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

In summary, the problem involves calculating the voltage V0(t) for t>0 after a switch is closed in a circuit with a discharging capacitor. The voltage from the capacitor can be found using the formula V*e^(-t/RC), where V is the stored voltage, R is the resistance, and C is the capacitance. There is a discrepancy between two answers given, one being (12e^(-t/0.02))-4 and the other being (6e^(-t/0.02)-6). Drawing a graph of v(t) and considering the initial voltages on either side of the capacitor, as well as the effect of closing the switch, can help in finding the correct answer.
  • #1
Sean Reed
4
0

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.
 

Attachments

  • question 3.JPG
    question 3.JPG
    11.1 KB · Views: 438
Physics news on Phys.org
  • #2
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...
 
  • #3



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.
 

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

1. How does an RC switch circuit work?

An RC switch circuit uses a combination of a resistor (R) and a capacitor (C) to control the flow of electricity. When a voltage is applied to the circuit, the capacitor charges up and allows current to flow. Once the capacitor is fully charged, it blocks any further flow of current. This creates a switching effect, allowing the circuit to be turned on and off.

2. What components are needed to build an RC switch circuit?

To build an RC switch circuit, you will need a resistor, a capacitor, a power source (such as a battery), and a switch. You may also need some wires and a breadboard to connect all the components together.

3. What is the purpose of using an RC switch circuit?

An RC switch circuit is commonly used to control the power supply to a device or system. It can be used to turn on and off electronic circuits, motors, lights, and other electrical components. It can also be used in remote control applications, where the switch can be activated from a distance.

4. How do I calculate the values of R and C for my RC switch circuit?

The values of R and C will depend on the specific application and the desired switching frequency. You can use the formula t = R x C to calculate the time constant (t) of the circuit. The time constant is the time it takes for the capacitor to charge or discharge to 63% of its maximum voltage. You can then adjust the values of R and C to achieve the desired time constant.

5. Are there any safety precautions I should take when working with an RC switch circuit?

Yes, it is important to handle electronic components with caution and always follow safety guidelines. When working with an RC switch circuit, make sure to disconnect the power source before making any changes to the circuit. Also, be mindful of the voltage and current ratings of the components you are using to avoid any potential hazards.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
0
Views
737
  • Engineering and Comp Sci Homework Help
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
2K
Back
Top