In this case, what you measure without any external resistance is the electromotive force of the battery, E. When current starts to flow, the terminal voltage becomes V which is lower than E. If Current I is flowing, R is external resistance and r is internal resistance, then,
E = I (R+r) = IR + Ir = V + Ir
therefore, r = (E-V)/I and again, I = V/R
Therefore, r = (E-V)*R/V.
In your case, (1.28-1.16)*8/1.16 ohm = 0.828 ohm
Madeleine L.
Thank you!!10/27/24