Help me solve an RLC task

Figure is attached

Plz solve this question for the attached figure:

iS=10u(-t) - 20u(t) A.
iL= current through inductor
vC= voltage across capacitor

Find
(a) iL(0-);
(b) vC(0+);
(c) vR(0+);
(d) iL(infinity);


Ans:
10 A; 200V; 200V; -20 A;

Plz solve and give reasons for these answers

Re: RLC

Hi,

Its very easy dear, u need to keep in mind following points, I know what the problem u r facing since i taught this course at least 5 times, and most of the students face the same problem.

1. Current through inductor can not change abruptly, i.e. I_L(0-) = I_L(0+).
2. Similarly voltage through capacitor cant change abruptly, i.e. V_C(0-)=V_C(0+).
3. Initially inductor acts as short circuit(SC) and capacitor acts as open circuit(OC) to a power source.
4. When inductor acts as SC, then all current will flow through the branch where inductor is, bcz capacitor is OC and hence offering infinity resistance.
5. U need to kow the current divider and voltage divider rule.


I am sure if u got these points in ur mind, then this circuit is piece of cake for u. I can give u exact solution and u can find thusands of such examples in the famous book of Nilson, but i will recoomend u to solveit by urself. I gave u all information that u need, and now u just have to apply these steps on this circuit.

Best of Luck.

BR,

Arif Khan,

Please press helped me!!

Re: RLC

sir can u plz provide me the exact solution
one of my friend said that these answers are not correct.
I also want to confirm the answers

Re: RLC

Hi Dear,


Look these answers are correct, ok now I will solve it for you.

You need to understand the Is (i.e. current source) its function is as

Is = 10u(-t) - 20u(t) , here u(t) is a Unit Step function which is defined as that it is 1 at time t ≥ 0 and zero when t < 0, but in this case as it is defined for both t < 0 and for t ≥ 0, so its values will be as follows:

i = 10 A when t < 0, i.e. when time is negative
i = -20 A when t ≥ 0. i.e. when time is 0 or +ve.

Ok , now as I told you in my previous reply that initially inductor acts as SC, so L and C are in parallel, OK, so when L is SC and C is OC, so all current will flow through the R and L and no current will flow through C branch.

Now at t < 0,

I_L = 10 A, L is SC, C is OC so no current through C, (this is you first answer).

As I told you that I_L(0-) = I_L(0+) = 10 A (current through L does not change abruplty,OK).

What to find Next? OK, V_C(0-).

Now C is OC, L is SC, as R & L branch is Parallel with C branch, so whatever Voltage is across R&L branch, will be the smae voltage across C, branch.
Now current passing through R & L branch is 10 A and R is 20 Ohms (remember L is SC and zero resistance at t < 0), so using Ohms law,

v = iR,
v = 10 * 20 = 200 volts
v_c(0-) = v_c(0+) = 200v, voltage across capacitor can not change abruptly,

I hope now u got it, I deliberatley did not solve the last part, and hope you can solve it by urself. If u need my help again let me know.

Best of Luck.

Arif Khan

Please press helped me!!!