measure ampere of battery in parallal?

Discussion in 'General Electronics Chat' started by sangpo, Sep 10, 2013.

  1. sangpo

    Thread Starter Active Member

    Aug 17, 2013
    I have been googling for more than one hour how to measure amperage of 3 batteries(1.5 v each) in parallel connection. Even than I did not get idea how to measure current with DMM. Please expaln in detail please
  2. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
  3. t06afre

    AAC Fanatic!

    May 11, 2009
    Why do you want to this? And also do you want to measure your battery capacity given in Ampere-hour Or just the the current from the battery at some given load. Anyway to short a battery to see how much current it can source is kind of a no-no
  4. sangpo

    Thread Starter Active Member

    Aug 17, 2013
    I am doing experiment to check if there is more current flow in parallel connection of bateries, as expalined in theory. But i did not know how to measure cuuurent
  5. MrChips


    Oct 2, 2009
    The simplest way to measure current with a voltmeter is to put a shunt resistor, say 1mΩ, in series with the battery and measure the voltage across the resistor.
  6. sangpo

    Thread Starter Active Member

    Aug 17, 2013
  7. sangpo

    Thread Starter Active Member

    Aug 17, 2013
    Let me explain the forum, how i analysed.

    In theory it says, if batteries connected in parallel, the total voltage equals to that of one battery. (referring aa battery) but the current will be more. To check that. i did the following.

    1. I have measure current of one aa battery(1.5 v)with resistor(as load) in series with red probe of DMM.(reading is 150mA)
    2. Then I have connected 2 aa battery(3v total) in parallel.
    3. I have measure voltage of that batteries in parallel and reading is 3v
    4. Then I measure the ampere of that 2 batteries in parrallled connection with the same resistor, (as load) Reading should be 300mA. But reading is totally different.

    I am sure, my method of measuring ampere of component in parrale is wrong.

    Please tell me the right way.
  8. MrChips


    Oct 2, 2009
    Use Ohm's Law:

    I = V/R

    V is the same. R is the same. Should I (current) be different?
  9. wayneh


    Sep 9, 2010
    What you described is putting batteries in series, plus to minus instead of parallel, plus to plus and minus to minus.

    Also your meter has a low but not zero resistance and will affect the measurement. Measuring voltage across the resistor is a better approach.

    Current thru the resistor should not change (much) by adding another cell in parallel. It has the same voltage across itself, and therefore nearly the same current will flow. The batteries, however, will last longer because they have more current capacity.
  10. PackratKing

    Well-Known Member

    Jul 13, 2008
    I made up this little jumper to attach to my DMM to get in between a battery positive and either a spring contact, or an adjacent negative, as a troubleshooter to measure what a particular battery powered device was drawing...

    It consists of a chunk of stick-on lead window tape for a burglar alarm system, some heavy insulation cardboard, and the wires...

    extremely handy little gizmo...
    Last edited: Sep 10, 2013
  11. WBahn


    Mar 31, 2012
    Would it really hurt that much to tell us what that totally different reading WAS?

    You are connecting the two batteries in series, not parallel. As you indicated, the voltage reading was 3V but the it should have remained unchanged at 1.5V.

    Putting batteries in parallel will allow for a greater current flow at the same voltage drop due to all of the various effects that get lumped together and called "internal resistance".

    You will only see this effect if you are drawing enough current for the internal resistance to be having a noticeable effect.

    Try this.

    Take three fresh AA batteries. Now use one of them and connect successively smaller resistances until the voltage across the battery terminals, under load, is 1V. Try not to keep the load connected to the batter any longer than necessary to minimize the drain.

    Now put two the other two batteries in parallel (negative terminals connected together and positive terminals connected together) and repeat the experiment. You should find that you can use roughly half the resistance of the single battery before the terminal voltage under load drops to 1V, which would mean that the batteries are supply about twice the current at the same voltage drop.
  12. vk6zgo

    Active Member

    Jul 21, 2012