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In this given circuit, how can we find \$ Vc(t) \$ that is voltage on capacitor?

\$ u(t) \$ is step that is 0 when \$ t<0 \$ and 1 when \$ t>0 $.

I want to find \$ Vc(t) \$ by 2nd order differential equation. Please help!!!

My Approach

I wrote KVL in bigger loop,

$$V_s = L \cdot \dfrac{\text{d}i_L}{\text{d}t} + R_1 \cdot i_L + V_c$$

and KCL in upper node,

$$i_L = \dfrac{V_c}{R_2} + C \cdot \dfrac{\text{d}V_c}{\text{d}t}$$

using \$ i_L \$ from KCL in KVL, I found equation

$$V_s = L \cdot C \cdot V_c'' + \left( \dfrac{L}{R_2}+R_1 \cdot C\right) \cdot V_c' + \left( 1 + \dfrac{R_1}{R_2}\right) \cdot V_c$$

solving this equation,

$$V_c = \dfrac{R_2}{R_1 + R_2} + A \cdot e^{r \cdot t} + B \cdot e^{s \cdot t}$$

where A,B are constants,

$$ a^2-b>0 \\ r = \dfrac{-a+\sqrt{a^2 - 4 \cdot b}}{2 \cdot a} \\ s = \dfrac{-a-\sqrt{a^2 - 4 \cdot b}}{2 \cdot a} \\ a = \dfrac{\dfrac{L}{R_2} + R_1 \cdot C}{L \cdot C}\\ b = \dfrac{1+\dfrac{R_1}{R_2}}{L \cdot C} $$

But this doesn't match with my teacher's sample circuit what's wrong??

  • \$\begingroup\$ Is this a homework problem? (Yes?) What approaches have you taken so far and what other resources (the textbook?) have you consulted? \$\endgroup\$ – user2943160 Jun 6 '16 at 16:08
  • \$\begingroup\$ Where are you struggling? Is it working out what the second order differential equation should be or solving the equation? There are plenty of resources on the web for this. There are plenty of us here who would be happy to help myself included but you need to show your effort. Your teacher does not want to know how clever we are on this site and you wont learn if one of us just gives you a textbook answer. \$\endgroup\$ – Warren Hill Jun 6 '16 at 17:55
  • \$\begingroup\$ thank you for your comments!! I am making this circuit on bread board and checking overdamped,critically damped,underdamped Vc output. I wrote KVL in bigger loop, Vs=LdiL/dt + R1iL + Vc and KCL in upper node, iL = Vc/R2 + CdVc/dt. using iL from KCL in KVL, I found equation Vs= LCVc'' + (L/R2 + R1C)Vc' +(1+R1/R2)Vc solving this equation, Vc=R2/(R1+R2) + Ae^rt + Be^st when A,B are constants, a^2-4b>0, r=(-a+《a^2-4b》)/2a, s=(-a-《a^2-4b》)/2a , (《》means root) a=(L/R2 + R1C)/LC , b=(1+R1/R2)/LC but this doesnt match with my teacher's sample circuit what's wrong?? \$\endgroup\$ – ggglemon Jun 6 '16 at 21:56
  • \$\begingroup\$ @ggglemon This should be part of the question not a comment. I've submitted an edit to do this for you. \$\endgroup\$ – Warren Hill Jun 7 '16 at 18:45
  1. Choose one node in the circuit to designate as the reference, or ground, node.

  2. Write equations based on KCL or KVL and the component properties to describe the circuit. You should have two equations.

  3. Reduce your two equations to one equation by algebra. You will now have a 2nd-order differential equation.

  4. Determine the general solution to this differential equation by inspection.

  5. Choose the parameters for the general solution to match the initial and final conditions (stored energy in the components is 0 at \$t=0\$ and the solution tends toward the steady state at \$t=\infty\$)

  • \$\begingroup\$ Stored energy is not zero at \$t=\infty\$ since there is current through the inductor and the capacitor is charged. \$\endgroup\$ – Chu Jun 6 '16 at 19:07
  • \$\begingroup\$ @chu, good point, edited. \$\endgroup\$ – The Photon Jun 6 '16 at 19:52

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