Voltage across a battery while discharging

[zorro]

What should be the voltage across a 9.7 volt battery pack when it is in use (discharging)?
I used a multimeter to check that and got the reading as 5.8 V. It reads 9.4V when not in use. From my knowledge, a battery is a constant source of e.m.f. But the above observation is weird.
What is the problem?

 

[Proton Soup]

batteries are not perfect emf sources. there is also internal resistance. for heavier loads, you draw more current, and lose voltage to the internal resistance.

http://en.wikipedia.org/wiki/Internal_resistance

 

[zorro]

The voltage drop due to internal resistance (it is of the order of milli ohms) should be very less. May be about 0.2V in my case. A drop of 3.6V is unacceptable.

 

[Dadface]

The voltage delivered to a resistive load depends on the load resistance and internal resistance of the supply:

V=ER/(R+r)

R= load resistance
r= internal resistance
E= EMF

 

[zorro]

The question is about the voltage across the battery not across a resistive load.

EDIT: I got your point there. Is it not possible that the battery is unable to handle high current withdrawal?

 

[Proton Soup]

i'm not really sure why it is, but i got annoyed enough with the question to try and find out what the open circuit characteristics might be at near zero charge. search google (and maybe check out images) for 'open circuit voltage residual capacity' and i think that you will see that, depending on the battery type, open circuit voltage can be pretty high in a battery that is more or less dead. so, possibilities may be that you have far exceeded the normal current capacity of your battery, the battery needs charging, or the battery is failing.

 

[Dadface]

The question is about the voltage across the battery not across a resistive load.

EDIT: I got your point there. Is it not possible that the battery is unable to handle high current withdrawal?

The higher the current withdrawal the greater the voltage lost across the internal resistance of the battery.Don't forget that withdrawing too high a current can prove to be expensive and dangerous.

 

[wazani]

hi
you may find the answer at "http://answers.yahoo.com/question/index;_ylt=AuunKfjGksf89bitoh96qNT_xQt.;_ylv=3?qid=20071007013353AAIQvZZ"
the voltage of electrochemical cell ,as you know from nernst equation is concentration dependent and when it used for a long time it will consumed and the voltage will dropped.