capacitances in parallel

Hello guys, It is seen that the capacitances in parallel add up. the equations are

Q= Q1/V +Q2/ V

= (Q1+Q2)/V

Being Q=CV

Ceq=(C1+C2)

But how do the charges in parallel add up.What is its physical meaning or implication.

Regards, Robotnik

Reply to
fathahrahman
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You just added up the charges Q ! They add up arithmetically like your sums show.

Graham

Reply to
Eeyore

But how do they add up? In the case of resistors and inductors, when they are connected in series only they add up. Hoew is it different in capacitances.Does charge make the difference.Plz make it more clear.

Robotnik

Reply to
Robotnik
** Groper alert !

** Your Q lacks context and hence is not defined.

Are you asking about connecting two CHARGED capacitors in parallel ?

If so, the Qt ( total charge ) afterwards is the sum of the two Q values.

Also, Qt = ( C1 +C2) x V

where V is a new voltage of the combined capacitor.

....... Phil

Reply to
Phil Allison

One litre plus one litre equals two litres. Just like that. In SERIES you have to use reciprocals.

Reply to
Homer J Simpson

The 'Q' mentioned was the charge inside the capacitor There is a small coorection to the first equation I provided.

Ceq=Q/V=Q1/V+Q2/V

where Q1 and Q2 arethe charges inside capacitors C1 and C2 and V is the voltage.

I am not concerned about the equation side. I am concerned about how the capacitances in parallel in add up.For comparison I shall say, the value of two resistances in series is the sum of te two resistance since the voltage supplied to the resistors will distribute itself between both of them.

Equationwise, V=V1+ V2= iR1 + iR2=i(R1+R2)

But how does it materialize in Cap. in II. Plz. give the reason .. Charge distribution or voltage distribution or anything else? How does it affect the value of the capacitance.

Regards, Robotnik

Reply to
Robotnik

One useful water analog of a capacitor is a rigid vessel with a rubber membrane dividing it into two halves, with a pipe connection on each side of the membrane. When you apply a pressure difference across the membrane (a voltage across the capacitor) a volume (charge) of water moves into one side of the vessel, and out of the other side, as the membrane develops stress (internal electric field) that resists the applied pressure. Double the pressure difference, and twice the water displaces across the vessel and twice the stress is stored in the membrane.

Pipe two equal tanks in parallel, and you double the volume of water stored against the same membrane force.

Reply to
John Popelish

"Robotnik"

** Forget making false comparisons with resistors or inductors.

Caps have the unique ability to store energy in isolation.

** Please use normal English, not your own invented symbols.
  1. You need to find out what " charge " really is.
  2. Then how caps store it.

Google & Wiki will help.

...... Phil

Reply to
Phil Allison

It's really a matter of the conventional units that are popular. People use capacitance to measure capacitors, and the corresponding unit for resistors should be conductance. Parallel capacitors add, and parallel conductances add. But, for some obscure reason, we like to buy resistors in units of ohms, "resistance", which is the reciprocal of conductance. So series resistors add in value. It's sort of an arithmetic illusion.

Both a big capacitance and a big conductance approach a short circuit, and both a small capacitance and a small conductance approximate an open circuit.

John

Reply to
John Larkin

And inductors don't?

John

Reply to
John Larkin

Take two sheets of aluminum foil, and put a sheet of newspaper between them. Use sheets of aluminum foil that are bigger, and capacitance will increase.

There really isn't any difference between that and putting capacitors in parallel. When increasing the size of the foil, you increase the capacitance of the whole thing. The same thing if you parallel capacitors.

This is no different from other passive components. If you need a larger value, you either increase the component, or connect more than one so their values add up.

Michael

Reply to
Michael Black

"John Farkin Idiot "

** Yawn - what a f****it.

....... Phil

Reply to
Phil Allison

How do you maintain the magnetic field when the power's removed ?

Graham

Reply to
Eeyore

--- Not after the charging source is disconnected, I think.

Version 4 SHEET 1 880 680 WIRE -64 160 -224 160 WIRE 128 160 16 160 WIRE 128 224 128 160 WIRE -224 240 -224 160 WIRE -48 240 -48 208 WIRE -224 352 -224 320 WIRE -48 352 -48 320 WIRE -48 352 -224 352 WIRE 0 352 0 208 WIRE 0 352 -48 352 WIRE 128 352 128 304 WIRE 128 352 0 352 WIRE -224 384 -224 352 FLAG -224 384 0 SYMBOL ind 112 208 R0 SYMATTR InstName L1 SYMATTR Value 1 SYMBOL voltage -224 224 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 12 SYMBOL voltage -48 224 R0 WINDOW 0 -53 5 Left 0 WINDOW 3 -242 110 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value PULSE(0 1 .001 1E-6 1e-6 .01) SYMBOL sw 32 160 M270 WINDOW 0 32 15 Left 0 WINDOW 3 32 44 Left 0 SYMATTR InstName S1 TEXT -192 384 Left 0 !.model SW SW(Ron=1 Roff=10Meg Vt=0.5Vh=0) TEXT -186 418 Left 0 !.tran .1

-- JF

Reply to
John Fields

with resistors and inductors the numbers reflect how well the impede the flow of current, capacitors are measured with inverse units the numbers reflect how well the permit current to flow,

hence the formulae for parallel and series combnation have the opposite sense to the resistor ones.

Bye. Jasen

Reply to
jasen

Super conductor? Now if we can just get 'em to work at 25C.

Reply to
James Beck

Imagine you have two cake pans and a pitcher of water. The water is the charge, the depth of water in the pan is the voltage, and the area is the area of the capacitor's plate. Oh, yeah - the current is how fast you're pouring the charge(water) onto the plate(cake pan).

Well, if you add a cake pan with the same area and depth of water, obviously the two will hold twice as much as the one. Charge works pretty much the same way. And usually, at least in wires, charge is carried by the electrons, somehow or another - probably black magic. ;-)

Hope This Helps! Rich

Reply to
Rich Grise

You're tying yourself in knots trying to derive all these formulas - capacitance is merely directly proportional to plate area, and 2

1" x 1" plates have the same area as one 1" x 2" plate. They simply add.

Hope This Helps! Rich

Reply to
Rich Grise

Short the terminals. Losses, in most inductors or capacitors, will eventually burn off the stored energy.

A low-field superconductive inductor will store energy almost forever, as will a properly insulated capacitor. I suppose the only dissipative mechanisms are cosmic rays bashing Cooper pairs in the L, or cosmic rays knocking charges off the plates for a C.

High-field superconductive magnets have other loss mechanisms, filament migration or something, that tend to be in the ppm per day sort of range.

John

Reply to
John Larkin

NMR/MRI magnets are installed, cooled down, and charged with a big power supply. The installer then shorts the magnet, puts the power supply into his truck, and drives away. As long as the helium is kept topped off, every couple of months, the current circulates and the field stays.

And inductor is the dual of a capacitor. Both store energy.

John

Reply to
John Larkin

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