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In article <1992Oct30.193947.16978@sco.COM>, abhinava@sco.COM (Abhinav Aggarwal) writes:
> I have a rechargeable Ni-Cad battery. I would like to drain it
> to about 1-2 V (from 10V ) before I recharge it.
> 
> I plan to use a 500K ohm resistor to discharge a battery.
> Is this ok in terms of current and loads ?
> 
> Any other simple, cheap, and effective solutions ?
> 
> Please email me at aggarwal@sunee.uwaterloo.ca 

Erm, not exactly....
   500K is a tad high as the current would only be 20 microamps. At that 
rate the battery would discharge faster from leakage effects!
  Normally, a 1.25V battery is considered discharged at anywhere below 
1.0V. Discharging farther can damage the battery. A needle on a voltmeter 
will show the battery voltage (under load) drop very quickly from 1.25V to
under 0.8V. The battery is then considered discharged.
  I assume you are discharging in order to 'rejuvinate' the rechargable.
The rate of discharge for doing this should be between 0.5C and 1C where
C is the capacity of your battery. eg. If the capacity of the battery is 
(say) 500mAH. You would discharge at 250-500mA until drop-off.
  In your case you have a 10V battery (1.25V x 8) so if the capacity is
1000mAH, you would connect a:
    V=IR => R=V/I => i] 10/1 =  10 ohms for discharge at 1A    (C)
                    ii] 10/.5 = 20 ohms for discharge at 500mA (0.5C)

Watch the battery, (interesting isn't it) for the drop in voltage.

Hope this helps you.

 Conor.

Ps.  I am assuming you have a voltmeter to monitor.
PPS  Watching the battery is a tad uninteresting. I am currently piecing
     together a simple circuit to check for the drop and shut off the
     discharge. (opamp to transistor etc....). It's pretty easy....


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