I think you are mixing too many things up,
1. Check your conversion factors between Wh and kWh.
2. You are leaving out the effects of power factor (reactive loads) and
form factor (nonlinearity from rectification).
Try again/

--

Sam

Conservatives are against Darwinism but for natural selection.

Power factor isn't all that important here. He's looking at the label on the
wall wart and assuming the WW really draws that much for the entire charge
cycle. If he measured the power he'll likely see an order of magnitude or
three less power. Something the size of a cell phone dissipating 240W for an
hour will get rather toasty.
A closer estimate would be to take the AH rating of the battery (usually 1.5
to 2AH) times the voltage (4V) and figure the energy in the battery (6-8Wh)
and add a fudge factor (50%, say) for efficiency. That would be more like
10Wh or a tenth of a penny. Even that assumes a *full* charge (unlikely);
leaving the WW plugged in when not charging will likely cost more.

With my lousy eyes, I read .24kWh as 24kEh. That is why I make it
standard practice to write .24 as 0.24. It is just too easy to miss the
decimal point. Even the European practice of using a comma for a decimal
point is not ideal either. There have been many severe economic and
safety consequences over the years because of missed decimal points.
I do, however disagree with the comment on power and form factors. Many
cheap and not so cheap ac meters are really dc meters that use a
rectifier to get peak waveform voltage (or current). Many of these warts
will charge batteries over a small fraction of a cycle. This leads to
high peak currents. These meters are calibrated to give the correct
current assuming a sine waveform rather than spikes during the fraction
of the cycle when the power supply waveform exceeds the battery voltage
(+ diode drop).

--

Sam

Conservatives are against Darwinism but for natural selection.

I was kinda surprised you made the mistake but I've done worse.

That's still not a PF issue or even a metering issue. An RMS meter (not rare
anymore) will still get the right answer and a PF would have to be *REALLY*
bad to be off by a couple of orders of magnitude. BTW, so would a meter. No,
this is much more likely a case of seeing a number and not understanding its
meaning. The "rating" of the WW is likely what its fused at, not what it
actually consumes.

My main point is that you should understand your instrumentation. In
these days, extreme specialization has resulted in too many engineers
hardly knowing anything outside their specialty. The spikes have more
to do with form factor rather than power factor.
At one time I was building power supplies for charging large capacitor
banks. Now, using high frequency chopping supplies that can be done
efficiently without using much iron. Form factor and power factor
problems were severe. RMS current in the lines could easily exceed
circuit breaker trip levels at levels much lower than the POWER capacity
of the line for unit power factor. It was especially bad for
ferroresonant transformers that limited inrush current as well as
limiting maximum output voltage near maximum charge.

--

Sam

Conservatives are against Darwinism but for natural selection.

I don't know the Optimus, but I suspect the 2 amps you refer to is the
output current of the charger, not its input. Assuming the charger
provides, say. 6 volts at 2 amps, that is 12 watts output. Input can be
guesstimated at 20 watts, or 0.02kW
John

On Dec 2, 12:10 pm, "John Nice" <johnDOTniceATbtinternetDOTcom> wrote:

Here's my charger.

http://i1190.photobucket.com/albums/z449/m75214/charger.jpg
Looks like the input is .3A.
.3 * 120 = 36 Watts
For 1 hour, .036 kWh
So, is it basically 36% of one cent to charge my phone?

http://i1190.photobucket.com/albums/z449/m75214/charger.jpg
Looks like the input is .3A.
.3 * 120 = 36 Watts
For 1 hour, .036 kWh
So, is it basically 36% of one cent to charge my phone?
If you're able to type you should be able to do your own simple arithmetic
(look the word up on Google if you're not sure what it means)
John

On Sat, 3 Dec 2011 23:20:16 -0000, "John Nice"
<johnDOTniceATbtinternetDOTcom> wrote:>

The picture shows:
VIZIO
MODEL:DSA-10PFD-05 FUS 050200
INPUT:100-240V~ 50/60Hz 0.3A
OUTPUT:+5V{DC symbol}2A
EFFICIENCY LEVEL:(V)
I take this to mean the MAXIMUM current draw is 0.3A,
for a maximum at of 36W at 120V. However, there is
no statement as to if the 0.3A is an instantaneous
load, perhaps at startup, or what could be expected
if the long term output were the maximum permitted
value.
The maximum allowed output is 2A (at 5V, or 10W)
The average load by the device is not stated.
If I read it correctly
http://dkc1.digikey.com/us/en/tod/CUI/NewLevelVEfficiency_NoAudio/NewLevelVEfficiency_NoAudio.html
page 4 of 15 says since the output power is between 1 and 49 Watts the
efficiency is at least:
([0.0750*{natural logarithm}(10 Watts) +0.561)
I get about 73% for this.
{The page says the formula give a percentage, but it
seems to give a number, since 0.860 is need for output of
more than 49 Watts, and surely they mean 86%, not 0.86%)

As I state above, I think you have done the calculation
correctly, but 36 Watts may just be an instantaneous load
(leaving out power factor effects, as also is stated.)
I think you should use (10 Watts)/0.73 = 14 Watts for
the maximum average load

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