Understanding Capacitor Transients: Solving for Initial Conditions

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In summary, after switching in steady state, the current will be zero since the capacitor becomes an open circuit. However, if you want to know what happens immediately after switching, you need to solve for i(0+)
  • #1
R0CC0
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Homework Statement


Ok so I have attached the question I am working on.
Capacitor transient question.jpg

Basically when I have the circuit before or after switching in steady state I believe the current will equal zero. However if this is the case, how do I determine what happens immediately after switching?

Homework Equations


i(t) = A + Be-t/tau
tau = RC
(capacitor in steady state becomes an open circuit)

The Attempt at a Solution


tau = 1000x1000x10-6 = 1
i(0-) = 0 = A+B
i(0+) = 0 = B
therefore A = 0?
 
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  • #2
R0CC0 said:

Homework Statement


Ok so I have attached the question I am working on. View attachment 52539
Basically when I have the circuit before or after switching in steady state I believe the current will equal zero. However if this is the case, how do I determine what happens immediately after switching?

Homework Equations


i(t) = A + Be-t/tau
tau = RC
(capacitor in steady state becomes an open circuit)

The Attempt at a Solution


tau = 1000x1000x10-6 = 1
i(0-) = 0 = A+B
i(0+) = 0 = B
therefore A = 0?

i(0+) is not zero. You have a fully charged capacitor connected to a resistor at that time. You have the correct overall equation:

i(t) = A + Be-t/tau

Now you just need to solve it given the initial condition of the charged capacitor which starts to discharge through the resistor after t=0...
 
  • #3
berkeman said:
i(0+) is not zero. You have a fully charged capacitor connected to a resistor at that time.
sorry that was meant to be i(∞) = 0 = B

i(0+) is what I want to know how to work out.
 
  • #4
R0CC0 said:
sorry that was meant to be i(∞) = 0 = B

i(0+) is what I want to know how to work out.

That may be true, but it's not relevant to the solution really. What is i(0+)?

Can you visualize what happens? Some initial current flows based on the initial voltage and the resistance, and that current decays according to the equation. And since you correctly calculated tau as 1 second, and they ask for something after 1 second, can you guess the answer?
 
  • #5
Ignore this post I did it before I saw your last one sorry
Actually sorry I see where I went wrong that should be i(∞) = 0 = A
But that will still give me A+B=0 initially thus giving B = 0??
 
  • #6
R0CC0 said:
Ignore this post I did it before I saw your last one sorry
Actually sorry I see where I went wrong that should be i(∞) = 0 = A
But that will still give me A+B=0 initially thus giving B = 0??

No. What is the initial current? It's the initial cap voltage divided by the resistance, right? In fact, the current at any time is the cap voltage divided by the resistance, right?

Use that and e0=____ (fill in the blank) to figure out the constants...
 
  • #7
Ok so the initial current immediately after switching will be 12mA?
Therefore B will equal 12x10-3?
Thus i(t) = 12x10-3e-t?
 
  • #8
R0CC0 said:
Ok so the initial current immediately after switching will be 12mA?
Therefore B will equal 12x10-3?
Thus i(t) = 12x10-3e-t?

You left the time constant divider out of the exponent. Even though it is "1", you should still show something there to show that the units work. So you could put a 1s in the denominator of the exponential term, for example. Other than that, it looks good. What is the answer at 1 second?
 
  • #9
Ok, I'll remember that for next time.
I got i(1) = 4.415mA
 
  • #10
R0CC0 said:
Ok, I'll remember that for next time.
I got i(1) = 4.415mA

Good! And notice how after 1 second, the value has decayed to 1/e of its initial value. That's how those exponential decays work. :smile:
 
  • #11
Ahh ok that's actually really easy now that I see how it works. I never understood how the initial (0+) condition worked before.
Thanks for all the help :)
 

1. What is a capacitor transient problem?

A capacitor transient problem refers to a situation where there is a sudden change in voltage or current in a circuit due to the presence of a capacitor. This can cause fluctuations or spikes in the circuit, which can lead to issues such as circuit damage or unstable operation.

2. How does a capacitor transient problem occur?

A capacitor transient problem can occur due to various reasons, such as sudden changes in the input voltage, switching of the circuit, or the charging and discharging of the capacitor. It can also be caused by external factors such as electromagnetic interference or circuit layout issues.

3. What are the effects of a capacitor transient problem?

The effects of a capacitor transient problem can range from minor fluctuations in the circuit to more serious issues such as circuit damage or malfunctioning. It can also cause interference with other components in the circuit and affect the overall performance of the system.

4. How can a capacitor transient problem be prevented?

To prevent capacitor transient problems, proper circuit design and layout are crucial. This includes using appropriate bypass capacitors, minimizing the length of high-frequency traces, and avoiding ground loops. Adding surge protection devices and using capacitors with higher voltage ratings can also help prevent transient issues.

5. How is a capacitor transient problem diagnosed and resolved?

Diagnosing a capacitor transient problem can be challenging as it may not always be evident in the circuit. Using specialized equipment such as oscilloscopes and spectrum analyzers can help identify and locate the issue. Once the problem is identified, it can be resolved by implementing appropriate circuit design changes or adding additional components such as capacitors or inductors to mitigate the transient effects.

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