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Underanding voltage and electron's energy loss

  1. Jun 12, 2012 #1

    i keep getting conflicting information on these and it confuses me

    1) First my book says "In series circuit the voltage supplied by the cell is "shared" between all the components, so the more bulbs you add to a series circuit the dimmer they all become."

    If you have a circuit with 2 components on it, called bulb 1 and bulb 2. Why would bulb 2 which is further along than bulb 1 lower the voltage bulb 1 receives? And why would you put a voltmeter in parallel with the object you are measuring if voltage is the same across the entire circuit, in that case you could just put it anywhere?

    This says the opposite.. i cant post the link it wont let me.

    "However, the voltage at different points changes, as the electromotive force or pressure, drops from a potential difference of 12 volts as it leaves the battery, to virtually no difference, no voltage at all, as it returns.

    This is called voltage drop. It is caused by the pressure lost in driving the current through the resistor.

    After the first resistor, voltage has dropped from 12 to 8 volts. After the second, it’s down to 4 volts. After the third, zero. "

    Which suggests you could have a bright bulb and then a dim bulb further on in the series when the voltage drops lower. But this is the opposite of what the first paragraph said.

    So either everything receives the same voltage or the voltage drops past each component, it cant be both

    2) What energy does the power station give electrons which is then taken by appliances and turned into other forms? I read it was kinetic energy but then someone else told me it's not so im wondering what it is because I can't find information about this anywhere.
  2. jcsd
  3. Jun 13, 2012 #2


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    They say the same thing... thats is that the voltage drops as you go through each component in the series circuit .... lets just look at a simple resistor circuit.


    OK a very simple 1 resistor circuit ( same thing would happen for one of your light bulbs)
    one side of the battery is 100V the other side ( the negative) is 0V that means there is 100V being dropped across the resistor.

    now a series circuit with 2 resistors.....


    With reference to the 0V line ( the battery negative terminal) what do you think the voltages are at point A, point B and point C ?
    will give you a hint .... the 2 resistors are a voltage divider and because they are the same value its a divide by 2

    Because the voltage you read across the component will be the voltage drop across that component ... It WONT be the same value as you measure across the battery, unless its the ONLY component in the circuit


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    Last edited: Jun 13, 2012
  4. Jun 13, 2012 #3


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    OK, hopefully I have this correct. you comment infers 3 resistors.....


    so what can you see is happening to the voltage across each resistor ?

    now using Ohms law V = I x R and knowing the voltage drop across each resistor and the current flowing in the circuit, we can work out the value of the 3 resistors.
    Have you used Ohms law yet ?


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  5. Jun 13, 2012 #4


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    in that last post we will compare that to the comment in your first post.....

    no thats not a correct statement you have suggested.
    look at that last post and imagine those 3 resistors are 3 bulbs, all identical 12V ones so they all have the 100 Ohms resistance shown in the diagram.
    when you work out what the voltage is across each resistor, it would be the same as if they were bulbs.
    1) how much voltage is each bulb getting ?
    2) how much voltage are they supposed to get to light correctly?
    3) if there is just 1 bulb and it lights to 100% brightness at 12V, how bright each would 2 bulbs in series be ( % wise) ?
    4) how bright would each bulb be (% wise) if there were 3 bulbs in series ?

    and the $6 million question for an extra 100 points ;) Why do you get these results ?

  6. Jun 13, 2012 #5
    For the second post I think:

    Before A: 100 volts
    After A: 100 - (100 * 1) = 0

    Before B: 0 volts
    After B: 0 volts

    Before C: 0 volts
    After C: 0 volts

    For the third post:
    It lost no voltage at A so no resistance
    It lost 4 volts at B so 4/1 = 4 ohms
    It lost 4 volts at C so 4/1 = 4 ohms
    It lost 4 volts at D so 4/1 = 4 ohms

    I think i see what is happening for the 4th one

    if the voltage is 12v and each bulb has 100 ohms of resistance then the current is 0.04A
    12 volts flows into the first bulb but 8 leave so it only gets 4V, I was thinking it would take the whole 12 if it needed it, the second gets 4V and the third gets 4V so they are all dim. This is due to the current dropping. I guess they would be at 33.3% normal brightness
    Last edited: Jun 13, 2012
  7. Jun 13, 2012 #6


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    Hi again Steve

    OK in my first post, second diagram with 2 resistors, you didnt take the hint
    the 2 resistors are a voltage divider and because they are the same value, its a divide by 2

    ..... at point B it will be 50V 100V divided by 2
    if the resistors happened to be different values, then the value of the drop across each resistor would be different
    You got points A and C correct :)

    In my 3rd post, the one without the diagram, we subsituted the 3 100 Ohm resistors for 3 bulbs of the same resistance.
    You should be able to see from the second post, with the 3 resistors that there is a stedy reduction of voltage from 12V down to 0V. Each resistor drops the voltage by 4 volts....
    see the voltage readings at point A = 12V, B= 8V, C=4V and finally D=0V

    So if you substituted the resistors for your 12V bulbs it means that each bulb is only going to see 4V across it rather than 12V so each of them is only going to be 1/3 (33%) as bright as what they would be if there was just 1 of them across the 12V supply.
    AGAIN Note... they will all be the same brightness

    If you are at school or university? maybe you have access to an electronics lab with a power supply and some bulbs and you will prove this for yourself :)

  8. Jun 13, 2012 #7
    Oh, I think I see what you mean for the second image. I took the current as 1A but that circuit can't exist unless it is 0.5A which makes each resistor eat 50V. I think the previous examples I mentioned can't exist in the real world then

    I think I understand this now, thank you.

    The only thing I am left wondering is what type of energy (kinetic, thermal, etc) electrons gain from power stations and lose when they go through an appliance.
  9. Jun 13, 2012 #8


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    ahhh hah

    you have caught me out :)

    that will teach me for just copying and pasting my first cct without checking the values!!
    well done on seeing that :)
    I will alter it
    Yes I have doubled the resistance from 1 x 100 Ohm resistor to 2 x 100 Ohm resistors and with the same voltage there will be 1/2 the current flowing.
    I better sort out my other 3 resistor ones as well !! or... of you like you tell me what the current should be in the circuirt with 3 resistors :)
    for working out the current flow just add up ALL the resistors for a total resistance and use Ohms law with the quoted voltage

    Remember in a series circuit like this, the current will be the same whereever its measured in the cct


    PS OK thats better :) the 3 resistor circuit is OK cuz we have named a voltage and current but in that cct we havent named the resistances. You did work that out earlier to be 4 Ohms each
    Last edited: Jun 13, 2012
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