Understanding Capacitor Transients: Solving for Initial Conditions

  • Thread starter Thread starter R0CC0
  • Start date Start date
  • Tags Tags
    Transient
Click For Summary

Discussion Overview

The discussion revolves around solving for initial conditions in capacitor transient circuits, specifically focusing on the behavior of current immediately after switching. Participants explore the implications of steady-state conditions and the equations governing capacitor discharge through a resistor.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants assert that in steady state, the current equals zero, raising questions about the behavior immediately after switching.
  • One participant proposes the equation i(t) = A + Be-t/tau, with tau defined as RC, and attempts to apply it to the problem.
  • Another participant challenges the assumption that i(0+) is zero, suggesting that the initial current should be based on the voltage across the capacitor divided by the resistance.
  • There is a discussion about the correct interpretation of initial conditions, with one participant initially misidentifying values for A and B.
  • A later reply clarifies that the initial current can be calculated as 12mA, leading to a proposed equation for i(t) that includes this value.
  • Participants discuss the importance of including the time constant in the exponential term to ensure dimensional consistency.
  • One participant calculates the current at t=1 second, arriving at a value of 4.415mA, and notes the decay characteristic of the exponential function.

Areas of Agreement / Disagreement

Participants express differing views on the initial conditions and the interpretation of the equations involved. While some progress is made in understanding the calculations, no consensus is reached on the initial assumptions regarding current values.

Contextual Notes

Participants navigate through various assumptions about initial conditions and the behavior of the circuit, indicating potential limitations in their understanding of the transient response of capacitors.

Who May Find This Useful

Students and individuals studying electrical engineering or circuit analysis, particularly those interested in transient response and capacitor behavior in circuits.

R0CC0
Messages
18
Reaction score
0

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?
 
Physics news on Phys.org
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...
 
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.
 
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?
 
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??
 
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...
 
Ok so the initial current immediately after switching will be 12mA?
Therefore B will equal 12x10-3?
Thus i(t) = 12x10-3e-t?
 
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?
 
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 :)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K