And the damping factor alpha is out over 12 and the resonant frequency Omega. Naught is 1 over under root LC. If you have difficulty following I recommend that you check the previous video, and then we are also in the previous video found out that there could be 3 current. Equations for the circuit, if alpha is greater than Omega, which is an over damped case. Then we use this formula.

If alpha is equal to Omega, which is a critically damped case. Then this is the second formula. And if alpha is less than Omega dot, which is an under damped case. In this case, we get oscillations. We have this formula with cosine and sine. And we had also discussed problem-solving strategy. We'll follow the same strategy.

First they'll find a 0 and V 0 for T less than 0. Then for T greater.Then 0 will find DI DT 0 also for T greater than 0 will calculate alpha and omega and based on Alpha and Omega we'll, select the current equation. And any unknown in the current equation like s1 or s2 RWD. We will calculate that, and then put get a general equation from general equation will to get a1 and a2 v2 Imams. So let's see, this is the issue.

In this case we have to find IT Keira Knightley through the circuit assume that the circuit has been has reached a steady state at T zero, minus 30 zero. Minus means before opening of the switch, the circuit has reached a steady state objects for the first step in the for T less than 0. We have to find y 0 and V zero. So this is the circuit for T less than 0 means. This is connected. This will open at T is equal to 0.

So this is our circuit this one connected it. And since this circuit was in a steady state, and therefore, we can say that the capacitor is fully charged. And it is now behaving like open and inductor is behaving like a short circuit. So the.

Next circuit will draw will be something like this. The capacitor is open charged to an initial voltage. We call that as v-0 and the inductor behaves like a short circuit. And the current through the inductor at T is equal to 0 or T equal to 0 minus is i0.

So from here we can calculate a 0 because there is no current in this. So this is the only loop available to us. So 10, volt supply in 4 ohm sections. So I is 10 divided by 4, plus 6, 1 ampere. And now what is Final now you have to be careful to.

Understand this any voltage present here will be present here. And since this is open, so that we'll be present at this opening. And so that voltage will be the voltage appearing across six chrome. We know the current is 1 ampere. So 6 multiplied, by 1 will be 6, volt current, which is equal to V naught. Okay. Now, the second step we have to write the KVL equation, and then I've got T is equal to zero.

So let's see, this is the circuit before now, when we open it, this is the circuit, and we can see that the. Current in the circuit is flowing like this. That means same current is flowing through six form and through three ones.

So we can add them. We can say that these two are in series. We have added them in the total resistance is 9 ohms.

And then we have inductor and the capacitor. And now we can write the KVL equations in this loop. So KVL we can write from here.

It is RI, L, DI, DT and minus V naught or VC. So at T is equal to zero I T will become a zero. So RI 0, plus L, DI, 0 DT and minus VC T is actually V. Naught which we have found earlier, the value add I T is equal to zero. And we had found out I naught as 1 ampere and V naught as 6 for remember, this capital, v, naught and small V, naught they are same.

So we plug in the value of the current. Our is nine multiplied by current 1l value. And this is minus six from here. We calculate that DI DT is equal to minus six amperes per second. The third step is that we calculate alpha and omega for t greater than zero circuit. So again, this is the T greater than. Zero circuit, and this was what we found out.

So from here we calculate alpha is our over 2l. Our is nine and putting in the value of L. So alpha is 9. Similarly, Omega naught is 1 over under root LC and putting the value of L and C. The value comes to be checked now based on the value of alpha and omega. We choose the equation for current alpha is 9 and omega is 10.

That means Omega is greater than alpha or alpha is less than Omega. So when alpha is less than Omega, this is an under damped case. And we. Had given in the previous video, this type of example that the man is alpha and the female is omega.

So when alpha is light or less, then she can play around or dance with this gentleman. So we get an oscillation case that is sine and cosine terms. So this is the equation that we will be using yet also. Now, in this case, we would like to know the value of WD. So we have to calculate did is given by this formula for putting in the values WD is four point, three, five nine.

And so from here. Now we can. Write the IT equation complete. He raised to the power minus alpha T alpha is nine. So for nine year, ninety a cos Omega D is four point, three, five, nine and similarly a to sign for 45 90. So this is the general equation. And now we will try to find the value of a 1 and a 2 from this equation.

Okay, now we come to step number five. We have to find the value of a 1 and a 2. So this was the equation. Now here also we have to do two things.

First for T is equal to zero to find one of the equations. So we. Put in T is equal to zero. We get this term.

Costume cos, 0 becomes 1 sine, zero becomes 0. So we get a 0 is equal to t1. And now we have already calculated i1 to be 1. So we can say that a 1 is equal to 1 now to find a 2, we have to differentiate this with respect of T. And then for T is equal to 0. So four different sessions will follow the product rule because there are two terms. This is one term here. And this whole is second term.

So this is what you get. We have first differentiated. The first. Term for a raise to the power, minus 390 will give you minus nine years to the power, minus T. The second term remains as it is and plus, we keep the first term and differentiate the second term. So you get answer like this, and now we'll put zero.

So just watch the answers, and we put T is equal to zero. So the sine times cosine terms here will become 1. This will become 0.

This will become 0, and this will become 1. So we get this term. And now we had also calculated the value of DI DT at T is equal to 0. For the 0 condition that was minus 6 for DI DT, minus 6 for plug in here, minus 6 and on the right-hand side of this equation.

We know a 1 to be 1 say 1 was 1 plugging in that. So we get a 2 to be 0, 4 6, 8 8 2. And from here, we put in the values of T 1, which was born here and a 2, which is 0 4 6, 8 2.

So this is the final answer for the complete solution. Now let's do the practice problem. Also, again, this circuit has reached steady state at T 0 minus and before make breaks, which moves to position B. At key is equal to 0 calculate, a 0 t for t, greater than zero. Human Anna may not have heard of this make before brakes, which this has actually two leaves when we are switching this here.

So there will be some time or some gap between switching. We want to avoid that. So we have two leaves one remains here. The other leaves move here. The woman, the other leaf touches this one, then the second also moves there. So that is the condition case for make before break see.

So this will help the continuity and. No wastage of energy because the energy is stored in the inductor and very quickly, it can get a third can become diminished. Okay.

So for T less than 0. We have to find a 0 V 0. So this is the circuit for T less than 0. The inductor will behave like a short circuit because it is in position there for a very long time. So the current in this case will be 10 divided by the sternum resistor. So 10 ampere. So i0 will be 10 amperes.

And since the capacitor is not connected, therefore the voltage across. Capacitor which we call v-0 is equal to 0. Okay. Now, step two, we have to write the KVL and then find the ID T. So let's go to the circuit here again. We just follow the procedure that is followed before will write KVL here. So since the current is following like this, we mark the polarities, positive negative, positive negative, first, negative, positive negative.

And now it will be easy for us to write the KVL. We started from here to write a KVL equation. So minus L DI, DI, DT, minus VC, T and minus IT after. Manipulation we can write it like this. And now T is equal to zero.

So it will be our. I 0 l, the, DK, zero, plus V, zero. And we know I salute an ampersand V zero or capital. V. Zero there same is zero volt so plug in the values.

Now here plugging in the values, you find the ID PT to be minus 50 amperes per second. So second is we have done third step for T greater than zero, calculate alpha and omega. So for this circuit I'll, find Omega Alpha is a power over twelve. You know, R is 5/2.

So two point five is Alpha. And Omega is three now here, if you can see, okay, step, four based on Alpha and Omega. We choose the equation current equation. Now alpha is less than Omega alpha is less than Omega means again, and you have discussed alpha is less than Omega.

So she is dancing, and we have the cosine sine terms for the ID equation. And as before we calculate WT and plugging in the values in the current equation, this equation, this is the general equation for current I T. Now we need to find a1 and a2. So we were here.

They. Put T is equal to 0. And this equation put in K is equal to 0. We get a 0 is equal to t1 I.

Hope you can follow T is equal to 0 cos, 0 there's, a sine 0. And we had also found earlier. They want to meet our sorry, eyes ERO to be 10 amperes. Therefore, we can say that a 1 is equal to 10 for the second part or a 2. We have to take derivative again, following the product rule, taking the derivative just like before, you can pause the video and do these steps yourself here. So over here now I will put T is equal. To 0 put in K is equal to 0.

We get this equation, and we had already calculated DI DT of so 4, minus 50s will plug in this value here, minus 50 is equal to this. We know the value of a 110 plugging in you get a 2 to be minus 15 point, 0 7. And so our final equation will plug in the values of a 1 and a 2. So this is the final equation, I hope. This gives you an understanding as to how to solve this problem, this type of problem. Thank you.