How Does Closing a Switch Affect the Voltage Across an RC Circuit?

[Sean Reed]

Homework Statement

The switch has been open for a vary long time, and is then at time T=0 closed. Determine the voltage V0(t) for t>0

Homework Equations

voltage from a discharging capacitor is V*e^(-t/RC) Where V is the voltage stored in the capacitor, R is the resistance and C is the capacitance.

The Attempt at a Solution

1) My understanding is that the capacitor will charge up to 12 volts, however I have been told that it only charges up to 6 because the two 2kohm resisters make a voltage divider. I did not think the location in terms of other resisters made a difference for the voltage that a capacitor will charge to but I may be wrong.
2) Once the switch is closed I thought the voltage at the node where the switch is would be 4 volts because it would act as a voltage divider between the first 2kohm resister and the two 2kohm resisters in parallel. My friend says it will be 6 volts because the capacitor acts as an open circuit so it is only a voltage divider between two 2kohm resisters. Now if this is the case then would it matter as the only voltage at V0 is the decaying voltage from the capacitor?
3) For the resistance in the time constant will it just be the 2kohm resister at V0 plus the 2kohm resister in series at the switch, or will it be something else?

Thus my answer is (12e^(-t/.02))-4 While his answer is (6e^(-t/.02))-6 I have no idea who is right or if we are both wrong, so any help is appreciated.

 

[berkeman]

Here's a hint to get you going -- draw a graph of v(t) using the following info...

The voltage on the - side of the cap initially is ground, right? And the voltage on the + side of the cap is initially what? Hint -- there is no voltage divider yet with the switch open. When the switch is closed, what does that instantaneously do to the voltage on the + side of the cap? Remember that the cap looks like a short circuit for instantaneous changes in voltage on one side of it...

 

[piercas]

I would recommend consulting a textbook or other reliable source for information on RC circuits and voltage dividers. It is important to have a clear understanding of the principles involved in order to accurately solve problems like this one. Additionally, it may be helpful to draw out a circuit diagram and label all the components and nodes to better visualize the flow of current and voltage.

In regards to the specific questions asked, the location of resistors in a circuit can affect the voltage that a capacitor charges to, as it determines the path of current flow. In this case, the two 2kohm resistors in parallel will act as a voltage divider and limit the voltage across the capacitor to 6 volts.

Once the switch is closed, the voltage at the node where the switch is will indeed be 4 volts, as it will act as a voltage divider between the 2kohm resistor and the two 2kohm resistors in parallel. The capacitor does not act as an open circuit, but rather as a voltage source in parallel with the resistors.

For the time constant, the effective resistance will be the sum of all resistors in the circuit, as they are in series. So in this case, the time constant would be 4kohm*0.01uF = 0.04 seconds.

In conclusion, it is important to have a clear understanding of the principles involved in order to accurately solve problems like this one. It may also be helpful to discuss with a professor or tutor for further clarification and guidance.