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enter image description here

I am struggling with c) part these differential equations, I am not sure i obtained it correctly or not, but i solved in matlab also, and i dont think it is correct answer.

My work: $$-50q_1'-100000q_1+100(I-q_1')=0$$ $$-50q_2'-100000q_2+150(I-q_2')=0$$ $$-50q_1'-100000q_1-50q_2'-100000q_2+75=0$$

Can you check my equations also? Here $I$ is the function of time $t$ enter image description here

If R1 and R3 are in parallel and R2 and R3 in parallel always for any time t, then this question is a piece of cake for me. Can u confirm are they in parallel ? ?

I solved it using laplace and both MATLAB now, Thanks if anyone else has tried

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  • $\begingroup$ I checked my answers with LTspice, and mine are right. $\endgroup$ May 26 '20 at 9:18
  • $\begingroup$ Okay...so you mean first of all I divide by s and then take inverse Laplace transform?? $\endgroup$ May 27 '20 at 3:59
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    $\begingroup$ yes indeed!!!!!!!! $\endgroup$ May 27 '20 at 6:18
  • $\begingroup$ okay i will try at home and let u know :D $\endgroup$ May 28 '20 at 10:52
  • $\begingroup$ Nice, I am curious if you now get it. If not, I am willing to help you further. $\endgroup$ May 28 '20 at 11:05
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Well, we are trying to analyze the following circuit:

schematic

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{i}_\text{i}=\text{i}_1+\text{i}_3\\ \\ \text{i}_1=\text{i}_2+\text{i}_5\\ \\ \text{i}_4=\text{i}_3+\text{i}_5\\ \\ \text{i}_\text{i}=\text{i}_2+\text{i}_4 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{i}_1=\frac{\text{v}_\text{i}-\text{v}_1}{\text{R}_1}\\ \\ \text{i}_2=\frac{\text{v}_1}{\text{R}_2}\\ \\ \text{i}_3=\frac{\text{v}_\text{i}-\text{v}_2}{\text{R}_3}\\ \\ \text{i}_3=\left(\text{v}_2-\text{v}_3\right)\text{sC}_1\\ \\ \text{i}_4=\frac{\text{v}_3-\text{v}_4}{\text{R}_4}\\ \\ \text{i}_4=\text{v}_4\text{sC}_2 \end{cases}\tag2 $$

Here I used small letters to denote the s-domain (using the Laplace transform). And notice that $\text{v}_1=\text{v}_3$.

Using $\text{i}\left(\text{s}\right)=\text{s}\cdot\text{q}\left(\text{s}\right)$ it is possible to solve your problem. Using my equations you can solve for all voltages and currents. To do this fast you can use Mathematica for example.


Using this Mathematica code:

In[1]:=Vi = 75/s;
R1 = 100;
R2 = 150;
R3 = 50;
R4 = 50;
C1 = 10*10^(-6);
C2 = 10*10^(-6); FullSimplify[
 Solve[{V1 == V3, Ii == I1 + I3, I1 == I2 + I5, I4 == I3 + I5, 
   Ii == I2 + I4, I1 == (Vi - V1)/R1, I2 == V1/R2, I3 == (Vi - V2)/R3,
    I3 == (V2 - V3)*s*C1, I4 == (V3 - V4)/R4, I4 == V4*s*C2}, {Ii, I1,
    I2, I3, I4, I5, V1, V2, V3, V4}]]

I found:

Out[1]={{Ii -> (6 (500 + s) (2000 + 3 s))/(s (2000 + s) (10000 + 17 s)), 
  I1 -> (6 (500 + s))/(s (10000 + 17 s)), 
  I2 -> (6000 + 9 s)/(20000 s + 34 s^2), 
  I3 -> (12 (500 + s))/((2000 + s) (10000 + 17 s)), 
  I4 -> 3/4 (1/(2000 + s) + 1/(10000 + 17 s)), I5 -> 3/(20000 + 34 s),
   V1 -> 45 (1/s - 2/(10000 + 17 s)), 
  V2 -> 75/2 (2/s - 1/(2000 + s) + 1/(10000 + 17 s)), 
  V3 -> 45 (1/s - 2/(10000 + 17 s)), 
  V4 -> (450000 (2000 + 3 s))/(s (2000 + s) (10000 + 17 s))}}

To find $\text{q}\left(\text{s}\right)$, you can use:

$$\text{i}\left(\text{s}\right)=\text{s}\cdot\text{q}\left(\text{s}\right)\space\Longleftrightarrow\space\text{q}\left(\text{s}\right)=\frac{\text{i}\left(\text{s}\right)}{\text{s}}\tag3$$

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  • $\begingroup$ Why the downvote? $\endgroup$ May 25 '20 at 15:06
  • $\begingroup$ Hey thanks for the solution , I don't know who have that thumbs down let me check your solution and after that I will vote up , I need charge I guess you didnt do the charge $\endgroup$ May 25 '20 at 16:00
  • $\begingroup$ @learningstudent true I see my mistake I will fix it as as soon as possible. $\endgroup$ May 25 '20 at 17:43
  • $\begingroup$ @learningstudent Look at my edit. Now it is correct. $\endgroup$ May 25 '20 at 18:15
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    $\begingroup$ I checked your work and it is correct ! +1 $\endgroup$ Jun 3 '20 at 14:26

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