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Here is the question I am trying to solve:

ques

I didn't find Vo(t). What way should I go to find this value? ( Vo(t) is a voltage on 40k resistance after switch moves b.)

Here is my attempt: ( It looks complicated but I explained what I did step by step on my attempt below)

attempt

I gave a number for each of my steps.

  1. I found Vc(0)

  2. I found Vc(∞)

  3. I found Vc(t)

  4. I tried to find Vo(t)

In 4th step, I thought that Ic(t) (which is the current on the capacitor) is equal to İ1+ İ2 (which are shown on drawing left) and these currents may be equal to each other because they are on the same resistance value (40k 40k).

And then, I tried to find İc(t) in 2 ways ;

  1. with derivation formula

  2. with the Thevenin circuit that I drew as well.

After that, I divided İc(t) by 2 to find İ1. Finally, I applied V=I.R to find Vo(t)

Another theory that I thought ( which I showed below on paper with "or") Vo(t) may be equal to directly Vc(t) because they are parallel.

Or I have a completely another theory. I wrote it below (again in the right side).

enter image description here

Which one is the correct solution? Or what is the solution? Can someone give me an idea?

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  • \$\begingroup\$ +1 for following site rules \$\endgroup\$ Jun 15, 2023 at 21:50
  • \$\begingroup\$ what do you mean ? \$\endgroup\$ Jun 15, 2023 at 21:54
  • \$\begingroup\$ You showed your attempt at solution, rather than simply presenting an academic problem and asking others to solve it for you. \$\endgroup\$ Jun 15, 2023 at 22:03
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    \$\begingroup\$ Yes, because this is what is asked of me while I am uploading a question on this website \$\endgroup\$ Jun 15, 2023 at 22:07
  • \$\begingroup\$ in step 1 and 3, Vc(0) should be -60V. Therefore: Vc(t) = -50 + [-60 - (-50)]exp(-4t) = -10 ( 5 - exp(-4t) ) V \$\endgroup\$ Jun 16, 2023 at 2:23

1 Answer 1

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I'm going to keep this quick. Which means I'm not going to read your photo and work. You'll have to make comparisons and figure out differences, where they may be.

Your schematic works out to:

schematic

simulate this circuit – Schematic created using CircuitLab

So at \$t=0\$ then \$V_{_\text{C}}=-60\:\text{V}\$. That's the initial condition prior to the right-side schematic taking over.

Noting that in the case where \$t\gt 0\$, \$R_{_\text{TH}}=50\:\text{k}\Omega\$ and \$V_{_\text{TH}}=-50\:\text{V}\$, the following KCL applies:

$$\frac{V_{_\text{C}}-V_{_\text{TH}}}{R_{_\text{TH}}}+C_1\cdot\frac{\text{d}}{\text{d}t}V_{_\text{C}}=0\:\text{A}$$

It's almost the same KCL as you've already experienced in earlier questions.

Re-arranging:

$$\frac{\text{d}}{\text{d}t}V_{_\text{C}} + \frac{1}{R_{_\text{TH}}\,C_1}V_{_\text{C}}=\frac{V_{_\text{TH}}}{R_{_\text{TH}}\,C_1}$$

This is a non-homogeneous equation. Do you know how to approach it?

Added:

$$\begin{align*} \left(\frac{\text{d}}{\text{d}t} + \frac{1}{R_{_\text{TH}}\,C_1}\right)V_{_\text{C}}&=\frac{V_{_\text{TH}}}{R_{_\text{TH}}\,C_1} \\\\ \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}}{\text{d}t} + \frac{1}{R_{_\text{TH}}\,C_1}\right)V_{_\text{C}}&=\frac{\text{d}}{\text{d}t}\frac{V_{_\text{TH}}}{R_{_\text{TH}}\,C_1} \\\\ \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}}{\text{d}t} + \frac{1}{R_{_\text{TH}}\,C_1}\right)V_{_\text{C}}&=0 \\\\ \left(\frac{\text{d}}{\text{d}t}+0\right)\left(\frac{\text{d}}{\text{d}t} + \frac{1}{R_{_\text{TH}}\,C_1}\right)V_{_\text{C}}&=0 \end{align*}$$

Which requires the sum of two, not one, separate exponentials:

$$\begin{align*} V_{_\text{C}}&=A_1\cdot\exp\left(-\frac1{R_{_\text{TH}}\,C_1}\cdot t\right)+A_2\cdot\exp\left(\vphantom{\frac1{R_{_\text{TH}}\,C_1}}0\cdot t\right) \\\\ V_{_\text{C}}&=A_1\cdot\exp\left(-\frac1{R_{_\text{TH}}\,C_1}\cdot t\right)+A_2 \end{align*}$$

Use initial conditions to resolve it into a specific solution. At \$t=0\$ then \$A_1+A_2=-60\:\text{V}\$ and at \$t=\infty\$ then \$A_2=-50\:\text{V}\$. So \$A_1=-10\:\text{V}\$.

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    \$\begingroup\$ This helped me a lot \$\endgroup\$ Jun 16, 2023 at 22:54
  • \$\begingroup\$ @Computengineering Except I mis-wrote. Sometimes I'm just not thinking. Admittedly, I was in a rush. But I should have skimmed back. Hopefully, what I did at the end is now just a little clearer than before. My apologies. (Annihilators are a helpful technique, at times, to create a homogeneous equation to solve.) \$\endgroup\$ Jun 16, 2023 at 23:26

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