3
\$\begingroup\$

enter image description here

enter image description here

I have attached my attempt and the solution provided by my instructor. I am confused on 2 things

  1. what it means by "the circuit is in quiescent state" ?
  2. why is he getting sqrt(8) in the current equation ?

Edit1: I have also attempted the same problem using Laplace

enter image description here

\$\endgroup\$
  • \$\begingroup\$ R u sure given ans is right? Yours seems to be perfectly fine \$\endgroup\$ – Deep Jan 11 '18 at 17:13
  • \$\begingroup\$ Yes, it was asked in ESE 2018 held on 7th jan 2018. \$\endgroup\$ – Nikhil Kashyap Jan 11 '18 at 18:57
  • \$\begingroup\$ @Deep electronics.stackexchange.com/questions/349472/… please take a look \$\endgroup\$ – Nikhil Kashyap Jan 11 '18 at 19:05
  • \$\begingroup\$ I'm not eligible enough yet to ans that question, thanks for A2A though. (It seems Andy Aka commented that given ans is wrong (in link of question you pointed me to) , I've already found enough mistakes in Gateforum's material (though I've ready only one book yet) to tell that it's not at all trustworthy source). \$\endgroup\$ – Deep Jan 12 '18 at 15:11
  • \$\begingroup\$ Thanks @Deep, btw it is a previously asked problem in gate EE. \$\endgroup\$ – Nikhil Kashyap Jan 12 '18 at 17:30
2
\$\begingroup\$

The problem happened when you differentiated the both sides of the equation. 100 Volts, which is a constant, became 0 in the left hand side of equation after differentiation. Don't you think there is a problem in it ? Cz even if it was any other number there, it would go zero and you will still end up in the same solution. Say, 25 Volts. You will get the same solution. But in the circuit , current has indeed changed. In fact you can never always differentiate both sides of some f(i) = g(i) equation and say the relation will still hold. Because you can differentiate the both sides ONLY if the equality f(i) = g(i) holds true for all real values of i. This is a heterogenous integro-differential equation. In this case the complete solution = CF + PI, mathematically. Both CF and PI have to be calculated. You have calculated only CF by neglecting LHS constant, and hence it is only partial solution. Maybe math stack exchange can provide more insight into this on how to solve this. Otherwise it's better to solve the equation in Laplace domain rather than in time domain. It is straight forward and easy to solve.

EDIT:

Your laplace solution is correct. So I think the book is wrong. If you take inverse transform of their solution, you will end up in a wrong RLC equation for that circuit. The correct series RLC ckt equations' deno should look like : $$s^2 + 2s(R/2L)+ (1/LC) = s^2 + 4s + 8 $$

While in their solution, it will evaluate to a wrong one: $$ s^2 + 4s + 12 $$

\$\endgroup\$
  • \$\begingroup\$ Please check my attempt using laplace \$\endgroup\$ – Nikhil Kashyap Jan 11 '18 at 16:12
  • 1
    \$\begingroup\$ Yes, I differentiated and lost 100 but that information is there in form of variable c2. I don't think there is anything wrong by differentiating the equation. \$\endgroup\$ – Nikhil Kashyap Jan 11 '18 at 16:26
  • 1
    \$\begingroup\$ But your laplace solution is correct. So I think the book is wrong. If you take inverse transform of their solution, you will end up in an invalid RLC equation. Series RLC ckt equations' deno should look like : s^2 + 2s(R/2L)+ 1/LC. Which is not in their solution. \$\endgroup\$ – Mitu Raj Jan 11 '18 at 17:26
  • 1
    \$\begingroup\$ Actually RHS will be always 100 for all values of i - says KVL. \$\endgroup\$ – Deep Jan 11 '18 at 17:33
  • 1
    \$\begingroup\$ Oh yes, PI has to be added into CF that's for sure. That's why I love Laplace 😛 \$\endgroup\$ – Deep Jan 11 '18 at 19:15
2
\$\begingroup\$

what it means by "the ckt is in quiescent state" ?

It means that there was no energy in either the inductor or capacitor prior to the voltage being applied.

why is he getting sqrt(8) in the current equation ?

With an inductor of 0.5 henries and a capacitor of 0.25 farads, the natural resonant frequency is \$\sqrt8\$ radians per second.

This is from the equation \$\omega_n = \dfrac{1}{\sqrt{LC}} \$

In other words there will be a damped oscillation in the current starting at a peak and decaying with a natural frequency of \$\sqrt8\$ radians per second.

\$\endgroup\$
  • \$\begingroup\$ where I did mistake in my calculation ? \$\endgroup\$ – Nikhil Kashyap Jan 9 '18 at 15:45
  • \$\begingroup\$ @NikhilKashyap look at example 1 on this page: tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx and see if it clears things up. I think the problem is how you translated a 2nd order diff equation with complex roots to sin and cosine terms. \$\endgroup\$ – Andy aka Jan 9 '18 at 16:06
  • \$\begingroup\$ No, He did the same thing as me in example 1. Please check once more. \$\endgroup\$ – Nikhil Kashyap Jan 9 '18 at 17:37
  • \$\begingroup\$ I think it will be wd rather than wn & wd=wn*sqrt(1-zeta^2)=2. So, it should be 2 not sqrt(8). \$\endgroup\$ – Nikhil Kashyap Jan 9 '18 at 17:48
  • \$\begingroup\$ Maybe try asking on maths stack exchange if all the doubts are about the math. As far as I'm concerned in analysing the frequency domain, the natural frequency is \$\sqrt8\$ rads per second and that will be the decaying waveform in the transient response. Maybe use LTSpice to simulate it? \$\endgroup\$ – Andy aka Jan 9 '18 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.