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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.I do not understand the question. Why should there be any voltage when the DC is blocked by C1?
I do not understand the question. Why should there be any voltage when the DC is blocked by C1?
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.
@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.
View attachment 38362
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.
There is no need for a little resistance or a lot of time...
All you need to understand is what DC voltage appears at the point between two arbitrary capacitors (initially at 0V) when a DC voltage exists at the ends.
They form a voltage divider producing the same voltage as a pair of resistors with a resistances inversely proportional to the capacitances.
So, 1 uf and 2 uf could be replaced with resistors 1 ohm and 0.5 ohms.
Then work out the voltage on the nodes. It will be the same as in the capacitance problem.
In this particular case it probably doesn't require you to solve any equations (least of all differential equations) if you don't want to because it can be treated stepwise as resistors in series and parallel.
Yes, that will work if there is no real resistance in the problem
The problem is not properly specified.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
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.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.
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
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.
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 !
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.
... but we can make reasonable assumptions - and we did - to make the problem solvable.The problem is not properly specified.
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
What were the assumptions made?... but we can make reasonable assumptions - and we did - to make the problem solvable.
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