Re: HELP!! Need answers for test!!!

["petem" <petem001@xxxxxxxxx>]

Frank he is right that if you change the value of a the total resistance in
circuit the current wont be the same...

don't you remember those formula?

I=I/R I mean current E mean Voltage R mean resistance..

so if I have a variable resistance ( a pot) is series with a fix load, lets
say 100 ohms fixed load connected to pot that varied from 0 to 1000 ohms on
a 10 volt supply

at minimum resistance of the pot the current will be 10/100 or 0.1 amp
at maximum resistance of the pot the current will be 10/100+1000 or 10/1100
or 0.09 (rounded) amps
so its about 1/10 of the max current..

the thing you describe about limiting power from a pot in a normal every day
situation is not simple resistor in series of the load (light of fan) its in
fact cutting parts of the ac phase of the load..so when on 50% of the time
power is feed to the load the average power seen at the load is 50% lower
then max but it still can be 110 volts across the load from time to time.(
at 60 cycle a second this come fast) ;-)



"Frank Olson" <Use_the_email_links@xxxxxxxxxxxxxxxxxxxxxx> a écrit dans le
message de news: wjImf.82200$ki.61030@xxxxxxxxxxx
> Robertm wrote:
>> "Frank Olson" <Use_the_email_links@xxxxxxxxxxxxxxxxxxxxxx> wrote in
>> message news:YVGmf.86520$Gd6.37113@xxxxxxxxxxx
>>
>>>petem wrote:
>>>
>>>>I do not agree and this link will say the same
>>>>
>>>>http://www.answers.com/topic/resistor
>>>>
>>>>a resistor is limiting current not voltage
>>>
>>>
>>>They're wrong.  A resistor drops voltage and is a voltage limiting
>>>device. If you (for instance) use a variable resistor to control the
>>>speed of a fan (or light), the current remains the same regardless of the
>>>setting. The only thing you're doing is reducing the voltage to the load.
>>
>>
>> We are also reducing the current if the resistance of the load remains
>> fixed. A lamp and fan would be more difficult to calculate because
>> impedance enters into fan calculations, and lamp resistance changes
>> dependent on element temperature which is dependent on applied voltage
>> and subsequent current. If we were to assume a resistive load of 12 ohms
>> across a 12 volt supply, we have one amp flowing. Add a 12 ohm resistor
>> in series with the load and we now have 6 volts across the limiting
>> resistor and 6 volts across the load giving us a current of 500 ma
>> through the load.
>
> What???  Current in a series circuit stays the same.  Halving the voltage
> through the load in the series circuit you describe above doesn't reduce
> the amount of current drawn from the source.  Where did you go to skool??