I've spent sometime learning about step responses for series/parallel RLC circuits, but this problem takes things a step further, since C seems to be in parallel with a series combination of RL and a voltage source. I'm trying to find out what i(t) is for t≥0, since the switch has been closed for a long time before opening at t =0. enter image description here My understanding is that for the step response, the solution calls for a second-order differential equation that takes into account both the forced response and the function for natural response. So when I focused on the t=0- condition, I set the inductor as a short circuit and the capacitor as a open circuit. From there, I also set i(t=0-) = i_L(t=0-). I used KVL to solve for i_L(t=0-):

i_L(t=0-) = (20V + 10V)/(5Ω + 5Ω) = 3 A = i_L(t=0+) <-- (I believe this is the initial current across inductor and R2?)

As for finding v_c(0-), I did a voltage divider to end up with: v_c(0-) = (5Ω)/(5Ω + 5Ω) = 5V = v_c(0+) <-- (I'm assuming this is the initial voltage of capacitor?) enter image description here Now what I don't completely understand is the t≥0 condition. With the switch open, the left side of the circuit is ignored, leaving only the right side of the RLC circuit as the circuit reaches steady state. The problem is:

  • From the way I modified the circuit, it seems to represent a series RLC circuit. Which would lead to the neper frequency (α = (R/2L) = 5 and resonant radian frequency, ω = (1/sqrt(LC)) = 5, leading to a critically damped response. From here, the general solution of the step response for series RLC circuit is: x(t) = Xf + D'1te^-at + D'2e^-at, with x either being v or i, and "Xf" being either the final voltage or final current

  • At t=∞, it's readily known that capacitor acts as a open circuit again, and the inductor as a short circuit, and because V2 is included in the reduced circuit, my understanding is that the final voltage, or v_c(t=∞), is 10V, and that the final current across R2 is ~ 0 A at the steady state. I even verified this with the use of the Falstad online circuit simulator shown below: enter image description here

I am quite lost on this issue on finding i(t). What I do know is that I am dealing with a series RLC circuit for t≥0, and that the general solution for i(t) takes the form of:

iL(t) = I_f + D'1*t*e^(-at) + D'2*e^(-at).

Does anyone have any ideas as to how to solve for i(t) for t≥0?

  • \$\begingroup\$ When the switch is opened it's a series circuit because the same current flows through each component.You will then need to derive and solve a 2nd order differential equation or analyse via Laplace transform. \$\endgroup\$ – Chu Dec 14 '18 at 18:44
  • \$\begingroup\$ @Chu So my suspicions were correct then. So, in regards to deriving and solving the 2nd-order diff. equation, would I just need to sum up the voltages around the closed path (when switch is open), and then differentiate with respect to time? I'd like to try to Laplace transform but my class never really had the opportunity to utilize that approach, but I'd like to know about that too. \$\endgroup\$ – Retrovert Dec 14 '18 at 19:01
  • \$\begingroup\$ Yes, use \$v_L=L\frac{di_L}{dt}\$ and, \$ v_C=\frac{1}{C}\int{i_C \: dt}\$, and then differentiate the KVL equation to get rid of the integral term. Don't forget to include the initial conditions on the capacitor and inductor. \$\endgroup\$ – Chu Dec 15 '18 at 1:45
  • \$\begingroup\$ @Chu Hmm, so for the initial conditions of both the inductor and capacitor, I calculated those values, but I'm not seeing how they play a role in solving the 2nd ODE. How exactly do those come into play? \$\endgroup\$ – Retrovert Dec 15 '18 at 5:17
  • \$\begingroup\$ @Chu Forgive me for asking this as well, but since the single loop involves a single current, can't we just idenitfy i_L = i_c, since they run along the same direction? Or am I wrong to assume this? \$\endgroup\$ – Retrovert Dec 15 '18 at 5:28

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