- #1

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## Homework Statement

**Relevant infor**

V(out) stands for the voltage divider I believe

## The Attempt at a Solution

I am just going to get the equation first.

For the first part, applying the loop rule I got

[tex]12 - 10^5q' - 10^7 q = 0[/tex] with q(0) = 0. Solving I get

[tex]Q = \frac{3}{25000}(1 - e^{-100t})[/tex]

Since V = IR = q'R = q'(10

^{5}) I should get [tex]V = 12e^{-100t}[/tex]. So the plot would be a exp curve, curving down.

For the question that follows after. The loop rule is

[tex]100 - q'(10^6) - q/C = 0[/tex]. Solving, I get

[tex]q(t) = 100C - 100Ce^{-10^{-6}t}[/tex]

Since V = IR = q'R = q'(10

^{6}) I should get [tex]V = 100e^{-10^{-6}t/C}[/tex]

Now [tex]V_{out} = 70 = 100e^{-10^{-5}/C}[/tex], solving I get C = 2.8

Here is the problem, the answer is supposed to be 8.3μF. All the ODEs were solved with Maple

What did I do wrong?