Hi vh2q, sorry for the delay in responding - I've been diverted as life is 'interesting' around here...
Good that your four cells measure an identical 3.56v. That means that the four-cell module is sitting at 14.24vdc.
Since the four cells are already 'balanced', let's see if we can charge or discharge this new module without disturbing the individual cells.
Step Zero: measure the voltage of a module (any good module), presently inside your i-MiEV, with the same DVM. Be careful, as the voltages inside the i-MiEV pack are LETHAL.
Your objective is to have your new module's voltage match the existing modules' voltage, EXACTLY, before installing the new module into the i-MiEV.
By far the best way of doing this is as kiev suggested: use a power supply that you simply dial in the voltage you need to achieve and the current you wish to charge at. Simply turn it on and let it take its time to get there and you're all done.
Alternatively, if your target voltage is around a14.4v then you can simply use a conventional 12v battery charger to bring up your module's voltage.
Whereabouts are you located? Perhaps one of us can come by and help you out?
Now, a dangerous alternative is to use a couple of 12v batteries in series as the voltage source and feed your new module THROUGH A RESISTOR. A vice-grip is NOT a resistor, as its resistance is probably lower than 0.000001 ohms and terrible things could happen...
The purpose of the resistor in series with your voltage source is to reduce the current flowing from the source into your module. Without a resistor, things burn and smoke happens (ouch, those are scary words right now in California).
Let's say you measure and find that each module inside your existing i-MiEV pack is sitting at, say 15.6vdc. For reference, that means that each cell is sitting at 3.900vdc.
Let's now take two plain old flooded-lead-acid 12v batteries (12.65vdc fully charged) and connect them in series. That give you 25.3vdc as a voltage source which you want to use to charge your pack.
Let's be conservative and say that we want to limit the current flow from your source into your four-cell pack to 2A, for example.
Recall that R = E/I, where R is resistance in ohms, E is voltage across the resistor in volts, and I is current in amps.
The size of the resistor you want to put in series is:
(25.3 - 15.6)/2 = 4.85ohms
Thus a 5ohm resistor will do the job.
Now, this resistor will be dissipating a significant amount of power as heat.
P= I^2 * R, where P is Power in watts
P = (2 * 2) * 5 = 20 watts
It's going to get hot, so I would use, for example, maybe a 5 ohm 100-watt resistor (funny, as that's what sitting on my bench as we speak..)
2A will only put in 2Ah in one hour into your 45Ah pack, so this could take a long time in this example. Depending on your target voltage you may want to speed things up and increase the current flow. Just follow the math above to get your new resistor values for maybe 5A or 10A (I would not go higher than 10A). Maybe somebody can suggest what to use for a resistor substitute?
I've sent you a PM with my contact information and I can email you a diagram of how to set this up. Feel free to call me directly.
What is VERY IMPORTANT is to NOT leave this setup unattended after you have attached the wires and are charging your new module. You must monitor the module voltage and disconnect the setup as soon as your target voltage is reached.
Sorry if I'm being too basic about this with you, but this is a completely manual setup I've described which can destroy your new module if not controlled carefully. Once again, it would be best to use a regulated power supply for charging this module.
EVs: 2 Wht/Blu SE Prem., '13 Tesla MS85, 3 156v CorbinSparrows (2 Li-ion), 24v EcoScoot(LiFePO4)
EV Conv: 156v '86 Ram PU, 144v '65 Saab 96
Hybrids: 48v1kW bike
ICE: '88 Isuzu Trooper. Mothballed: '67 Saab (orig.owner), '76 MBZ L206D RHD RV