- #1

Neutrinogun

- 9

- 0

## Homework Statement

Specifically #13 only.

## Homework Equations

Same image as above.

## The Attempt at a Solution

__Loop equations__

10 - 500,000* i

_{1}- 500,000 * i

_{2}- 500,000 * i

_{3}= 0

10 - 500,000* i

_{1}- q/ 5 * 10^-6 - 500,000 * i

_{3}= 0

__Junction equations__

i

_{1}= i

_{2}+ dq/dt

i

_{1}= i

_{3}

Solving for i

_{1}in the second loop equation, you get

i

_{1}= 1 * 10^-5 - 0.2q

Substituting second junction and above i

_{1}into the first loop equation, you get

i

_{2}= 2 * 10^-5 - 2*10^-5 + .4q = .4q

Substituting into the first junction,

1*10^-5 - .6q - dq/dt = 0

Using the equation i = E/R * exp(-t/RC) to solve for time, I used

(2/3)(8) uA = (1*10^-5 / 1)(exp(-t/1.6667))

However, the correct answer is using 4 uA with the above equation. My main question is why do we use 4 uA as the current going through the capacitor instead of 16/3 uA. The overall resistance of the path through the switch would be 1,000,000Ω while the resistance of the capacitor path is 500,000Ω. Doesn't this mean that 2/3 of the current would flow through the capacitor, so 2/3 of 8 uA?

Thanks.

(PS. I realized afterwards that you don't need to use the current equation, but can use the given charge equation once you figure out the charge on the capacitors when 8 uA goes through the capacitors, which is 10 uC. However, I still think my way should work, or want to understand why it doesn't.)