Olin,

 You are correct sir.  I got my tongue wrapped around my eye teeth, and
 couldn't see what I was saying.  I stand corrected.  But either way, a
 5 watt resistor would handle it.

                                                   Thanks and Regards,

                                                          Jim



>>  My suggestion would be to place a small value resistor (resistance)
>>  in series with the battery pack, and measure the voltage across it.
>>  Then calculate the current through the resistor (resistance).  Since
>>  the current is the same in the components of a series circuit, that
>>  will give you the charging current.   Say you used a 1 ohm resistor.
>>  And lets also say you measure 1 volt across that 1 ohm resistor.
>>  Current then would be 1 Amp.  If you measured 40mV across the
>>  resistor, your current would be 40mA.
>
> Or better yet use an ammeter.  This is exactly how multimeters work in
> current mode.
>
>>  This will be pretty accurate even with the voltmeter (DMM)
>>  connected because the input impedance of the DMM is probably at least
>>  1 MEG and more like to be closer to 10 or 20 Megohms.  And 1 ohm in
> parallel
>>  with >1Megohm is still 1 ohm.   Anyway, it's worth a try.   Just make
>>  sure your resistor can handle the power to be dissipated.  ie 2 watts
>>  should handle any current up to about 2 amps.
>
> No 2 amps would result in 4 watts using the 1 ohm resistor in your
> example. Power = voltage x current = V**2 / R = I**2 * R.
>
>
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