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The loads are in parallel. The frequency is 60hts. The parallel loads are identical with 32ohms. There are 2 of them. With that configuration, what is the voltage on the back end? Hope I'm not confusing the issue.
tl

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Originally Posted by trout lake
The loads are in parallel. The frequency is 60hts. The parallel loads are identical with 32ohms. There are 2 of them. With that configuration, what is the voltage on the back end??? Hope I'm not confusing the issue.
tl
1/[(1/32) + (1/32)] = 16 ohms.
We speak Schematic on this forum; please post one.

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Originally Posted by trout lake
The loads are in parallel. The frequency is 60hts. The parallel loads are identical with 32ohms. There are 2 of them. With that configuration, what is the voltage on the back end? Hope I'm not confusing the issue.
tl
The "back end" of what? You are confusing the issue. On a coil, there is no voltage out. It's all being turned into current, magnetic energy, and heat if the coil is energized. The only voltage "out" of a 115 volt ac coil is if you close the control side, open the neutral side, and measure from the neutral side of the coil to main neutral or ground, in which case you'll have 115 volts, because none of the potential is being used. And there's no such thing as a "hts". There are hertz, and there are hz.

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Klove
my hts abbr was for hertz. should have been htz.....sorry

Lets say you have 4 heating elements from a commercial toaster, call it a Hobart for lack of a better name, and the 4 elements are in parallel. Lets say the unit is operating on 220vts (volts). The resistance of the 4 elements is 32ohms.(and no.....not johnny). All i was wondering was based on the resistance of the elements, what would be the voltage across the terminals (t1-t2) of the forth (last element).

Now I know from ohms law, lenz law and every other kind of law you can think of, that there "has to be a voltage drop".
My original question was "what is the correlation between the resistance of the plates and the voltage drop. If the back end voltage(the final voltage read across the plates of the 4th element) is x with 32ohms of resistance on each plate, what is voltage if the resistance was 25 or 15 or 10 for example.
I know for sure if there is no resistance,(impossible) there would be no drop.

I was simply asking the learned scholars on the site if there is a "simple" way that they know of, through experience, to come up with an answer.
The ie thing devided by r times the square root of p gives me a headache.

Thanks for your patients and thoughts back

tl

5. Ohms law is used for a purely resistive circuit "heating coil". When you mention coil which is an inductor it changes. Adding a capacitor changes more. If you want to know the V.D without using the equations put a volt meter across a load this will give the voltage drop.

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But aren't the elements in fact heating coils? If not, what do you mean by the term?

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also......It's easy enough to put the meter across the terminals and take the reading. But then what? All I can say is "yup, i got a drop". To the guys that commented to the question by making lite of it,and seem to know tonns about this stuff, I would only ask of them to answer the question with an answer. Or, perhaps with the 4 in parallel and the voltage in a given and the resistance a given, there really is no answer other than the meter solution.

thanks
tl

8. If you were in Windsor I would say you must work for Cardinal...lol

Only time I ever had to replace the elements on a Holman was when some idiot decided to get too rough with it when he was cleaning it and busted the glass tubes that encased the heating elements.

9. Heating coils are nothing more than resistors , A coil or inductor would be a xmfr or motor. Your answer for 2 resistors in parallel 32 ohms = 32/2= 16ohms @115v= 7.19a Since were talking parallel amperage splits voltage remains the same. 4 resistors in parallel 32/4=8ohms @115v=14.37a All is derived from Ohms law I X R= E . As stated before when adding other components to a circuit you will change how to solve the problem.

10. Your confusing the uses of ohm's laws.

First.
On the heating elements. If they are 220 volt. You'll read 220 volts across t1 and t2.

Second.
On heating elements. You measure the voltage under load, and then the amp draw.
Then multiply the 2 together to get wattage. And see if that equals their wattage rating.

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When the heating elements are connected in parallel the only change in voltage would be caused by the voltage drop in the wires connecting the elements. If the connections are good and the wire properly sized this should be a very small amount. Ohms law would be applied to the resistance of the wires between the elements (not the resistance of the heating element. Also be aware that for a heating element the resistance of the element will increase as the element heats up (technical term: positive temperature coefficient). The actual amount would depend on the material used for the heating element and the actual temperature rise. It would probably be in the 4-8% range. This is a case where actual measurements will be better than any theory.

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Look at the last two posts. They're dead on correct. That's why I said you are confusing things. T1-T2 on parallel loads in your example is 220 volts.

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Allen j
that was the explanation that i needed to get me over the brink. thanks
thanks to all of you for your input and concerns.

How would that scenario change in series allen?

same 4 elements, same 220 in. In that case what would be the reading at t1 and t2 of the last element?

tl

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