MRAl

VERY, VERY , VERY, GOOD explanations.

 

Hello again,


Well thank you 12345...etc. I hope that helped some.

 

What do you mean by “electronic inventor”?? Are you defining that as someone who is developing leading edge technology or someone who is trying to come up with just another “me too mouse trap”?? Those are two very different environments. Development and general design work just aren’t the same and never will be.

Now, as for, What's obvious is that you don’t know me, what I do for a living or for how long I’ve been doing it, so that is a rather arrogant comment on your part.

 

Ok, what about this simple one?

Can we find the final voltage of the capacitor after the Switch of 180 ohms Resistor is closed.

Attached Thumbnails

 

 

Can somebody suggest me how to solve the above circuit?
It is not a simple RC circuit at some times (when we close the SW of 180Ohms resistor)

 

Hi,

Are you saying that the first lower switch is closed already and then you close the
upper switch and you want to know the voltage across the capacitor after a
very long time has passed?
If so, it is just the voltage across the resistor as long as both switches stay closed.


Just to note, it helps if you label the switches too so we can refer to them
as SW1, SW2, etc.

 

I think I will end up with just a voltage divider via resistors, and the cap will have the volatge across the 180R resistor and has nothing to do even if it has alery been charged via the 400R resistor?

Can we find any time constant for the cap and see whenit reachs to voltage of 180R?

besides what about when the cap is charging but not full charged then how can I calculate the time constant for this circuit?

Thanks

 

Hello again,


The time constant of this circuit isnt too hard to calculate, but you have to realize
that the more complicated the circuit gets the harder it starts to become to calculate
the time constant and you eventually have to resort to more sophisticated circuit
analysis techniques that you would have to learn. Some of the techniques are not
that difficult however but they do take a little while to learn.

As you know, for a single resistor and capacitor R and C the time constant is
simply R*C. For this simple circuit the time constant can still be calculated by
multiplying R times C, but before that we have to calculate what R is.
For this circuit, R is simply the parallel combination of R1 and R2:
R=R1*R2/(R1+R2)
and so again the time constant is:
R*C.

As you also know the voltage across the capacitor in a single RC circuit is:
Vc=Vs*(1-e^(-t/RC))

but there is another thing to think about with the two resistor circuit, and that is the
source voltage level Vs.

With a single R and C we usually take the voltage to be the full battery voltage,
but for this circuit because there is a divider we also have to calculate a new
voltage value. Since we want the voltage across the capacitor, we first calculate
the divider voltage Vd:
Vd=Vbatt*R1/(R1+R2)
so
Vd=12*180/(180+400)

and now we have the right voltage to use for the equation:

Vc=Vd*(1-e^(-t/RC))

with R calculated above as the parallel combination of R1 and R2.

Note that Vc is now the voltage across the capacitor, with the reference
node being the junction of R1 and R2. Thus if we used a meter to try to see
this change we would put the black lead on the junction of R1 and R2 and the
red lead on the +12v power supply positive terminal.

If you want to calculate the voltage at the node junction of R1 and R2 with
reference to ground, then you would change Vd to be the voltage across
the 400 ohm resistor due to the voltage divider instead of across the 180
ohm resistor.

Note the above analysis is for when both switches are turned on at the same time.
If you want to calculate what happens when switch 1 is turned on first and switch
2 is turned on later, you have to first calculate the single RC circuit and then add
the second resistor later when switch 2 is turned on. When switch 1 is turned on
the cap starts to charge up with full battery voltage until it reaches the full 12v.
Later when switch 2 is turned on the cap starts to discharge until it reaches the
final value determined by the voltage divider.
The first part is easy because it's just a single RC circuit, but with the second
part (switch 2 turned on) you would have to calculate the difference between the
12v supply and the R1,R2 node (voltage across C1) and so Vd would come out to
the difference between 12 and the voltage divider 3.72v (final value of node R1,R2)
and so the cap voltage then would be:

Vc=12-(12-3.72)*(1-e^(-t/RC))

where again R is the parallel combination of R1 and R2 and the time constant again
is R*C.

The above assumes that the capacitor C1 has been allowed to charge to the full voltage
of 12v before the second switch was closed. If the capacitor was not allowed to charge
for at least 5 time constants or so, then the initial cap voltage is not 12v when switch
2 is closed so we have to modify the formula to account for this by changing the 12v to
whatever the cap voltage was just before switch 2 is closed:

Vc=Vc0-(Vc0-3.72)*(1-e^(-t/RC))

so if the initial cap voltage is 6v then:

Vc=6-(6-3.72)*(1-e^(-t/RC))

Some simplification of this last formula might be possible also.

 

Hello And thanks a bunch for your very very helpful posts.

Very complete explanation thanks for it too, but one thing that I am not able to understand is Why there is a PARALLEL combination of R1 and R2 in this circuit? I am not able to see any though!?

 

Hello again,


That's a good question and one that is a little hard to understand at first, but
a simple application of superposition helps to explain a little bit.

One way to fully analyze this circuit is to use superposition where the two
sources to be 'killed' are actually one and the same source, the 12v supply.
That's what we will attempt now...

The first step would be to realize that the 12v source feeds BOTH the resistor
R1 and the capacitor C1. Simply put, they both connect to that 12v supply.
To make this circuit simpler to analyze, we can use superposition by performing
a few simple manipulations on the circuit to allow the individual analysis of two
parts of the circuit (which are simpler than the whole) and then add the results
together to get the final analysis. This means there will be two steps where
we manipulate the circuit and one final extra step where we simply add the two
results together:

[1]
Disconnect the left side of C1 and short it to ground (which BTW puts it in parallel
with R2) and take the response at the R1,R2 node. This gives us a voltage we will
call simply Va. So far though the analysis is incomplete, but we have simplifed the
circuit a little.
[2]
Reconnect the capacitor as it was before to the 12v supply, then disconnect the
left side of R1 and short that lead of R1 to ground instead of to the 12v supply.
Note that doing this puts R1 in parallel with R2, which is the key to this question of
how R1 gets in parallel to R2. Then, find the response at the R2,C1 node (same
node as before but now the left side of R1 is shorted to ground). We'll call this
new voltage response Vb.
[3] (final step)
The only thing left to do now is to add Va plus Vb, and that gives us the total
response, but note that in order to get Vb we had to put R1 in parallel with R2.

Another way to find out how R1 gets in parallel with R2 is to do a full analysis
of the circuit and look at the time constant in the resulting formula. It's the
denominator D of the exponent of e^(-t/D) which results. From that we can
get the value of the equivalent resistor R in the exponential part of the formula
which written out that way comes out to e^(-t/RC). If we did this numerically
we could simply divide the denominator by the capacitance C, but if we did this
algebraically we would actually end up with R*C in the denominator, and R would
come out equal to the parallel combination of R1 and R2.

I know this sounds a little strange at first, but after you do a few problems like
this on paper it starts to make a lot of sense.




If instead of using superposition we could do a full analysis like this...


First, transform all the elements into their impedance equivalents:
R1=>R1
R2=>R2
zC1=1/(s*C1)

Next, calculate the parallel combination of R1 and zC1 and call this
new impedance Zu:
Zu=R1/(s*C1*R1+1)

Next, calculate the voltage divider effect of combined impedance Zu and R2:
V=E*R2/(Zu+R2);

which comes out to:
V=E*(s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)

Now we want the step response so multiply that by 1/s:
V=(1/s)*E*(s*C1*R1*R2+R2)/(s*C1*R1*R2+R2+R1)

which comes out to:
V=E*(s*C1*R1*R2+R2)/(s^2*C1*R1*R2+s*R2+s*R1)

Now transform to the time domain using the inverse Laplace transform, and we get:
V=E*(R1*e^(-t*(R2+R1)/(C1*R1*R2)))/(R2+R1)+E*R2/(R2+R1)

and now we concentrate on the exponential part:
e^(-t*(R2+R1)/(C1*R1*R2))

and note that this is the same as:
e^(-t/(R*C1))

where R*C1 is the time constant if we substitute:
1/((R1+R2)/(R2*R1))

with R, and that:
1/((R1+R2)/(R2*R1))=(R1*R2)/(R1+R2)

and finally we note that that last result is equal to R:
R=(R1*R2)/(R1+R2)

which is again the parallel combination of R1 and R2. Thus, R*C1 is the time constant
of the circuit and R equals the parallel combination of R1 and R2.

 

Thanks a lot sir, you are a genius,

Just another question is why we want to do so? is that due to this fact that the capacitor has a charge voltage and then a discharge voltage or based upon having 2 power sources in the circuit..??

Please tell me if we have to use superposition law for all firs order circuits base up on capacitors ( due to existance 2 sources: a, power supply and b, the capacitor itself)?

Thanks a bunch

 

Superposition is for when we have 2 or more power sources, why could you use it here in this circuit too?

 

Hi again,


We are not forced to use superposition, but for the question of how R1 gets in parallel
with R2 to form the time constant R*C1 can be visualized a little easier when looking
at what happens when we do use superposition. See the attached drawing for a
graphical view on this and note the last circuit how R1 and R2 ended up in parallel.

You can use superposition for any circuit like this really. It's a more general principle
that pertains to linear circuits.

Here's the drawing:

Attached Thumbnails

 

 

Oh, now I understand what you are talking about, you have used the source transfer or something so to make the single sourece to be converted in to 2 and then use superposition law. good job.

sorry but 2 other questions.
Suppose the SW2 (the above switch) is open for hours, and I turn the left SW on (while the capacitor has no charge) and want to calculate the time constant for the capacitor for while the 9V is across it ( 6V is across the R2), Then I am not able to use the the formula (t=RC) so I have to use the first order formula I,e
Vc=Vb(1-e^(-T/RC)) so: 9=15(1-e^(-T/(400x1000uF)),
If I find "T" then It is my time constant i.e when the capacitor reaches to "9V" right?
1: How to convert the above formula so that I could calculate "T"? I know that I have to use "Ln" but can you let me the right formula to do it so that I do not make mistake (sorry my poor math)?
2: do I need to multiple the calculated "T" by 5 after I calcualte it by the formula I asked you?

 

Hi again,

If you start with a formula like:
Vc=Vs*(1-e^(-t/RC))

and you want to solve for t, you have to manipulate the formula a bit using
algebra and logarithms.

Starting with:
Vc=Vs*(1-e^(-t/RC))

we can first distribute the Vs to get:
Vc=(Vs*1-Vs*e^(-t/RC))

then simplify and drop the outer parens and we get:
Vc=Vs-Vs*e^(-t/RC)

and then subtract Vs from both sides and we get:
Vc-Vs=-Vs*e^(-t/RC)

and then divide both sides by Vs and we get:
(Vc-Vs)/Vs=-e^(-t/RC)

and then multiply both sides by -1 and we get:
(Vs-Vc)/Vs=e^(-t/RC)


and now that we have the exponential part isolated we can take the natural log
of both sides and we get:
ln((Vs-Vc)/Vs)=(-t/RC)

drop the right side outer parens and we get:
ln((Vs-Vc)/Vs)=-t/RC

multiply both sides by -1 and we get:
-ln((Vs-Vc)/Vs)=t/RC

and finally multiply both sides by RC and we get:
-RC*ln((Vs-Vc)/Vs)=t

and swap sides to get the final formula:
t=-RC*ln((Vs-Vc)/Vs)

So now knowing R,C,Vc and Vs, we can calculate t instead of calculating Vc as before.

When you calculate t in this way you do not need to multiply that t by 5 because
that is the actual t that would cause the voltage Vc.