I am doing an experiment where I measure the internal resistance of two battery cells. I've measured the internal resistance of both of the cells, and when they were in series. The series internal resistance was the resistances summed. However, when I connected them in parallel, the result was unexpected.

By connecting two battery cells in parallel you should be able to calculate the internal resistance to the equivalent battery if you know the internal resistance of both of the battery cells by $\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}$? Am I wrong here? Because the internal resistance I measured was higher than the single battery cells' resistances.


The discrepancy is possibly related to different internal voltages $V_1$ and $V_2$ of the cells (e.g., by one being older than the other) which, when putting the batteries in parallel, already produces a current $\frac {V_1-V_2}{r_1+r_2}$ without a load, so that the equivalent Thevenin voltage $V_{th}$ of the parallel battery circuit is lower than the battery voltage $V_1$ and is given by $$$V_{th}=V_1-r_1\frac {V_1-V_2}{r_1+r_2}$$ Thus, when in the determination of the equivalent Thevenin resistance $R_{th}=(1/r_1)+(1/r_2)$ of the parallel battery circuit (e.g. by measuring the short circuit current $I_s=V_{th}/R_{th}$) you erroneously assume that both batteries have the same Thevenin voltage $V_b=V_1=V_2$ which is higher than the correct $V_{th}$, you would get the wrong Thevenin resistance $$R_{th, wrong}=\frac{V_b}{I_s}>R_{th}$$ which is higher than the actual Thevenin resistance $R_{th}=(1/r_1)+(1/r_2)$.

  • $\begingroup$ Thank you, however, how did you get the $V_{th}$ to be $V_2 - r_2\frac{V_1 - V_2}{r_1+r_2}$? I understand how you get the current, but I am not sure how you got the Thevenin voltage $\endgroup$ – Matty111 Dec 13 '16 at 22:43
  • $\begingroup$ @Matty111 - Sorry, there was a index error in the formula. I corrected it. (1) What values did you get for the internal resistances and voltages of the single batteries? (2) How did you determine the internal resistance of the batteries in parallel? (3) Did you also measure the internal voltage of the parallel circuit of the batteries or did you assume an internal voltage measured on one of the single batteries? (4) There could also be additional (chemical) effect changing the internal voltages of the batteries with different internal voltages when a large current flows in the open case. $\endgroup$ – freecharly Dec 14 '16 at 0:16
  • $\begingroup$ (1) Battery 1: 4.042 V (open-circuit) and 0.7157 ohm. Battery 2: 3.997 V and 0.8589 ohm. (2) I measured, in short intervals (2-3 seconds) , the voltage of the parallel circuit over different currents by changing load and found the slope of the function. Outside of the short intervals the parallel circuit was open so no current could flow between the batteries. (3) I measured the voltage over the whole parallel circuit. (4) Isn't the voltage difference too small and the internal resistances too high for large currents? $\endgroup$ – Matty111 Dec 14 '16 at 0:35
  • $\begingroup$ @Matty111 - Thats the correct way to measure the Thevenin voltage and resistance of the parallel combination of the batteries. According to the single battery values you should have a current 28.6mA in the parallel connection. According to this, the Thevenin voltage of the parallel connection should be $V_{th}= 4.042V-0.0286·0.7157V=4.022V$ which is only marginally lower than the first battery's open voltage. What were your measurement results for the Thevenin voltage and resistance in the parallel case? Small deviation from the calculated value could be within the errors of the measurement. $\endgroup$ – freecharly Dec 14 '16 at 1:04
  • $\begingroup$ Thevenin voltage measured to 4.014V. This seems pretty fair, but the Thevenin internal resistance was 1,414 ohms, which is what bothers me. FYI the measured internal resistance for the same batteries in SERIES was 1,717 ohms, which seems fine. $\endgroup$ – Matty111 Dec 14 '16 at 1:35

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