question

C

curt

Guest
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
 
"curt" <hungry1@fastmail.fm> wrote in message
news:e8d8f7c6.0312131100.4955a288@posting.google.com...
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??
Click to expand...
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.
 
"curt" <hungry1@fastmail.fm> wrote in message
news:e8d8f7c6.0312131100.4955a288@posting.google.com...
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
Click to expand...
NO: it's right, there are 2(pi)turns unaccounted for in the formula... and
then squared!
 
In article <e8d8f7c6.0312131100.4955a288@posting.google.com>,
hungry1@fastmail.fm mentioned...
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
Click to expand...
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?


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <S2MCb.76388$zE5.3080303@phobos.telenet-ops.be>,
colin.watters@pandoraBOX.be mentioned...
[snip]

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.
Click to expand...
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.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
"Watson A.Name - Watt Sun, Dark Remover"
<alondra101@hotmail.com> wrote in message
news:MPG.1a454e8d9fd29a7d989a16@news.dslextreme.com...
In article
e8d8f7c6.0312131100.4955a288@posting.google.com>,
hungry1@fastmail.fm mentioned...
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

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?


--
Actually if you apply 240 volts to a 120 volt lamp the
Click to expand...
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.
--
John G

Wot's Your Real Problem?
 
"curt" <hungry1@fastmail.fm> wrote in message
news:e8d8f7c6.0312131100.4955a288@posting.google.com...
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.
Click to expand...
You need to apply the right formula to your observation:

Power = Voltage * Current

Am I asleep or why am I seeing the current increase with voltage
increase (or do I have the formula wrong??
Click to expand...
The formula is correct, but does not apply to your product observation
in this case, as the product is NOT a pure resistance.

Thanks,
curt
Click to expand...

--
*
| __O Thomas C. Sefranek WA1RHP@ARRL.net
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org
 
"curt" wrote ...
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??
Click to expand...
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.
 
"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a4550f2487cc803989a17@news.dslextreme.com...
In article <S2MCb.76388$zE5.3080303@phobos.telenet-ops.be>,
colin.watters@pandoraBOX.be mentioned...
[snip]

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.

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.
Click to expand...

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.


"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:aC2Db.77959$IS6.3260443@phobos.telenet-ops.be...
"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a4550f2487cc803989a17@news.dslextreme.com...
In article <S2MCb.76388$zE5.3080303@phobos.telenet-ops.be>,
colin.watters@pandoraBOX.be mentioned...
[snip]

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.

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.


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?
Click to expand...
 
In article <aC2Db.77959$IS6.3260443@phobos.telenet-ops.be>,
colin.watters@pandoraBOX.be mentioned...
"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a4550f2487cc803989a17@news.dslextreme.com...
In article <S2MCb.76388$zE5.3080303@phobos.telenet-ops.be>,
colin.watters@pandoraBOX.be mentioned...
[snip]

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.

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.


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?
Click to expand...
Or Pop Tarts. :eek:)



--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <3fdbf741$0$18690$afc38c87@news.optusnet.com.au>, "John G"
<greentestatoptusnet.com.au> mentioned...
"Watson A.Name - Watt Sun, Dark Remover"
alondra101@hotmail.com> wrote in message
news:MPG.1a454e8d9fd29a7d989a16@news.dslextreme.com...
In article
e8d8f7c6.0312131100.4955a288@posting.google.com>,
hungry1@fastmail.fm mentioned...
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

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?


--
Actually if you apply 240 volts to a 120 volt lamp the
current
will double for a short time till the lamp disintergrates.
Click to expand...
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.

What you tried to say, but not very clearly, was that a 240
Click to expand...
Mea culpa.

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.
Click to expand...
Zackly.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <r5OdnUdGYsk_okGiRVn-sw@comcast.com>, WA1RHP@ARRL.NET
mentioned...
"curt" <hungry1@fastmail.fm> wrote in message
news:e8d8f7c6.0312131100.4955a288@posting.google.com...
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.

You need to apply the right formula to your observation:

Power = Voltage * Current

Am I asleep or why am I seeing the current increase with voltage
increase (or do I have the formula wrong??

The formula is correct, but does not apply to your product observation
in this case, as the product is NOT a pure resistance.
Click to expand...
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.. ;-)


Thanks,
curt
Click to expand...

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
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