RC Circuit current through capacitor question

• Neutrinogun
In summary, the problem involves solving loop and junction equations to find the current through a circuit with a capacitor. The main question is about the current through the capacitor and why it is not equal to 2/3 of the initial current. This is because the current through the capacitor is a decreasing function of time and can be found using the charge equation.
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.)

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.

1. What is an RC circuit?

An RC circuit is an electrical circuit that contains a resistor (R) and a capacitor (C) connected in series or parallel. The RC circuit is used to control the flow of electric current and is commonly found in electronic devices such as radios, televisions, and computers.

2. How does current flow through a capacitor in an RC circuit?

In an RC circuit, current flows through the capacitor when the circuit is first energized. As the capacitor charges, the current decreases until it reaches zero. When the circuit is then de-energized, the capacitor discharges and current flows in the opposite direction until it reaches zero again.

3. What is the relationship between the current, voltage, and resistance in an RC circuit?

According to Ohm's Law, the current (I) in an RC circuit is equal to the voltage (V) divided by the resistance (R). In other words, I = V/R. The voltage across the capacitor (Vc) is equal to the current (I) multiplied by the capacitance (C), or Vc = I * C.

4. How does the time constant affect the current in an RC circuit?

The time constant (τ) is the time it takes for the capacitor to charge or discharge to 63.2% of its maximum voltage. As the time constant increases, the current in the circuit decreases, and vice versa. This means that the higher the time constant, the longer it takes for the capacitor to charge or discharge, and the lower the current will be.

5. How can I calculate the current through a capacitor in an RC circuit?

To calculate the current through a capacitor in an RC circuit, you can use the formula I = V/R * (1-e^(-t/RC)), where V is the voltage, R is the resistance, C is the capacitance, and t is the time. Alternatively, you can use a circuit simulator or an oscilloscope to measure the current in the circuit.

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