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At first I ran into the same issue. Considered statically you're right. But considering that the circuit is built from the capacitors and the voltage source is applied as the last component, a charge current will flow into the capacitors and a DC voltage will develop accordingly across the different capacitors.
Alternatively one may replace the DC voltage source by an AC voltage source. Then it is obvious that AC current will flow. The voltages across the capacitors will be the same as in the DC case.
It is unfortunate to put this question in this form to beginners, I think.
It has nothing to do with DC blocking.
All the capacitors will conduct current till they are "charged-up" to a final DC voltage.
In the steady state (t--> infinity) there will be no more current in the circuit and capacitors ,
each capacitor will have a voltage depending on it's value and the way the capacitors are connected in the circuit.
It is sometimes called a capacitor voltage divider.
Without any resistance in the circuit, the current will be at an infinite value for an infinitesimal amount of time to fully energize the capacitors. I would include a little resistance in the voltage source so that the calculations don't "blow up".
Ratch
@Ratch,
That is true,but for beginners it is a simple steady state homework question,no need to complicate the issue.
And since it is homework,we only need to guide, not solve it for the OP.
@duke37,
here is a simpler question,solvable as a capacitor voltage divider:
Find the voltages and charges on caps(in Farads),try it out.
That is true,but for beginners it is a simple steady state homework question,no need to complicate the issue.
And since it is homework,we only need to guide, not solve it for the OP.
@duke37,
here is a simpler question,solvable as a capacitor voltage divider:
Find the voltages and charges on caps(in Farads),try it out.
After an infinitesimal time and an infinite current spike, the voltage at the junction of c2 and c4 will be 4 volts. The voltage at the junction of c3 and c4 will be 2 volts. The charge imbalance is Q = C E for every capacitor. And the point is?
Ratch
So, you would start with no voltage on any capacitor and then input a step voltage change. In other words, an AC component would be added to the voltage source. This is not shown in the diagram.
In practise, the voltages will be dependant on resistance leakages and the ionisation of the air.
In practise, the voltages will be dependant on resistance leakages and the ionisation of the air.
A step voltage is what a switch does. I don't see any resistance leakages documented in the problem or schematic. Not much ionization at 12 volts. I am just doing what the problem specifies.
Ratch
Yes, that will work if there is no real resistance in the problem and the only interest is the voltage after a long time. But if RC is present in the circuit, that method breaks down.
Ratch
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It works for an idealised problem where the voltage source can supply the infinite current for zero time to capacitors with zero leakage and series resistance.
In short, it will work in the theoretical case which is what these problems represent.
The problem is not properly specified.
The voltage source is 20V not 12V, not that it makes any difference to the problem.
No switch is shown so it is not clear what the initial conditions are.
There is plenty of ionisation here if I skuff my feet over the nylon carpet.
Agreed. That is what I more or less said. It will also work if only the voltage source has some resistance and the time after activation is long.
Ratch
No one asked or is interested in the current(at any time/state of the circuit),
it is totally irrelevant to the question.period!
Why are you needlessly complicating things?
It is a simple Voltage Capacitor Divider question for a novice in the stage of learning DC circuits(who probably didn't even start learning about AC at all) ,the answer is in voltage on the caps,that is all !
Yes, I was thinking of the 12 volt problem Dorke submitted to me. Well, the connection or switch on had to be made at some specific time. Your feet are not in the circuit.
Ratch
It is a DC problem, not an AC problem. The time for final values of voltage will depend on the resistance in the circuit. That is a fact, not a complication.
Ratch
Duke,
I have changed the values intentionally, (assume steady state and ideal battery and caps which is obvious for these questions).
If you can't answer the simple question at #14 ,it is back to school amigo
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... but we can make reasonable assumptions - and we did - to make the problem solvable.
Duke,
I have changed the values intentionally, (assume steady state and ideal battery and caps which is obvious for these questions).
If you can't answer the simple question at #14 ,it is back to school amigo
I can solve the problem in #14 with an AC input but cannot do it without initial conditions, resistance or other current input.
LTspice is much cleverer than I but it cannot solve it, it just goes into a sulk.
Ratch
My feet are not shown in the circuit but I often put my foot in it.
What were the assumptions made?
Trevor
I can solve the problem in #14 with an AC input but cannot do it without initial conditions, resistance or other current input.
LTspice is much cleverer than I but it cannot solve it, it just goes into a sulk.
Ratch
My feet are not shown in the circuit but I often put my foot in it.
What were the assumptions made?
Trevor
My assumptions were that there was no initial voltage on any of the capacitors, and only a DC voltage was applied. I also assumed the circuit did not contain any resistance. If resistance was present, then a full solution would have to show the voltage varying according to time. Resistance or not, the final voltage will converge to the same value after a long interval.
Ratch
I can solve the problem in #14 with an AC input but cannot do it without initial conditions, resistance or other current input.
LTspice is much cleverer than I but it cannot solve it, it just goes into a sulk.
Ratch
My feet are not shown in the circuit but I often put my foot in it.
What were the assumptions made?
Trevor
It is astonishing,truly so!
What the hack do you need LTspice for?
If it were resistors only ,not capacitors, would you use LTspice?
Assumptions?trivial really.
All components (and wires) are ideal.
If you must, then initial conditions :all voltages are 0 volts.
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