capacitors in parallel

wmodavis

Joined Oct 23, 2010
739
Only the Eigen value of the watts saturates when two capacitors of half the resistance are added in series parallel when measuring the ohms.
 

Hypatia's Protege

Joined Mar 1, 2015
3,228
thankyou - does watts increase when voltage saturates?
'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
 
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Hypatia's Protege

Joined Mar 1, 2015
3,228
Only the Eigen value of the watts saturates when two capacitors of half the resistance are added in series parallel when measuring the ohms.
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
 

Thread Starter

ranch vermin

Joined May 20, 2015
85
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?
 

Hypatia's Protege

Joined Mar 1, 2015
3,228
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?
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)

If watts doesnt go up, then their must be resistance involved when the voltage goes up...
For static (or non-degenerate) current, power will always increase with EMF in circuits exhibiting finite, non-zero resistance...

Best regards
HP
 
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Thread Starter

ranch vermin

Joined May 20, 2015
85
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???!??
 
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what does the hypotenuse and angle from origin stand for?
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
 
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ok i dont get that.

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

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

Then ive got another problem. I seem to think impedance is amperes.
Impedance is not current! -- What else can I say?

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

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

Joined Oct 23, 2010
739
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."
 

dl324

Joined Mar 30, 2015
18,326
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???!??
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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