Voltages - continuation of the discussion

[Harald Kapp]

LTspice is much cleverer than I but it cannot solve it, it just goes into a sulk.

Not if you ramp the input voltage up from 0 V to 20 V over say e.g. 1 ms. This allows the simulation of the charging current [removed reference to post #9 as it belonged to the original thread].

 

[duke37]

What the hack do you need LTspice for?

I have used LTspice to simulate audio filters and discriminators, also RF filters and replacement components to eliminate the dropper resistor in AC/DC valve radios. The last job I did was to find the value of the gimmic capacitor for top coupling an IF transformer. Why do you need to know?

LTspice will not accept floating nodes in the circuits I have simulated and I usualy ground them through a 1E9Ω resistor. Doing this in the present case will obviousy give an incorrect value.

A politician will change the question to suit the answer he wants to give, will divert the question to avoid an answer he cannot give and, if all else fails, will imply that the competance of the antagonist is suspect. There has been some of this in this thread which is below the normal standard of electronicspoint.

 

[dorke]

Dear duke,

Some problems ,like this one,are so basic and trivial.
This is a school question,and as such, all parts(and conditions) are assumed to be ideal,
that goes without saying!

They should be answered with paper and pencil and should take you less than a minute to complete.
The fact that you take it to the "simulation world" with LTspice is simply astonishing to me.

Can you do it with paper and pencil?

 

[duke37]

Dear dorke, you seem to make unwarranted assumptions.
I did not take it to LTspice since I had experience with it in other applications and know that floating nodes are not accepted. Harald has described how LTspice can be made to accept them by adding a varying voltage giving an AC component.


I cannot do it with paper and pencil. Xc=1/(wC) which is infinity. How are a couple of these added or proportions calculated?
Could you show how to calculate the voltage at the junction of two capacitors with a DC battery supply?
Please give assumptions and whether an AC component is used.

The trivial answer is that it cannot be done since capacitors block DC. Perhaps it is a trick question.

 

[Harald Kapp]

I cannot do it with paper and pencil. Xc=1/(wC) which is infinity.

Paper and pencil work, your method doesn't.

Assume you have reduced the circuit to a single equivalent capacitor (cf. post #6 here). Then you can calculate the charge in that capacitor from Q = C*V.
Then go one step back to where two capacitors are present (the two which had been condensed into one equivalent capacitor in te previous steps). From the fact that the charge calculated in step one is equally distributed between these two capacitors you can calculate the voltage across them (V = Q/C).
Repeat these steps until you arrive at the original circuit with all the single capacitors present. Tedious but doable.

 

[Ratch]

Dear dorke, you seem to make unwarranted assumptions.
I did not take it to LTspice since I had experience with it in other applications and know that floating nodes are not accepted. Harald has described how LTspice can be made to accept them by adding a varying voltage giving an AC component.


I cannot do it with paper and pencil. Xc=1/(wC) which is infinity. How are a couple of these added or proportions calculated?
Could you show how to calculate the voltage at the junction of two capacitors with a DC battery supply?
Please give assumptions and whether an AC component is used.

The trivial answer is that it cannot be done since capacitors block DC. Perhaps it is a trick question.

Assume three caps designated c1 = 1 farad, c2 = 2 farads, and c5 = 5 farads are connected in series across 8 volts. The caps have no initial voltage. Since Q = C V, and the same imbalance of Q is present in all caps in series, the voltage equation will then be;
x/1+x/2+x/5==8 -----> 10x+5x+2x = 80 ----->x = 80/17 coulombs, where x is the number of coulombs imbalanced on each cap. Therefore, the volts across c1,c2,c5 will be 80/17, 40/17, and 16/17 respectively. As you can see, it all adds up to 8 volts.

Caps are different than resistors in that they store energy instead of dissipating it. Therefore, when the voltage source is disconnected, the energy is still in the circuit instead of being gone like it would be for resistors.

Caps don't really block DC. They neutralize it by building up a reverse voltage.

Ratch

 

[dorke]

O.K Duke,
Let's take it step by step to solve this problem,from #5,the answer was given by Ratch at #6

We can treat this simple circuit as we do with resistors only.

Equivalent cap and resistors values:
The calculations of capacitors in parallel is like that of resistors in series(addition).
The calculations of capacitors in series is like that of resistors in parallel.

Voltage divider for caps and for resistors:
For 2 resistors in series we have Vr1=VxR1/(r1+r2)
For 2 capacitors in series we have Vc1=VxC2/(c1+c2)
Note the difference!!!
I don't want to go into how it is derived,if you need to,we can do that as well.

So Lets calculate,,
for C3 and C4 ,which are in series, we get C34=(C3xC4)/(C3+C4)=(2x2)/2+2=1.
C2 is in parallel with the above calculated C34 thus we get C234=C2+C34=1+1=2.


Calculating voltages :
on C1:
VC1 =VxC234/(C1+C234)=12x2/(1+2)=8V

on C2
VC2=VC234=12-8=4V

on C3,C4
since C3=C4 we get VC3=VC4=VC234/2=4/2=2V

Note also that the charge on each capacitor in a series connection is the same!

It takes much longer to write it down than to calculate it by heart,
that is the reason I changed the values and circuit to give an easy round numbers solution.

Hope it clarifies things.

 

[duke37]

Caps don't really block DC. They neutralize it by building up a reverse voltage.

Taps don;t really block water. They neutralise it by building up a reverse pressure.

When I saw a calculation containing 17, it reminded me of the camel problem where the question was altered to give a solution.

There is no point in continuing with this. I would like to know what the official answer is and how it was obtained.

 

[Ratch]

Taps don;t really block water. They neutralise it by building up a reverse pressure.

Not the same thing. A cap stores energy, a valve tap does not. Pressure if present before and after the tap is turned on or off. The effect of a cap takes a transient amount of time, a water tap's effect is immediate. A valve is not the same as a storage element.

When I saw a calculation containing 17, it reminded me of the camel problem where the question was altered to give a solution.

I wish I knew to what you are referencing.

There is no point in continuing with this. I would like to know what the official answer is and how it was obtained.

The point is understanding. Official answer to what?

Ratch

 

[duke37]

I wish I knew to what you are referencing.

I learned the reason for this when I went to school 70 years ago.
Just Google 17 camels.

There was a sheik with 17 camels and 3 sons. His will said 1/2 the camels should go to the first son, 1/3 the camels should go to the second son and 1/9 to the third son. Much argument ensued until a poor neighbour arrived with only one camel. He gave this camel to make the number up to 18. The first son had 18/2 = 9, the second 18/3 = 6 and the third 18/9 = 2. They sons were happy and there was a spare camel which they gave to the poor neighbour.
This is a case of changing the question to get an answer. In the capacitor problem, you have effectively used a frequency w = 1 to get 1Ω equivalent to 1F instead of w = 0.

There must be an official answer to the capacitor problem to set the marks.

 

[Ratch]

I learned the reason for this when I went to school 70 years ago.
Just Google 17 camels.

There was a sheik with 17 camels and 3 sons. His will said 1/2 the camels should go to the first son, 1/3 the camels should go to the second son and 1/9 to the third son. Much argument ensued until a poor neighbour arrived with only one camel. He gave this camel to make the number up to 18. The first son had 18/2 = 9, the second 18/3 = 6 and the third 18/9 = 2. They sons were happy and there was a spare camel which they gave to the poor neighbour.

I am sure there are many such problems that suggest you try to divide something into shares that are not aliquot parts of the whole. The variation I remember is an estate of a rancher whose horse herd was not divisible by integers according to the fractions specified by his will. The slick lawyer threw his broken-down nag (BDN) into the herd and settled the estate. Only the BDN was left, which the slick lawyer took back.

This is a case of changing the question to get an answer. In the capacitor problem, you have effectively used a frequency w = 1 to get 1Ω equivalent to 1F instead of w = 0.

There must be an official answer to the capacitor problem to set the marks.

There is no "official" answer, only a right or wrong answer.

This is not an AC problem, so forget about omega and impedance. The same current will be present in all capacitors in series, and the sum of the capacitor voltages will add up to the value of the voltage source. The current will cause an equal charge imbalance in each capacitor, which in turn will make a voltage appear across each capacitor. The voltage will be inversely proportional to the value of each capacitor (V = Q/C). That is the way I worked the problem in post #26. The capacitors will have infinite impedance at DC, but will still be energized to a voltage dependent the the source voltage and their capacitance.

Ratch

 

[Arouse1973]

Can someone link to the original thread please.
Thanks
Adam

 

[Harald Kapp]

It's a pleasure: https://www.electronicspoint.com/threads/voltages.286900/#post-1754378

Harald

 

[duke37]

The point is that the circuit is given with the battery already connected. There are no currents and no charge transfer. Numbers can be derived by building the circuit without the battery and then connecting it but this was not the original problem.

I may be pig headed but I think it is time to let this one go.

 

[Ratch]

The point is that the circuit is given with the battery already connected. There are no currents and no charge transfer. Numbers can be derived by building the circuit without the battery and then connecting it but this was not the original problem.

I may be pig headed but I think it is time to let this one go.

If the battery is already connected when you first become aware of this circuit, then the caps are already fully energized. No current is present, and the charge flow has already taken place. The only thing left to do is calculate the voltages across each cap.

Ratch

 

[Arouse1973]

It's a pleasure: https://www.electronicspoint.com/threads/voltages.286900/#post-1754378

Harald

Thanks Harald!