# capacitors in parallel

Discussion in 'General Electronics Chat' started by ranch vermin, May 23, 2015.

1. ### ranch vermin Thread Starter Member

May 20, 2015
85
2
2 capacitors in parallel is half the resistance right?

Last edited: May 24, 2015
2. ### crutschow Expert

Mar 14, 2008
13,501
3,375
No, unless you are talking about ESR.
Capacitors have capacitance, not resistance (except for the parasitics).
And the capacitances add together in a parallel connection.

3. ### ranch vermin Thread Starter Member

May 20, 2015
85
2
thankyou - does watts increase when voltage saturates?

Mar 31, 2012
18,087
4,917
Huh?

5. ### wmodavis Well-Known Member

Oct 23, 2010
737
150
Only the Eigen value of the watts saturates when two capacitors of half the resistance are added in series parallel when measuring the ohms.

6. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,847
1,317
'Saturation' in the context of Electrical Engineering is a term descriptive of disparate while analogous properties of magnetic and active devices --- It would not seem to be applicable to electrostatic components...

Generally speaking, power, in non-reactive or (CIP) 'pure' direct current circuits, is in direct proportion to EMF, current being [held] equal...

Best regards
HP

PS -- Clever profile name!

Last edited: May 24, 2015
7. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,847
1,317
I Dunno... It isn't my impression the OP is preforming Circuit Analysis --- This language barrier problem, NO thanks to 'Google Translate', is driving me 'round the bend

Best regards
HP

8. ### ranch vermin Thread Starter Member

May 20, 2015
85
2
Excuse my n00biness!!!
By saturation, I just mean hitting a square wave with the voltage.
Im confused... I always thought watts was impedance... but im probably wrong arent I!
If watts doesnt go up, then their must be resistance involved when the voltage goes up... or I=V/R doesnt make sense.
When you resist something, watts goes down doesnt it?

Whats the difference between watts and impedance?

9. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,847
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Impedance may be regarded as a vector comprised of resistance (Magnitude) and Reactance (direction[i.e. angle])

Or, should you favor 'rectangular notation':
Resistance= the 'Real' part of impedance
whereas
Reactance = the 'Imaginary' part of impedance

The Watt, on the other hand, is the unit of electrical power -- not to be confused with the 'VAR' (Volt-Ampere-Reactive)...

IOW

1 Watt = 1 Volt*1Ampere (Resistive)
1 VAR = 1Volt*1Ampere (Reactive)

For static (or non-degenerate) current, power will always increase with EMF in circuits exhibiting finite, non-zero resistance...

Best regards
HP

Last edited: May 24, 2015
10. ### ranch vermin Thread Starter Member

May 20, 2015
85
2
ok i dont get that.

Ive only got 1 current value, so how do I get resistance and reactance from it.

Then ive got another problem. I seem to think impedance is amperes.

Also, it seems wierd to go formulate C=V/R then go multiply it by the volts again.

WATTS=V/R*V ---> watts = resistance???!??

Last edited: May 24, 2015
11. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,847
1,317
The length of the hypotenuse represents impedance, you may interpret this numerically via: Z{s}=(R^2+X^2)^.5 or Z{p}=RX/((R^2+X^2)^.5)

Where Z{s} and Z{p} denote impedance in series or parallel circuits respectively

Note, that while the 'Z{p}' formula (above) is non-standard, it is valid! (I promise)

I don't know how to apply 'angle from origin' to this context

Best regards
HP

Last edited: May 24, 2015
12. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,847
1,317

Resistance=E/I

Reactance (case-in-point capacitive reactance) = 1/(2*Pi*F*C)
Where Pi= constant ~3.14159..., F=frequency (in Hertz), C=Capacitance (in Farads)

Or, more elegantly: Xc=1/(F*C) where frequency is denominated in radians-per-second

Impedance is not current! -- What else can I say?

Inasmuch as it is utter nonsense, It should seem weird

Among other misapprehensions, you seem to be confusing reactance units (i.e. ohms) with reactor value units (e.g. Farads)

Furthermore you are confusing capacitance with power, please consider the following:

I=E/R, Hence P=(E/R)*E => P=(E^2)/R

P=Power denominated in watts
E=EMF denominated in Volts
I=Current denominated in Amperes

No! To use 'your lingo' but 'my math' WATTS=(V^2)/R --- Hint: It's [V/R]*V NOT 1/(V/[R*V]) -- Get it???

Best regards
HP

Last edited: May 24, 2015
13. ### crutschow Expert

Mar 14, 2008
13,501
3,375
I can safely say I've never seen someone with a more mangled concept of electrical units.

Hypatia's Protege likes this.
14. ### wmodavis Well-Known Member

Oct 23, 2010
737
150
Like I said "Only the Eigen value of the watts saturates when two capacitors of half the resistance are added in series parallel when measuring the ohms."

15. ### dl324 Distinguished Member

Mar 30, 2015
3,379
651
It would be helpful if you gave a more detailed description of how the caps are being used so we can better understand your questions.