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Battery on a breadboard

Discussion in 'Power Electronics' started by Kenny_L, Nov 28, 2015.

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  1. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    Hi Everyone,
    I am a newbie to the electronic world so sorry for the stupid question. I have recently bought a breadboard and some electronic components. My questions are the following:

    1) I have a 9-V battery which is connected to the +/- rails of the breadboard. Now I connect a very simple series resistors circuit to the rails. My question is why did the voltage across the battery drop when I measured it under the connecting stage? My assumption was that because the battery can only provide 9 volts in total. When we connect it to some resistors, according to the voltage division law, the voltage of the battery will be dragged down. Not sure if i'm on the right track or not

    2) The second question is regarding to Arduino Uno board. The same setting as above. When I connected the battery directly to Arduino Uno which is currently not connected to any circuits, the voltage reading on the battery is 7.3V. But when I connected the 5V and Gnd to the +/- rails of the breadboard, the battery reading suddently became 2.8V (which caused Arduino Uno to be off) and the volt drops on the rails of the breadboard (currently still connected to 5V and Gnd of Arduino Uno) was down to some 15mV!
    Thank you for your time
     
    Last edited: Nov 28, 2015
  2. davenn

    davenn Moderator

    13,617
    1,881
    Sep 5, 2009
    hi there
    welcome :)



    show us your circuit/construction
     
  3. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    Thank you for your response :) Please have a look at the attached files :) Thank you soo much for your time :)
    Sure :) Can you please refer to the attachment :) I have renamed the file name to the description of the step :) Resistor used for Q1 is 100 ohms and Q2 is 10 ohms. The reason why I changed to 10 ohms in Q2 was that under 100 ohms, Uno could still operate. For better observation, I changed to 10 :)
    By the way, should I go and buy one of those machines which can provide a constant voltage :)
    Thank you sooo much for your time :) Have a good day buddy :)
     

    Attached Files:

  4. davenn

    davenn Moderator

    13,617
    1,881
    Sep 5, 2009
    OK well 100 Ω across the battery is very low resistance heading towards a short circuit
    a 10 Ω is pretty much going to be a short circuit as far as the battery is concerned
    either one will cause a significant voltage drop ... the poor battery !! ;)

    so why are you doing this with these resistors .... ie. what is it you are trying to achieve ?

    Dave
     
  5. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    oh haha i was just playing around with the breadboard when i discovered this incident. But why would there be a voltage drop on the battery when the circuit is simply a circuit without resistors( short circuits)? The battery is in series with the resistor right? So wouldn't it still remain at the original voltage?

    Sorry about that but one thing i cant get my head around is why can a volt drop on a battery change (i.e. become low) when it's simply connected to either the Uno or a simple series circuit.
     
  6. Colin Mitchell

    Colin Mitchell

    1,417
    312
    Aug 31, 2014
    Because the battery has a little resistor inside it and it is technically called INTERNAL RESISTANCE.
    Suppose this resistor is about 20 ohms. When you place a 10ohm resistor across the battery, the real circuit is a 9v supply and a 20 ohm resistor connected in series with a 10 ohm resistor and the other end of the 10 ohm resistor goes back to the negative pole of the battery.
    Now you can see the 10 ohm and 20 ohm resistors are in series and when you measure across the 10 ohm resistor you get a reading of about 3v.
     
  7. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    Hi Colin,
    but does the volt drop across the battery change? i constructed a circuit using a 100 ohm resistor. i observed that the volt drop across the battery changed. in your example, does that mean the resistor has 3 v and the battery has 6v if we started from 9v.
    thank you
     
  8. Colin Mitchell

    Colin Mitchell

    1,417
    312
    Aug 31, 2014
    The output voltage from the battery changes according to the value of the resistor placed across the battery.
    The INTERNAL RESISTANCE and the resistor across the battery create what is known as a VOLTAGE DIVIDER.
    10 ohm and 20 ohm produces 3v across the 10 ohm and 6v across the 20 ohm.
    100 ohm and 20 ohm produces 7.5v across the 100R and 1.5v across the 20 ohm - so the battery voltage appears as 7.5v under load.
     
    Kenny_L likes this.
  9. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    haha thanks guys i got it. by the way, justone trivial question, if i want to ensure that the input voltage is always constant, should i buy one of those machines that produce constant voltage output
    thanks again
     
  10. davenn

    davenn Moderator

    13,617
    1,881
    Sep 5, 2009
    yes, depending on the value of the resistor you put across it

    but lets back up a bit here as there are 2 voltage drops to consider

    firstly, the voltage drop across an resistor(s) in series and parallel configurations across a power supply
    consider these 2 circuits ....

    cct1.GIF

    DC circuit1.GIF
    the top circuit, we have a 10 Ω resistor across a 10V supply ( much like your situation)
    now using Ohms Law, we can work out the current flowing in the circuit

    Ohms Law Triangle.GIF
    put your finger over the unit you want to find .... V = Voltage, I = Current, R = Resistance
    so if we want to find I ( current) in that first circuit ... put your finger over I, That leaves
    V / R so ...
    I = 10V / 10 Ω
    I = 1 Amp

    Note: the voltage of the power supply is ALWAYS fully dropped across the total resistance of the circuit
    The voltage at point A at the top of the resistor in circuit 1 and 2 is 10V and 100V respectively relative
    to the negative (0V) terminal of the battery

    in circuit 1, ALL the voltage is dropped across just 1 resistor
    in circuit 2, ALL the voltage is dropped across the combination of 2 resistors
    and using Ohms Law again, we can work out how much voltage is dropped across each resistor

    --------------

    Now the second "voltage drop", as in what you are seeing is a result of the power supply's ability
    to hold its voltage as the resistance across it decreases
    ( decreasing resistance can be viewed also as an increasing load)

    Those small 9V batteries, the PP3, like you are using, cannot supply a large current, a few
    100 mA ( milliAmps) at best ( ~300 - 400mA)
    As we saw above, a 10 Ω resistor load across a 10V supply is going to have a 1 A current flowing,
    BUT those small 9V batteries cannot supply that 1 A, so as a result the battery's actual voltage drops
    as a response.

    If you had a more powerful supply, one that could supply more than 1A, then you would find that its
    output voltage wouldn't drop significantly as you increased the load ( lowered the resistance)


    digest that and see if you have any questions :)


    Dave
     
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  11. davenn

    davenn Moderator

    13,617
    1,881
    Sep 5, 2009
    read what I just posted :)
     
    Kenny_L likes this.
  12. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    Hi Dave,
    Thank you soooo much for your explanation. Just to consolidate my understanding. Can I say that since there's internal resistance for the battery, when we connect a resistor to it, by the voltage division principle, there will be a volt drop on the "internal" resistor and the real resistor. Now according to Kirchhoff's voltage law, the battery's voltage has to match the resistor's voltage. As a result of that, the voltage drop across the battery decreases. :)
    Thank you sooo much mate :)
     
  13. Kenny_L

    Kenny_L

    7
    1
    Nov 28, 2015
    Hey Colin and Dave, Thank you guys sooo much for your precious time :) Thank you for helping me out :) I really appreciate it :) :) :)
     
    davenn likes this.
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