RC Circuit current through capacitor question

[Neutrinogun]

Homework Statement

Specifically #13 only.

Homework Equations

Same image as above.

The Attempt at a Solution

Loop equations
10 - 500,000* i₁ - 500,000 * i₂ - 500,000 * i₃ = 0
10 - 500,000* i₁ - q/ 5 * 10^-6 - 500,000 * i₃ = 0
Junction equations
i₁ = i₂ + dq/dt
i₁ = i₃

Solving for i₁ in the second loop equation, you get

i₁ = 1 * 10^-5 - 0.2q

Substituting second junction and above i₁ into the first loop equation, you get

i₂ = 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.)

 

[mighty2000]

Hello,

Thank you for your post. It seems like you have a good understanding of how to approach this problem. However, there are a few things that may be causing confusion.

Firstly, when solving for i1 in the second loop equation, you should get i1 = 1 * 10^-5 + 0.2q, not i1 = 1 * 10^-5 - 0.2q. This is because the direction of current flow through the capacitor is opposite to that of i1.

Secondly, when substituting i1 and i2 into the first junction equation, you should get 1 * 10^-5 + 0.2q - 2 * 10^-5 + 0.4q = 0. This simplifies to 2 * 10^-5 + 0.6q = 0.

Now, to answer your main question, the reason why we use 4 uA as the current going through the capacitor instead of 16/3 uA is because the current through the capacitor does not remain constant throughout the charging process. As the capacitor charges, the current through it decreases, following the equation i = E/R * exp(-t/RC). Therefore, the current through the capacitor at any given time is not equal to 2/3 of the initial current of 8 uA. Instead, it is a decreasing function of time.

In order to find the correct current through the capacitor at a specific time, we need to use the charge equation, which is q = C * E * (1 - exp(-t/RC)). Substituting the values given in the problem, we get q = 5 * 10^-6 * 10 * (1 - exp(-t/1.6667)). Setting q = 10 uC, we can solve for t and get t = 2.197 seconds. Plugging this value into the current equation, we get i = 4 uA.

I hope this helps to clarify your confusion. Let me know if you have any further questions.