How long would it take the capacitor in the given circuit to completely discharge?

[charmcaster.engg]

How long would it take the capacitor in the given circuit to completely discharge if the switch was in position 2?

 

[Laplace]

What is the general exponential equation for capacitor discharge?

 

[hevans1944]

Answer this question first: How long did it take the capacitor in the given circuit to completely charge if the switch was in position 1?

 

[charmcaster.engg]

Answer this question first: How long did it take the capacitor in the given circuit to completely charge if the switch was in position 1?

I don't know. Can you provide me explanation/web link about the same?

 

[Laplace]

It is just a question about how long the switch has been in position #1. Has it been long enough for the capacitor to become fully charged? Is that question not relevant to the problem? What is the purpose of R1 if it is not relevant to the problem? Or is it a trick question where you are expected to realize what is relevant and what is not?

 

[davenn]

charmcaster, if you don't start making good attempts at finding the answers and showing us what you have done to find those answers,
I am going to give you a holiday from the forum

Dave

 

[hevans1944]

I don't know. Can you provide me explanation/web link about the same?

I provided you with this link in reply to a similar question you asked in another post. Clearly, you either did not follow the link or did not understand the content therein, for you replied not, either to thank me for my "answer" or indicate you have any understanding of the subject matter.

Therefore, I must personally conclude that you are either incapable of critical thinking, or not willing to do your own research, or possibly both. Alternatively, perhaps there is a language or cultural barrier, which means I am also part of the problem, as I have never been to your part of the world and do not know your native language. Either way, it is obviously a waste of my time to attempt to alleviate your learning deficiencies using American English and a Western approach to problem solving. I must put you on my "ignore" list and move on to other horizons. Good luck @charmcaster.engg with your personal efforts to master the deep mysteries of electricity and perhaps even electronics.

 

[charmcaster.engg]

Capacitor topic is difficult for me. It will take time for me to understand.

 

[charmcaster.engg]

If at position 1 capacitor is fully charged then Vc=30V, I=0 and Q=300u. how to solve it further?

 

[Laplace]

If the problem statement said nothing about how long the switch has been in position #1, then all you can do is assume it has been long enough for the capacitor to become fully charged. (hevans1944 asked what minimum time was long enough)

But you should not try to over-think the solution. My linear circuits textbook had a section on how to analyze switched (aka step response) RC & RL circuits with one energy storage element, and how to do it graphically because there are only two possibilities (and duality applies, i.e., capacitor current behaves the same as inductor voltage, and capacitor voltage behaves the same as inductor current.)

So the two possibilities are as follows for quantity 'X' when switched at t=0:
1: 'X' starts with an initial value, then decays to zero.
2: 'X' slowly rises from zero to its final value.

If #1, use X=(initial value)∙e^(-t/τ)
If #2, use X=(final value)∙(1-e^(-t/τ))
where τ is the time constant, either R∙C or L/R

Note that the initial slope of the exponential curve is such that 'X' would achieve its final value in one time constant. Also note that some circuits may have constant offsets involved but this discussion only applies to the transient behavior. You could generalize the equations for non-zero offsets. Thevenin equivalent of the resistor portion of the circuit may be necessary.

So for any switched RC or RL circuit where the 'X' is the capacitor or inductor voltage or current, you just need to decide whether the waveform rises or decays from t=0 and decide what the initial or the final value is. Here is where you need to focus on the simplicity of the waveforms, not on the complexity of all the situations this applies to.

 

[charmcaster.engg]

If the problem statement said nothing about how long the switch has been in position #1, then all you can do is assume it has been long enough for the capacitor to become fully charged. (hevans1944 asked what minimum time was long enough)

But you should not try to over-think the solution. My linear circuits textbook had a section on how to analyze switched (aka step response) RC & RL circuits with one energy storage element, and how to do it graphically because there are only two possibilities (and duality applies, i.e., capacitor current behaves the same as inductor voltage, and capacitor voltage behaves the same as inductor current.)

So the two possibilities are as follows for quantity 'X' when switched at t=0:
1: 'X' starts with an initial value, then decays to zero.
2: 'X' slowly rises from zero to its final value.

If #1, use X=(initial value)∙e^(-t/τ)
If #2, use X=(final value)∙(1-e^(-t/τ))
where τ is the time constant, either R∙C or L/R

Note that the initial slope of the exponential curve is such that 'X' would achieve its final value in one time constant. Also note that some circuits may have constant offsets involved but this discussion only applies to the transient behavior. You could generalize the equations for non-zero offsets. Thevenin equivalent of the resistor portion of the circuit may be necessary.

So for any switched RC or RL circuit where the 'X' is the capacitor or inductor voltage or current, you just need to decide whether the waveform rises or decays from t=0 and decide what the initial or the final value is. Here is where you need to focus on the simplicity of the waveforms, not on the complexity of all the situations this applies to.

Thanks!

 

[Merlin3189]

I'm a bit puzzled by this question. I'd have thought the obvious, simple answer is, it will never discharge completely. And the time the switch has been in position 1 is a red herring as it will never charge completely either!
The question may be a very simple one to ask that point.
Otherwise the question is inadequate for calculating a finite answer.

NB. the question says, "...completely discharge"

 

[Gryd3]

I'm a bit puzzled by this question. I'd have thought the obvious, simple answer is, it will never discharge completely. And the time the switch has been in position 1 is a red herring as it will never charge completely either!
The question may be a very simple one to ask that point.
Otherwise the question is inadequate for calculating a finite answer.

NB. the question says, "...completely discharge"

Well.. in a perfect world.. the capacitor will never discharge, but it will get infinitely close to 0V.. additionally, it would get infinitely close to the supply voltage. In the real world, leakage in the capacitor will actually allow it to completely discharge, and the infinitely close supply voltage will be a little lower. That said, It's my understanding that it takes 5 "Time Constants" to be considered as fully charged or discharged. At some point, you need to be able to say "Close Enough!" it's charged, or it's empty.
A single Time Constant can be calculated with "C * R"
In the provided diagram in post #1 R2, R3, and C1 would be used. (Calculate R would be the sum of R2 and R3)

 

[charmcaster.engg]

Well.. in a perfect world.. the capacitor will never discharge, but it will get infinitely close to 0V.. additionally, it would get infinitely close to the supply voltage. In the real world, leakage in the capacitor will actually allow it to completely discharge, and the infinitely close supply voltage will be a little lower. That said, It's my understanding that it takes 5 "Time Constants" to be considered as fully charged or discharged. At some point, you need to be able to say "Close Enough!" it's charged, or it's empty.
A single Time Constant can be calculated with "C * R"
In the provided diagram in post #1 R2, R3, and C1 would be used. (Calculate R would be the sum of R2 and R3)

Thanks!

 

[Merlin3189]

Yes, of course.
Perhaps this is assumed by the questioner, though I'd never seen 5 Tc in textbooks - just a rule of thumb for practical chaps. (And I do wonder why one would want to know the 5 Tc discharge time?)
Anyhow, the only thing to do is to calculate the time constant, then explain how you draw your conclusion based on that.
On Laplace's point about how long the capacitor has been charged before you flip the switch, from a few rough calcs I've done, it doesn't seem to make much difference. As you say, after 5 Tc, it's changing very little, 99.5% of the change has happened and the difference due to the starting point has been reduced to 0.5% of the initial difference. Discharging from 15 V instead of 30V doesn't make a difference of 1 Tc after 5 Tc. (attachment) Though it would make a difference to the level at 1 Tc.

The useful point to make for the OP here, is that capacitors are not much use for timing more than one Tc - the time error for any voltage threshold is too great. If you use a capacitor for timing, like in 555 circuits, you need to set your threshold voltage where the capacitor is (dis)charged to no more than about 50%.

And whatever one says in formulae / equations, I don't think you can beat looking at the exponential graph to get a feeling for what's happening. Just imagine trying to read off the voltage when the graphs reach 5 Tc.

 

[charmcaster.engg]

Yes, of course.
Perhaps this is assumed by the questioner, though I'd never seen 5 Tc in textbooks - just a rule of thumb for practical chaps. (And I do wonder why one would want to know the 5 Tc discharge time?)
Anyhow, the only thing to do is to calculate the time constant, then explain how you draw your conclusion based on that.
On Laplace's point about how long the capacitor has been charged before you flip the switch, from a few rough calcs I've done, it doesn't seem to make much difference. As you say, after 5 Tc, it's changing very little, 99.5% of the change has happened and the difference due to the starting point has been reduced to 0.5% of the initial difference. Discharging from 15 V instead of 30V doesn't make a difference of 1 Tc after 5 Tc. (attachment) Though it would make a difference to the level at 1 Tc.

The useful point to make for the OP here, is that capacitors are not much use for timing more than one Tc - the time error for any voltage threshold is too great. If you use a capacitor for timing, like in 555 circuits, you need to set your threshold voltage where the capacitor is (dis)charged to no more than about 50%.

And whatever one says in formulae / equations, I don't think you can beat looking at the exponential graph to get a feeling for what's happening. Just imagine trying to read off the voltage when the graphs reach 5 Tc.

Thanks!