I understand the formula for current is:I=E/R and I notice some products say the current for 240v is i/2 that of
120v. Am I asleep or why am I seeing the current increase with voltage increase (or do I have the formula wrong?? Thanks, curt
I understand the formula for current is:I=E/R and I notice some products say the current for 240v is i/2 that of
No but I think you might be confusing two different sets of equations....
1) If you have a fixed value resistor then the current flowing in the resistor is given by the equation: I = V/R Notice that if you double the voltage V you get double the current I. 2) The power (Watts) consumed by a piece of electric equipment is given by the equation W = I x V. Lets say you were designing two different toasters - one designed to work on 240V and one designed to work on 120V. You wouldn't want the one running on 120V to take twice as long would you? So you design them both to use the same amount of power. If one operates on half the voltage it needs twice the current to produce the same power. eg V x I = (V/2) x (Ix2)Look at it another way. A fridge needs to burn a ceratain amount of power to keep the contents cold. If you feed the fridge twice the voltage it only needs half the current to keep the power the same.
NO: it's right, there are 2(pi)turns unaccounted for in the formula... and then squared!
In article , snipped-for-privacy@fastmail.fm mentioned...
If a light is 120W for example, the light will take 1A at 120VAC or .5A at 240VAC. It's the power that stays the same as the voltage changes. So the resistance for the 240VAC light is four times the resistance of a 120VAC light.
Understand, rubber band?
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In article , snipped-for-privacy@pandoraBOX.be mentioned... [snip]
Kind of a sorry example. The fridge could run at four times the power, but for one fourth the time, since it's a duty cycle type of appliance.
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,
i/2 that of
with voltage
120VAC orvoltage
times the
Actually if you apply 240 volts to a 120 volt lamp the current will double for a short time till the lamp disintergrates. What you tried to say, but not very clearly, was that a 240 volt lamp designed to produce 120 watts will take 1/2 the current a 120 volt lamp will use to produce 120 watts. Very simple questions sometimes require very carefull answers because the questioner does not have a good understanding of the subject or he would not be asking the question.
You need to apply the right formula to your observation:
Power = Voltage * Current
The formula is correct, but does not apply to your product observation in this case, as the product is NOT a pure resistance.
The formula is correct for a simple load.
But equipment designed to operate on 120 and 240v has a switch (usually manual, sometimes automatic) that changes the "equivalent load" of the equipment so that it draws the equivalent amount of POWER.
A given load requires half the current at 240v as it does at 120v.
I = P/E
A 120 *WATT* load would draw 1 amp at 120 volts, but only 1/2 amp at 240 volts.
power to
Ok ok I didn't actually mean the _same_ fridge.
But if you want an argument.... I'm sure fridge pumps must have an optimum operating duty cycle at which the fridge overall is most efficient. In which case the fridge should be redesigned to operate close to that point regardless of the voltage that it's designed to run on.
I note you didn't propose a better example?
A toaster perhaps? Who wants to wait 4 times longer for their toast?
Oops sorry I see you did provide a better example. The light is better you win.
only
which
In article , snipped-for-privacy@pandoraBOX.be mentioned...
Or Pop Tarts. :o)
In article , "John G" mentioned...
Well, the current of a light bulb doesn't obey the straight line V-I characteristic of a resistor. But yeah, the 120V light would definitely burn out at 240VAC.
Mea culpa.
Zackly.
In article , snipped-for-privacy@ARRL.NET mentioned...
Pure resistance? What's that mean? I thinkk you mean the resistance of the light is non-linear.
Pure? As opposed to unpure? Maybe you need to filter the light to make sure it's pure.. ;-)
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