capacitors in parallel

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

  1. ranch vermin

    Thread Starter Member

    May 20, 2015
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    2 capacitors in parallel is half the resistance right?
     
    Last edited: May 24, 2015
  2. crutschow

    Expert

    Mar 14, 2008
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    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
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    thankyou - does watts increase when voltage saturates?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Huh? :confused::confused::confused:
     
  5. wmodavis

    Well-Known Member

    Oct 23, 2010
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    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
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    '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...

    Please pardon the implicit redundancy:oops:


    Best regards
    HP

    PS -- Clever profile name!o_O:D:D:D
     
    Last edited: May 24, 2015
  7. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    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:rolleyes:

    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
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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
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    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:confused:

    Best regards
    HP
     
    Last edited: May 24, 2015
  12. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    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:rolleyes:

    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
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    I can safely say I've never seen someone with a more mangled concept of electrical units. :rolleyes:
     
    Hypatia's Protege likes this.
  14. wmodavis

    Well-Known Member

    Oct 23, 2010
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    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
    3,242
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    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.
     
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