Battery Problem

Discussion in 'The Projects Forum' started by KC15, Dec 6, 2007.

  1. KC15

    Thread Starter New Member

    Dec 6, 2007
    1
    0
    Hello,
    i have a problem i need help with so here it is:
    I have a device that reguires 2V to operate, and has an internal resistance
    6*exp(6)\Omega. i also have 2 batteries with 6V and 3V respectively, and each can deliver a current up to 1 A. In the lab i found a 150ft roll wire rated with a resistance of 0.1\Omega/ft. Design a setup to operate my device.

    If anyone can help please do!
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Can you elaborate on the meaning of "6*exp(6)\Omega" in describing the internal resistance?

    hgmjr
     
  3. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    cud it be 6*10^6 ohms(6 megaohms)?
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Basically, this is a voltage divider problem.

    Ignore the 6V battery for the moment.

    Take the wire, and cut 100 feet from the spool. It's resistance will be 10 Ohms.
    Connect this 100' piece of wire from the power input (Vcc) terminal of the device to the ground or Vdd terminal of the device.

    Take the remaining 50' of wire, and cut it into two 25' lengths. Connect one 25' piece of wire from the negative terminal of the battery to the ground or Vdd terminal of the device. Connect the other 25' piece of wire from the positive terminal of the battery to the power input (Vcc) terminal of the device.

    Each of the 25' long pieces of wire will measure 2.5 Ohms. The 100' piece of wire will measure 10 Ohms. Resistance in series is additive; thus the two 25' sections together measure 5 Ohms, the three sections of wire together measure 15 Ohms.

    The load measures 6M Ohms, but it's resistance is so high in comparison to the 100' length of wire (10 Ohms) that it's negligible. Therefore, the 100' length of wire in parallel with the load will measure very slightly less than 10 Ohms, or about 9.999983.
    Resistance in parallel:
    Rt = 1 / ( (1/R1) + (1/R2) + ... (1/Rn) )
    or for just 2 resistors in parallel:
    Rt= ( R1 X R2 ) / ( R1 + R2 )

    Using Ohm's Law, we'll find the total circuit current:
    I=E/R (Current = Voltage / Resistance)
    I = 3/15
    I = 0.2
    Using Ohm's Law again, we'll determine voltage across the load (your device):
    E=I x R
    E=0.2A x 10 Ohms
    E=2V
    Your 3v battery would work just fine in this circuit.

    For the 6v battery, you would need 5 Ohms across your load, and 10 Ohms in series with the load. You'd need three 50' sections of wire instead of one 100' and two 25'.

    Keep your wire cutters handy ;)

    Now, you COULD simply wire one 50' piece across the load, and use it for both the 3v and 6v batteries.
    50' x 0.1 Ohms = 5 Ohms.
    Use a 5' section from the negative to the ground or Vdd terminal of the device
    Then for the 6V battery, you'd need the remaining 95' from the positive terminal of the battery to the positive/Vcc terminal of the device.
    5' + 95' x 0.1 = 10 Ohms; you would drop 4V across the 100' of wire, and 2V across the load; total circuit resistance = 15 Ohms
    Circuit current:
    I=E/R
    I=6V/15 Ohms
    I=0.4 Amperes

    For the 3V battery then, instead of the 95' wire you'd use 20'.
    5' + 20' x 0.1 = 2.5 Ohms, you would drop 1V across the 25' length, and 2v across the load.
    Total circuit resistance = 7.5 Ohms
    Circuit current:
    I=E/R
    I=3V/7.5 Ohms
    I=0.4 Amperes
    Not as efficient as before, as nearly twice as much power is being wasted in the divider network.
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    That is along the lines of my guess as well.

    Hopefully KC15 will help confirm our interpretation.

    hgmjr
     
  6. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Sounds like a #30AWG wire. It can safely handle about 860 mA maximum. Should work fine as long as you don't coil it up. If you coil it up, it can handle around 200 mA maximum
     
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