Analyzing an RC Circuit with a Sudden Switch Closure

In summary, the conversation discusses a circuit with a switch that has been open for a long time and is suddenly closed. The time constant before and after the switch is closed is determined, and the current in the switch as a function of time is also discussed. The equations used for charging a capacitor and the concept of capacitance are mentioned. The conversation also addresses using Kirchhoff's laws when the switch is closed and the resistors are in series.
  • #1
bassplayer142
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Homework Statement


In a circuit, the switch S has been open for a long time. It is then suddenly closed. Determine the time constant (a) before the switch is closed (b) after the switch is closed. (c) let the switch be closed at t=0. Determine the current in the switch as a function of time.


Homework Equations


for charging cap
q(t) = CE[1-e^(-t/RC)]
I(t) = E/Re^(-t/RC)
c=q/v v=IR

The Attempt at a Solution



I assume that since it has been open for a long time the cap is fully charged to 1E-6F. Using the first equation I can't find Q because if you plug in q=cv in the equation it is ln(1)=0 then the time is 0. I know that the answer to a is t=1.5s from the back of the book. I guess I just don't know how to find Q. Also, since the 50kohm and 100kohm resistors are in series but not next to each other can I used the series law to make it a 150kohm resistor. I guess I don't know since there is either a capacitor or a battery in between them.
 

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  • #2
Capacitors do not get charged up to a certain capacitance, They get charged until the voltage across them is so high that no more current can flow.
If you have a capacitor, a resistor and a battery in series, that point is reached once the voltage across the capacitor is equal to the voltage across the battery, so the voltage across the resistor is 0.

Unfortunately I can't see the picture yet.
 
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  • #3
If the switch is open the resistors and the capacitor will all have the same current passing through them. that means it's OK to swap two of them around. You can prove this with kirchhofs loop law. After the switch closes you have two loops with the current in them independent. The sum of those currents goes through the switch
 
  • #4
So basically I can treat the two resistors as being in series, but when the switch is closed I must use kirchhoff's laws? Sounds right to me.
 

Related to Analyzing an RC Circuit with a Sudden Switch Closure

1. What is an RC circuit?

An RC circuit is a type of electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series or parallel. It is commonly used in electronic devices for filtering, timing, and signal processing.

2. What is the time constant of an RC circuit?

The time constant of an RC circuit is a measure of how quickly the capacitor charges or discharges. It is calculated by multiplying the resistance (R) and the capacitance (C) of the circuit, and is represented by the symbol "τ" (tau).

3. How do you calculate the time required for a capacitor to fully charge in an RC circuit?

To calculate the time required for a capacitor to fully charge in an RC circuit, use the formula t = 5τ, where t is the time and τ is the time constant. This means that after 5 time constants, the capacitor will be approximately 99.3% charged.

4. What happens to the charging time if the resistance in an RC circuit is increased?

If the resistance in an RC circuit is increased, the charging time of the capacitor will also increase. This is because a higher resistance will result in a slower flow of current, thus taking more time for the capacitor to charge to its full capacity.

5. How does a DC voltage source affect the time constant of an RC circuit?

A DC voltage source does not affect the time constant of an RC circuit. The time constant is determined solely by the values of the resistance and capacitance in the circuit, and is independent of any external voltage source.

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