# cost of charging

• posted

I have an LG Optimus, and I want to compute the cost of charging it.

I have a charger that uses 2 amperes of current. It takes my phone about an hour to charge. Is this right?

120 Volts x 2 amperes = 240 watts

240 watts x 1 hour = .24 kWh

.24kWh x 10 cents/kWh = 2.4 cents

• posted

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

• posted

Is this for school?

• posted

I think you are mixing too many things up,

1. Check your conversion factors between Wh and kWh.

1. You are leaving out the effects of power factor (reactive loads) and form factor (nonlinearity from rectification).

Try again/

• posted

240Wh = .24kWh

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.
• posted

Here's my charger.

Looks like the input is .3A.

.3 * 120 =3D 36 Watts For 1 hour, .036 kWh

So, is it basically 36% of one cent to charge my phone?

• posted

Here's my charger.

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

• posted

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).

• posted

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.

• posted

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.

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

Similarly 0.014 kWh

• posted

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.

• posted

0.3A * 120V = 36 volt_amperes! That is very different.
• posted

My main point is that this doesn't appear to be an "instrumentation" issue at all. Really, can you imagine 2A@120V into a cell phone charger?

A cell phone charger is going to exceed a CB rating? Really? GMAB.

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