example problem about capacitor

[PG1995]

Hi

Please help me with the problem in the attachment. It would be kind of you. Thanks.

Regards
PG

 

[MrAl]

Hi PG,

Q1:
Im not sure if you are asking this so if not just reply again...
"Vo(t)" simply means "the output voltage as a function of time", or to put it another way, "the output voltage in the time domain".
Many times the 't' is dropped when it's obvious that we are working in the time domain so we would just have "Vo".

Other common terms like this would be:
"Vo(f)" the output voltage as a function of frequency,
"Vo(w)" the output voltage as a function of angular frequency (w=2*pi*f),
"Vo(s)" the output voltage in the complex frequency domain, or output voltage as a function of complex frequency, sometimes shortened to just "the output voltage in the frequency domain".

Another common term would be:
|Vo(s)|
or:
abs(Vo(s))
or:
ampl(Vo(s))
or:
norm(Vo(s))

where all these mean the amplitude of the signal disregarding the phase.

Q2:
The equivalent resistance across C is the parallel combination of the three resistors. The right two are in series, that makes one resistor then that is in parallel with the first resistor.
This is the resistance that is used as part of the time constant of the circuit. It happens this way because the source is considered a zero ohm impedance so effectively the two resulting resistances are in parallel.
The catch is that although they are in parallel for the time constant, they are in series for the max amplitude. Then you have the two on the right which change the amplitude once again.

So to do this the way i think you are approaching it, you have the two resistors on the right which are in series so equal 6k, then the one on the left which is 3k so the 3k and 6k are in parallel for the time constant.
For the amplitude, you have the 3k and 6k voltage divider, but then you have another voltage divider to get to the output which is the two resistors on the right the 4k and 2k which cause another voltage division.

For a charging capacitor however you dont use e^-t/RC, you use 1-e^-t/RC as the basic form of the equation. You also end up with a voltage multiplier so you get this form:
Vo(t)=Vin*A*(1-e^-t/RC)

where A is due to the series resistance voltage divider effects.

 

[PG1995]

Once again, many, many thanks, MrAl.

I have solved it but ended up with wrong answer. Please have a look on the attachment.

Best wishes
PG

 

[MrAl]

Hi,

In series you get 2/3 times 12v equals 8v as you already found.
In parallel you get 2k which you already found.
Therefore the time constant is 2000*C=2000*0.0001 (100uf cap) which equals 0.2 so the time constant RC=0.2 and that's easy enough.

Now that we know both the multiplier and the time constant all we have to do is stick it into the defining equation for a charging capacitor:
Vc=A*(1-e^(-t/RC))
and so we get:
Vc=8*(1-e^(-t/0.2))
or
Vc=8*(1-e^(-5*t))

Note the answer is not anything times e^(-t/0.2) it also contains the '1' because it is charging. If the answer really has the form e instead of 1-e then that means that they may be asking for the equation as the capacitor DISCHARGES. In that case we would assume the cap was charged to 8v before we start and then go from there.

 

[PG1995]

Hi,
Note the answer is not anything times e^(-t/0.2) it also contains the '1' because it is charging. If the answer really has the form e instead of 1-e then that means that they may be asking for the equation as the capacitor DISCHARGES. In that case we would assume the cap was charged to 8v before we start and then go from there.

Hi MrAl

Please check the attachment. Thanks.

Regards
PG

 

[MrAl]

Hi,

This is why it is important to post the original question rather than a redrawn version. In the original version they are clearly indicating that the switch is being OPENED at t=0, and this means the capacitor will be discharging therefore we will get a response in the form:
Vc=A*e^(-t/RC) and not the other form for a charging capacitor.

You you should end up with the same answer they got assuming they did it right.

 

[PG1995]

Sorry for the confusion, MrAl. Now I understand that that curved arrow shows that switch is being moved after t=0. But I'm curious about that Vo(t) shown on the drawing. Why is it there? The author is also using the same circuit diagram for the next problem given below. Please have a look on the attachment.

Best wishes
PG

 

[MrAl]

Hi,


Vo(t) is just the 'normal' output of the network that's all. They may or may not want you to solve for that in this particular problem. No big deal really.