>  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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