Need assistance making sense of my potato circuit findings

Discussion in 'General Electronics Chat' started by wiseachoo, Jan 21, 2012.

  1. wiseachoo

    Thread Starter New Member

    Jan 14, 2012
    I was just playing around with LTSpice (only started using it last week) and was trying to compare the benefits of arranging an array of potato cells in series versus parallel. Please keep in mind I'm a beginner hobbyist who's read up through middle of chapter 7, volume 1 thus far (had them all printed!), so try and keep your explanations somewhat simple if possible :).

    I set up two circuits (A and B) making use of the same 3 potatoes and same hypothetical load. I had anticipated that the parallel potato circuit (A) would produce more current to the load than the series potato circuit (B). This was indeed true (ALoad versus BLoad) but the improvement was only from 1.07m amp to 1.5m amp. Likewise, the voltage, comparing A1 to B1 seems off somehow also. I would have thought B1 would be 1.5v and A1 would have been .5v.

    I think the thing that's throwing me off here is having parallel/series power sources versus parallel/series loads off of a single power source. Likewise, I wasn't completely sure I set up the "internal resistance" of each of the potatoes correctly. I took a potato, stuck a zinc screw and penny into it, grabbed my multimeter, and measured roughly 400ohms of resistance, so I "think" I set that up correctly, but maybe not?

    Any insight/help you guys could offer would be much appreciated. I've attached both a picture showing both circuits and the results as well as the LTSpice file should anyone need to reference how I set things up.
  2. Adjuster

    Late Member

    Dec 26, 2010
    You can't measure the internal resistance of a cell (even a potato cell) with an ohmmeter directly: the cell emf interferes with the measurement.

    The internal resistance of a cell can be determined by measuring the cell voltage at at least two different load currents.

    The simplest way would be to measure the voltage V1 into the voltmeter alone (almost no current), and then measure the voltage V2 with a resistance of Rload ohms in parallel with the cell (current I = V2/Rload ).

    The internal resistance Rcell is then equal to the voltage change divided by the current Rcell = (V1-V2)/(V2/Rload) = Rload*(V1-V2)/V2

    You might try different values of Rload: you probably would not get terribly consistent results, but don't worry about it.
    wiseachoo likes this.
  3. Adjuster

    Late Member

    Dec 26, 2010
    Whether cells in parallel or series will provide more current depends on the internal resistances and the load resistance.

    A high load resistance will get more current from a series stack, as internal resistance is negligible and voltage dominates.

    A lower load resistance will get more with the batteries in parallel, because the effects of the internal resistance are more important.

    With a little algebra, you might be able to find the relationship between Rload and Rcell which gives the same current for series and parallel connection of a three-cell battery.
    wiseachoo likes this.
  4. crutschow


    Mar 14, 2008
    You are seeing correct results. Since the internal resistance of the battery is higher than the load, there is a lot of voltage drop across the internal resistance. And remember when the batteries are in series the internal resistances add, so the total battery resistance for case (B) is 400 x 3 = 1200Ω. If you use a higher load resistance you will see voltage and current changes between (A) and (B) more like what you expected.
    wiseachoo likes this.