I have an old calculator which contains 4 2/3 AA NiCd batteries in series. These batteries are very old and need replacing, so I'm going to replace them with 4 2/3 AA NiMH batteries in series, which have a capacity of 600mAh. The power supply is 6VDC and I want to charge the cells at C/10. What value of resistor would I need to use?
Sounds like you want to connect 6VDC charger directly across a series string of the cells plus a resistor, limiting current to C/10, or 600/10 = 60 mA.
Highest current will be when battery is at lowest voltage. If the cells have not been over-discharged, this will be around 0.8V per cell, or 4 x 0.8V = 3.2V for the 'battery pack'. V = IR, so R = V/I, R = (6.0 - 3.2)/.06 = 46.6 ohms. 47 or 51 ohms for a standard value resistor.
P = V^2/R, so resistor power rating should be at least (6.0 - 3.2)^2/47 = 0.16W. To be more safe use full DC supply voltage (as if battery voltage could be 0V), for power rating (6.0^2)/47 = 0.77W. Pragmatically a 1W or 5W power resistor would work.
Note that while NiCd typically maxes out at 1.5V per cell during charge, NiMH may go as high as 1.8V per cell. You may need a higher power supply voltage to fully charge the battery.
This is not at all a recommended way to charge NiMH safely! C/10 is a bit high for "trickle charge" and might be reduced to more like C/20. You have no protections for faster charge, such as full charge detection, time limit, or over-temperature, that a battery charger IC would provide.
If you're only looking to keep an old calculator working, keep a timer on how long the battery is charging and maybe don't charge all the way, say 8 hours at C/10. This will help with keeping the cells from dying early due to overcharge, which is likely with this charging scheme if left unattended for too long.