quick resistance/current calculation question

[moon lander]

(Vs-Vr) / I = R

Where:

Vs = Voltage supplied
Vr = Voltage required
I = Current draw of component in Amps
R = Resistance needed

all above from here: http://flashlightreviews.com/qa/resist.htm

the question is: is Vs supposed to be the supplied voltage including voltage sag (in the case of using batteries)? for example, when i test a pair of cr3023s in series, they say 6 volts. if i do the same test while they are driving an led (under load) they read a total of 4 volts. do i use 6v or 4v as my Vs? also, do all battery types experience voltage sag?
thanks,

 

[LEDite]

Lucas;

The voltage sag is due to the resistance of the batteries and wiring.

All batteries have some series resistance.

The drop from 6 to 4 volts seems pretty large to me, but I guess it's not too extreme for that size battery.

The #18650 lithium-ions I use typically have 100-200 milliohms resistance at the 700ma current load I operate at.

Larry Cobb

 

[moon lander]

Lucas;

The voltage sag is due to the resistance of the batteries and wiring.

All batteries have some series resistance.

The drop from 6 to 4 volts seems pretty large to me, but I guess it's not too extreme for that size battery.

The #18650 lithium-ions I use typically have 100-200 milliohms resistance at the 700ma current load I operate at.

Larry Cobb

thanks larry,

so based on your answer i assume i should use the original voltage in the calculation, not the sagged voltage, since the sagged voltage is dependant on the rest of the circuit.
-luke