I just bought an amwatt thp103 http://www.reliancecontrols.com/Products.aspx?pg=search&q=amwatt to test the electrical load on some appliances like a toaster etc. When I hooked it up to a cfl, a sylvania cf23el/minitwist 23w 120v 60hz. .34a 3000k, it told me it was using 100 watts. Whats up with that? Is there something I dont know about? maybe these dont work well with cfl's. It seams to work fine with a 60 watt incandescent bulb using 56 watts , an 800 watt toaster, 1100 watt microwave, a 300 watt work light using only 245 watts, and a 150 watt incandescent bulb, and 2 fans using up to 45 and 150 watts.
AMWATT
- Thread starter chewy78
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kramer5150
Flashaholic
That is strange...
Furthermore .34A x 120V = 40 Watts
Is it the only bulb in the pack that performs like this?
Defective ballast or voltage step up circuitry?
Mis-labeled product?
Is your wall socket on a dimmer rheostat? What is the voltage of the wall AC outlet? I have had those things wreck all kinds of havoc with my tube amp circuits.
Furthermore .34A x 120V = 40 Watts
Is it the only bulb in the pack that performs like this?
Defective ballast or voltage step up circuitry?
Mis-labeled product?
Is your wall socket on a dimmer rheostat? What is the voltage of the wall AC outlet? I have had those things wreck all kinds of havoc with my tube amp circuits.
dano
Flashlight Enthusiast
Flashlight?
Morelite
Flashlight Enthusiast
From the product specs,
"Accurate digital readout from 1.0 - 15.0 Amps and 125 - 1875 Watts".
Your 23watt CFL at .34A is below the working range of the device so it probably is not a good reading. The 100 watt reading is probably the lowest it can display.
"Accurate digital readout from 1.0 - 15.0 Amps and 125 - 1875 Watts".
Your 23watt CFL at .34A is below the working range of the device so it probably is not a good reading. The 100 watt reading is probably the lowest it can display.
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Morelite
Flashlight Enthusiast
No, they just use less, see my post above this one.
Yes, I guess you could say so. Cheap electronic ballasts are likely to have real bad nonlinear power factor, while an incan is not only a nice linear load, it also has unity power factor.
That seams to make more sense now
Morelite
Flashlight Enthusiast
Is it possible that the 100watt reading was the peak, at turn-on the CFL ballast would have need more current to fire up. Just a guess.
bluepilgrim
Enlightened
Yeah -- sounds like power factor -- it's not a resistive load.
http://en.wikipedia.org/wiki/Power_factor
The power factor of an AC electric power system is defined as the ratio of the real power flowing to the load to the apparent power,[1][2] and is a number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf). Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power.
also: http://www.allaboutcircuits.com/vol_2/chpt_11/3.html
and
http://en.wikipedia.org/wiki/Volt-ampere
It's a little bit like if you fill a water tank which then feeds your faucet, it looks like you have a lot of water running and going down the drain, but you aren't because you have the water stored in the tank. That's like how electricity can be stored in an inductive or capacitive load.
If you know the voltage, switch the meter to measure amps and see if that works (and you can figure power as watts = volts X amps).
http://en.wikipedia.org/wiki/Power_factor
The power factor of an AC electric power system is defined as the ratio of the real power flowing to the load to the apparent power,[1][2] and is a number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf). Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power.
also: http://www.allaboutcircuits.com/vol_2/chpt_11/3.html
and
http://en.wikipedia.org/wiki/Volt-ampere
It's a little bit like if you fill a water tank which then feeds your faucet, it looks like you have a lot of water running and going down the drain, but you aren't because you have the water stored in the tank. That's like how electricity can be stored in an inductive or capacitive load.
If you know the voltage, switch the meter to measure amps and see if that works (and you can figure power as watts = volts X amps).
DM51
Flashaholic
As far as I can see, this has nothing whatever to do with flashlights. Moving it to non-flashlight electronics.
ok thanks
