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Re: Ceramic core resistors
Andres wasn't quite right here..
The power dissipation capability is simply the sum of the capabilities of
each of the components, regardless of the arrangement. What you have to
watch out for is what the parts are actually dissipating.
Trivial case #1. 3 resistors, each 120 ohms, each rated at 200 Watts.
Arrange them in parallel across 120V line. The total resistance is 40 ohms,
so there is a total of 3 amps through the source. I^2R is 9*40 or 360 watts.
You can also realize that each resistor will have a current of 1
Amp(=120V/120 ohms), and will dissipate 120 watts, for a total of 360 Watts.
Trivial case #2, 3 identical resistors, just like above. Arrange them in
series, across the same 120V source. Now, the resistance is 360 ohms, so
the current flow will be 1/3 amp. Total power dissipated is 40 Watts (=120
* 1/3) (distributed as 13.3=40/3 watts per resistor) or, using I^2R, 1/9*360
Non trivial cases. Assume that one of the resistors is 60 ohms, and the
other two are 120 ohms, as before. Hook them in parallel. Parallel
resistance is 30 ohms, so the current from the 120V source will be 4 amps.
TOTAL power dissipated in the resistors is 480W, but let's look at each of
the resistors: The 120 ohm resistors have 120V across them, so an amp
flows, and they each dissipate 120W. The 60 Ohm resistor has 2 amps flowing
through it, so it is now dissipating 240W (and getting very hot in the
Hook the same 3 resistors in series. The total resistance is 300 ohms, so
the current flow is 0.4 amps, dissipating 48 watts in all the resistors
together. The voltages don't divide evenly: the 60 ohm takes 24V, and the
pair of 120 ohm resistors has 48 volts across it. The 60 ohm resistor
dissipates 9.6 watts (=24*0.4) and the 120 ohm resistors each dissipate
twice that (19.2 watts).
Current through each resistor
From: Tesla List <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com <tesla-at-pupman-dot-com>
Date: Thursday, June 08, 2000 9:17 AM
Subject: RE: Ceramic core resistors
>Original Poster: "Andres Vogel" <dreman-at-techie-dot-com>
>>Series = more ohms ?
>>Parallel = more watts ?
>Absolutely right! In parallel it would disribute the I2R losses more
>arross each resistor. Remember that in parrallel branches, the current
>usage must total zero. (K's Current Law) So if I have 3 resitors in
>parrallel, all of them must divide the current loss, each dissiopating I2R
>as heat. Whereas in series the current is constant across each resistor,
>meaning all the resistors must be able to dissapate the same number of
>| I(2) |
>| I(3) |
>I(total) = I(1) + I(2) + I(3)
>All the current drops must be equal to the total current.
>I(1), I(2), and I(3) must be less than I(total). Meaning each one
>dissapates power dependent upon I2R of the current across the resistor. In
>series its different
>I(1) I(2) I(3)
>I(total) = I(1) = I(2) = I(3)
>Note the equal signs, each reisitor draws the same current. But causes a
>voltage drop, dependent on V = IR.
>Hope this helps, if I'm wrong feel free to correct me.
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