Ok so we are given the capacitance and voltage of the entire circuit composed of only capacitors, the different capacitors are in series and parallel. So my question is, what are the rules/steps for solving for the capacitance, charge, and voltage across each capacitor in the circuit?
Solving for capacitance, charge and voltage in a capacitor schematic?
- Thread starter No0bert
- Start date
Well if you are given the voltages around the circuit, then you should be able to determine the voltage across each capacitor. Then using Q = CV, you should be able to determine the charge within each...
...I feel like I'm missing something - misinterpreting something you've said.
ylp88
...I feel like I'm missing something - misinterpreting something you've said.
ylp88
Right.
Except heres a scenario:
Given a chart with 9 different capacitor values solve for the Charge, and voltage of each.
You are only given the capacitors values and the voltage applied to the entire circuit.
For example, first you find the equivalent capacitance and apply that to the ones in series to figure out charge... etc....
thanks
All types of analysis require a schematic. Without knowing the connections you cannot determine any values. Caps will not act like resistors in a network, with voltage drops, etc. When you turn it on, all the caps will be zero volts, unless previously charged. If dc is applied, they will just charge up to some voltage, and stay there. Caps in series don't increase the capacitance, just the voltage ability.
nevermind -.-
Andres Rodriguez
New Member
Not nice when you are asking for help
How do you know the circuit configuration (which are in series and which are in parallel): you said that you are only given the values and the supply voltage...?
Capacitors in series will have the same charge, but not necessarily the same voltage across them, unless they have the same capacitance.
I still feel you're not telling us something - why so cryptic?
ylp88
Actually, if your entire circuit consists of capacitors, then calculating the voltage across them is the same as for the same network of resistors. Just replace every capacitor with a resistor of value 1/(2*pi*f*C), with pi=3.14..., f the frequency of the applied voltage, and C the capacitance of the particular capacitor. When doing the calculation you'll soon discover that the frequency cancels out everywhere (still assuming all you have is capacitors, and no inductors or resistors), so your solution is also valid at DC.
Wolfgang
Wolfgang
BeanieBots
Moderator
Um, hello.
This is a PICAXE forum, not a "please do my homework" forum.
You have been told about this before.
All you need to solve the problem has been given in the earlier replies.
Here's a little extra.
Capacitors in parallel, add the capacitances.
Capacitors in series is similar to resistors in parallel. 1/total = 1/C1 + 1/C2 etc.
There are many ways to solve a circuit with many capacitor values but I would suspect that your teacher is looking to see if you know and understand the method that was explained prior to setting the assignment.
Have you covered complex numbers yet?
This is a PICAXE forum, not a "please do my homework" forum.
You have been told about this before.
All you need to solve the problem has been given in the earlier replies.
Here's a little extra.
Capacitors in parallel, add the capacitances.
Capacitors in series is similar to resistors in parallel. 1/total = 1/C1 + 1/C2 etc.
There are many ways to solve a circuit with many capacitor values but I would suspect that your teacher is looking to see if you know and understand the method that was explained prior to setting the assignment.
Have you covered complex numbers yet?
The equation provided by womai describes the impedance of a capacitor:
Zc = 1 / jwC
where w is the angular frequency (i.e. 2*pi*f) and j is the imaginary number, sqrt(-1).
Kirchoff's Laws (KVL and KCL) still apply.
ylp88
Surely all this stuff is available at N0berts school library?
And I would have guessed that good pictorial tutorials are available on the nerdynet somehwere?
As said, this is a totally non-PICAXE question. So, its a little bit off that N0obert gets impatient when people don't immediately answer a non PICAXE question -especially as only half the info was supplied.
If the information supplied doesn't answer the question (or, at least, help you) then it would seem that the question is incomplete and you'd be better off asking your teacher. We don't know, for example, the circuit layout. And I don't suppose anyone will answer it now.
But, good luck nonetheless.
And I would have guessed that good pictorial tutorials are available on the nerdynet somehwere?
As said, this is a totally non-PICAXE question. So, its a little bit off that N0obert gets impatient when people don't immediately answer a non PICAXE question -especially as only half the info was supplied.
If the information supplied doesn't answer the question (or, at least, help you) then it would seem that the question is incomplete and you'd be better off asking your teacher. We don't know, for example, the circuit layout. And I don't suppose anyone will answer it now.
But, good luck nonetheless.
Yes, Kirchhoff's Law is perfectly applicable in that case:
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
I'm not going to do anybodies homework, but the posts so far should be more than sufficient for the original poster to solve the rest of the problem by himself.
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
I'm not going to do anybodies homework, but the posts so far should be more than sufficient for the original poster to solve the rest of the problem by himself.
BeanieBots
Moderator
ALL the laws are valid and applicable. That includes the all too often underestimated and over simplified Ohm's Law. V=I*R.
In this case however, the full vector representation of V,I & R needs to be used. I'm sure this was all explained in the lesson.
In this case however, the full vector representation of V,I & R needs to be used. I'm sure this was all explained in the lesson.
I think it was supposed to be a puzzle. But he did not supply enough info. Like
if the applied signal was dc or ac. If dc, a bunch of caps will not act like resistors, once they charge. In fact, capacitors block dc.
The way I see it, caps stabilize voltage, inductors stabilize current, and resistors just have a voltage across them from a current going thru them, and dissipate power, where the ideal cap or inductor doesn't ( or is not supposed to). Yes, a simplification without 2 pie f or phase effects. Here is a real puzzle, with resistors, maybe you heard of it, I had to solve on a real test:
Imagine a cube made of 1 ohm resistors, the edges. You have 2 connections
on opposite corners. Using delta Y calculations, determine the equivalent R.
This is actually hard to do that way, about 3 pages of equations. It was an
extra credit question. I had some time left, and tried to write a kirchoffs eq on the nodes, since there were 3 resistors, assuming a current of 3A would
divide equally 3 ways, it was easy to assign the values. Writing voltage loop equations then solved the overall resistance value. Puzzles can be valuable!
if the applied signal was dc or ac. If dc, a bunch of caps will not act like resistors, once they charge. In fact, capacitors block dc.
The way I see it, caps stabilize voltage, inductors stabilize current, and resistors just have a voltage across them from a current going thru them, and dissipate power, where the ideal cap or inductor doesn't ( or is not supposed to). Yes, a simplification without 2 pie f or phase effects. Here is a real puzzle, with resistors, maybe you heard of it, I had to solve on a real test:
Imagine a cube made of 1 ohm resistors, the edges. You have 2 connections
on opposite corners. Using delta Y calculations, determine the equivalent R.
This is actually hard to do that way, about 3 pages of equations. It was an
extra credit question. I had some time left, and tried to write a kirchoffs eq on the nodes, since there were 3 resistors, assuming a current of 3A would
divide equally 3 ways, it was easy to assign the values. Writing voltage loop equations then solved the overall resistance value. Puzzles can be valuable!