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Calculate average current flow in an electric eel (Simple Circuit Questions)

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data

    i) A particular source of potential difference is capable of providing a maximum potential difference of 2 Volts. Sketch how you would connect two such sources to provide a potential difference of 4 Volts.

    ii) If each source can provide a current of 20 mA before its potential drops to 1.5 Volts, sketch how you would connect them to provide a curent of 40 mA at 1.5 Volts.

    iii) A 10 micron length cell in an electric eel produces a potential of around 100 mV for 10 msec. Calculate the average current flowing into the cell during the action potential assuming the charge per unit length flowing into it during the action potential is 100mV (capacitance = 3 × 10−7 F/m)

    iv) Given the small voltage and current associated with a single cell, explain how an eel produces voltages of the order of 500 V and currents of 80 mA.

    2. Relevant equations

    I = V/R (ohms law)
    ∆Q = C ∆V

    3. The attempt at a solution

    i) IMG_0065.JPG
    As voltages sum in series circuits I assume this drawing is correct? Any part of the circuit not between the two 2 volt cells will measure 4 volts?.

    ii) IMG_0064.JPG
    As amperage sums across a parallel circuits and voltages do not I assume that my drawing is correct? I'm going off of the assumption that both cells are now producing 1.5volts due to the electricity having to pass through 2 cells now each lap of the circuit. Would the voltmeter give a reading of 1.5volts (and 40mA even though it shouldn't be able to measure amperage)?

    iii) 3 × 10−7 F/m x 0.1 V = 3x10^-8C/m. 10um = 1x10^-5m

    ∆Q= 1x10-5m X 3x10^-8C/m = 3x10^-13C.

    I also = Coloumbs over seconds so: I = 3x10^-13C over 0.01s = 3x10^-11 Amps = Average cell current?

    Iv) We want V = 500 and I = 80mA . In order for an eel to produce these numbers I assume that it needs a large mixture of compact series and parallel circuits (cells).

    First we need 100mV = 500V - Therefore we need 5000 cells in a series (500v/0.1v).
    In a series each one is still only producing 3x10^-11 Amps while we need 80mA.
    8x10^-2A/ 3x10^-11 A = 2666666667 cells in parallel? (Not sure about this calculation).
    Am I confusing cell number for chain number? I'm having a bit of trouble visualising the actual circuit. Is it 5000 cells one after an other (question i) with 2666666667 parallel circuits coming off of where A is in my drawing of i?
     

    Attached Files:

  2. jcsd
  3. Oct 18, 2014 #2

    gneill

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    Staff: Mentor

    Yes, connecting the batteries in series is the right idea, but you've drawn a short circuit across the result. Besides draining the batteries very quickly, anything connected in parallel with the short will see no potential difference.

    Instead of a short circuit, why don't you sketch in a resistor as a load?
    The parallel connection idea is right, but your reasoning for the drop in potential is not. Real batteries (as opposed to ideal ones) are not perfect. They have internal resistance that drops the potential seen across the battery terminals as the current drawn increases. "Laps of the circuit" are irrelevant. What matters is the amount of current drawn by the load.

    Your diagram shows two cells and a current source in parallel. Was that intentional? As it stands the current source will be driving current into the + terminals of the batteries, thus charging the cells rather than the cells discharging into a load. Note that the longer line segment of the cell symbol represents the + end of the cell.

    If you want to demonstrate the parallel connected cells delivering 40 mA at 1.5 V, why not draw in a resistor load and use Ohms law to give it an appropriate value? Then indicate on your circuit the current and voltage.
    Well, the units work out. Seems like a reasonable value for the current for a single (living) cell.
    Yeah. Imagine that one battery consists of 5000 (biological) cells stacked in series. Each of those batteries will produce a 500 V potential difference from end to end, but with a very tiny current capability. Then put about 2.7 billion of those batteries side by side to make up the required current.
     
  4. Oct 18, 2014 #3
    Okay so I redid i) http://i.imgur.com/sZAcDjk.jpg
    I put in a resistor and assigned it a random ohm value of 600 which would mean the current of the circuit would be 6.66mA (4volts/600ohm) (is giving the resistor a random value what you meant when you said load resistor)? As the question doesn't mention anything about current I assume it's okay to put in the resistor?

    For ii) http://imgur.com/Fa8jyqR
    I just replaced the V with a resistor with 37.5 ohm that I worked out using ohms law, I know this would work if I had just one 40mA cell but I'm not sure on how to account for the different currents in each branch. Would the circuit directly after the resistor be providing a 40mA current (at least before it reaches the first battery?). Sorry for all of the questions, I'm just finding this really hard to wrap my head around.

    iii) Glad I got one right!

    iv) http://i.imgur.com/lVX5OdS.jpg
    So each battery consists of 5000 cells and the eel has 2.7 billion of these in a parallel circuit to generate the necessary amperage? Like this? (I drew a cell by accident instead of a battery). Would this circuit also need a load resistor at the end?
     
  5. Oct 18, 2014 #4

    gneill

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    Staff: Mentor

    Yup. You don't even have to give it a value, really. It just indicates how the load is connected to the series-connected cells.

    Each cell will provide 20 mA for a total of 40 mA that passes through your load resistor. So something like this:
    Fig2.gif
    The yellow boxes represent the non-ideal cells with their internal resistances that cause the voltage drop down from the ideal 2V when current is drawn. You could work out a value for them, if you're keen :smile:
    No need for a load. But you could put together a diagram that gets across the idea that there are biological cells lined up and connected together. Maybe something like this?

    Fig3.gif
     
  6. Oct 18, 2014 #5
    Thanks for all of the help. So the load resistor is used to show the voltage and amperage passing through it so that I can answer the question? Also what are the black dots connecting the cells together in the eel diagram and how come some of them are going horizontal?
     
  7. Oct 18, 2014 #6

    gneill

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    Staff: Mentor

    It just makes clear how the batteries are connected and how they supply current to a load, and it does this by showing the question statement's voltage and current values where they would show up in the circuit.
    Those dots indicate the "missing" terms in the series of items. Just as in : 1,2,3, ... 100 .
     
  8. Oct 19, 2014 #7
    I'm a bit vary of including the load resistors (the markers are pretty strict), wouldn't their inclusion see a decrease in voltage after them? Or do they only decrease current. Could I get away with not joining the circuit and using an arrow to point out the voltage +amperage like you did for the eel question?
     
    Last edited: Oct 19, 2014
  9. Oct 19, 2014 #8

    gneill

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    Staff: Mentor

    You can do it either way. I don't think that the markers can be so strict that they would allow only one or the other.

    The important thing being taught in this lesson is the concept of series and parallel combining of voltage sources:
    • When do voltages add
    • When do currents add
    • How to combine sources to achieve a desired total voltage and current capability
    If your diagrams correctly show those connections for the each case then you should be good to go.
     
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