RC Circuits Voltage: Solving for V(out) with Loop Rule

[Dens]

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

$$12 - 10^5q' - 10^7 q = 0$$ with q(0) = 0. Solving I get

$$Q = \frac{3}{25000}(1 - e^{-100t})$$

Since V = IR = q'R = q'(10⁵) I should get $$V = 12e^{-100t}$$. So the plot would be a exp curve, curving down.

For the question that follows after. The loop rule is

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

$$q(t) = 100C - 100Ce^{-10^{-6}t}$$

Since V = IR = q'R = q'(10⁶) I should get $$V = 100e^{-10^{-6}t/C}$$

Now $$V_{out} = 70 = 100e^{-10^{-5}/C}$$, 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?

 

[cepheid]

Relevant infor
V(out) stands for the voltage divider I believe

Huh? What voltage divider? V_out is the output voltage, which is the voltage across the capacitor, in this circuit.

The Attempt at a Solution

I am just going to get the equation first.

For the first part, applying the loop rule I got

$$12 - 10^5q' - 10^7 q = 0$$ with q(0) = 0. Solving I get

$$Q = \frac{3}{25000}(1 - e^{-100t})$$

Since V = IR = q'R = q'(10⁵) I should get $$V = 12e^{-100t}$$. So the plot would be a exp curve, curving down.

You're solving for the WRONG V here. The voltage given by IR is the voltage across the resistor (let's call it V_R). You're looking for the voltage across the capacitor. To get the capacitor voltage (V_C), use the fact that q = CV_C for a capacitor. OR, use the loop rule to solve for V_C in terms of V_R and V₀. Either method should give you the same answer.

Also, whenever you get an answer to a problem, always ask yourself, "does this make any sense?" A decaying exponential for the capacitor voltage makes no sense, because after the switch closes, the capacitor is charging, not discharging.

EDIT: and you made the exact same mistake in part 2. Your equation for the capacitor voltage vs. time is just wrong.