Help needed : capacitor calculation

[Erasmus]

Hi there!

I'm helping my friend on a few questions for the preparation of the exam of physics in a few days. There's 2 questions she can't solve, so can any of the CPF members help us? Here we go :

1. A capacitor C is not charged. Calculate the current going through R1 and R2 and the potential difference across the capacitor [a] immediately after connecting the switch (t=0) and after the switch being connected for a long time (t=infinite). Here's the setup : http://img50.imageshack.us/my.php?image=dscf0806hp2.jpg

2. A capacitor with capacity C1=2microF is being charged until V0=15V. Then it is disconnected from the battery and with a resistor R=100KOhm connected to a non-charged capacitor with capacity C2=1,5microF. Calculate the current in the circuit and the potential difference across each capacitor in function of the elapsed time. Here's the setup : http://img50.imageshack.us/my.php?image=dscf0809zn3.jpg

I hope this post is not unappropriate, otherwise moderators please tell me so. I just hope someone of the great CPF community can help us out on this one! I bet there's enough knowledge about this matter to solve the questions Please include the way of calculating, that would be great!

Thanks in advance!!!

Erasmus.

 

[jtr1962]

1a) Voltage across capacitor = 0V, I (R1) = 1 mA, I (R2) = 0
1b) Voltage across capacitor = 3.3333V, I (R1) = 0.66667 mA, I (R2) = 0.66667 mA

When the switch is initially closed, since the capacitor is initially at 0V, it can be assumed to remain there. By inspection current through R1 is 10V/10K, or 1 mA, and through R2 is 0. After an infinite time the capacitor charges up to the voltage set by the resistor divider, or 10V*R2/(R1+R2) = 3.3333V. Current across R1 is (10-3.3333)/10K, or 0.66667mA, and through R1 is 3.3333V/5K, or 0.66667 mA.



2) V (C1) = 10 + 5e-0.45t and V(C2) = 10 - 10e-0.45t
I = 1.5e-0.45t mA

The final voltage across both capacitors at t=infinity is 15*C1/(C1+C2) or 10V. The voltage at C1 decreases exponentially from 15V to 10V with a time constant of 0.45 seconds (=R*(C1+C2)). The voltage at C2 increases exponentially from 0 to 10V with the same time constant. The current in the resistor decreases exponentially from 1.5 mA at the instant the switch is closed to 0 mA when C1 and C2 are at the same voltage. Note: C1 = 3 uF according to the diagram, not 2 uF. All calculations based on C1 = 3uF.

Hope this helps.


Not 100% sure about the last one, but I do think the capacitances add together when figuring the time constant.

 

[Erasmus]

Thank you sir! I will check the calculations this afternoon, if anyone has some more advice please post. Thank you!

 

[greg_in_canada]

For the second one the capacitors are in series so you don't add the capacitances together. The value to use is 1/(1/3 + 1/1.5).

You can see that they are in series by pretending the 1.5uF one is very small (say 1.5pF). It would charge up very quickly so the time constant would be very close to 100e3 * 1.5e-12 seconds, not the (approximately) 100e3 * 3e-6 seconds you would get if you added the capacitance together.

Greg

 

[jtr1962]

For the second one the capacitors are in series so you don't add the capacitances together. The value to use is 1/(1/3 + 1/1.5).

Yes, correct. Like I said, I wasn't entirely sure. So the time constant is 0.1 seconds, not 0.45 seconds.

 

[MrAl]

Hi there,


There is more than one way of solving these kinds of circuits.
Since this is a bit of a broad area which incorporates a rather
large amount of information i would post my reply in
a few posts rather than one single post.

Basically here is an outline of some methods:

#1 Knowing beforehand how simple cap and resistor circuits work.
#2 Calculating the frequency domain transfer function and then transforming
into the time domain using the Inverse Laplace Transform.
#3 Using State Vector Differential Equations.
#4 Using a circuit simulator.

Method #1 relies on experience, and is not as general as the other
methods, meaning once you get to a more complex circuit this method will
not be simple anymore. The solutions shown so far basically use this
approach. It works ok usually when only one cap is involved.

Method #2 relies mostly on algebra, which isnt too bad really, and is
still quite general in that you can solve many circuits like this.

Method #3 is probably the easiest when the circuit is complex, but you
usually have to use a computer program that can solve differential
equations numerically, unless you have a simple circuit. This turns out
to be not too hard if you already have the software, and there is already
free software to do this kind of thing out there.

Method #4 is quite simple because all you have to do is draw the circuit
into the programs schematic editor and then run a simulation. If you have
some idea what you are doing this can be very easy to do. If you care
to learn how to use a simulator you can solve some very complex circuits
even if you dont care to study any advanced math. I know people who
dont bother too much with math who still can use simulators successfully.



If you have any preference as to what method you would like to see
first now is the time to mention it

 

[Erasmus]

Thanks to everyone for the help, my friend understands it now.