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Zero voltage

  1. Jan 19, 2015 #1
    Hi all,

    I have a question about voltage drop of a battery. I understand Kirchoff's laws well, but it occurred to me that I did not understand intuitively why the "zero volts" at the end leg of a series circuit would still produce current. Theoretically you could have the majority of a series loop at zero volts but still carrying plenty of current. How is this possible?
     
  2. jcsd
  3. Jan 19, 2015 #2

    russ_watters

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    It is never exactly zero, just close to zero -- and with near zero resistance as well.
     
  4. Jan 19, 2015 #3
    Thanks. That makes sense. It also leads me to another mental breakdown. Is resistance before a resistor the same value as the resistor, and then no resistance (except for the wire's) is felt after the resistor?

    In other words, an initial circuit leg leading up to a 10 ohm resistor would feel 10 ohms resistance throughout the initial leg. After the resistor, resistance returns close to zero.
     
    Last edited: Jan 19, 2015
  5. Jan 19, 2015 #4

    russ_watters

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    No, the resistance is only in the resistor; that's why they get hot.
     
  6. Jan 19, 2015 #5

    sophiecentaur

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    You are putting a very mechanical slant on this. The resistor drops most of the volts (Nearly the full 12V) - because of its geometry (thickness / length) and what its made of (resistivity). Each charge that passes through it will transform a lot of energy into heat. The two joining lengths of wire, if equal lengths, will dissipate equal very small amounts of Power and (this is the same thing) have a very small voltage drop across each of them (perhaps 0.05V). The resistance 'before' the resistor has exactly the same significance in the circuit as the resistance 'after' it. How does the current 'know what's coming' on its path. Initially, it doesn't and only establishes itself at switch on when the initial EM pulse has travelled around the circuit and a steady value is established (takes just a few ns, perhaps) It is not like a journey in a car when you meet up with a traffic jam and slow down when you see brake lights in the distance or the signs tell you to, but shoot off as soon as you're past the hold up.
    It's a big help if you never get involved with the idea of electrons chugging around a circuit - like cars. Electrons should be banned from all electrical discussions until a person can demonstrate that they can use the conventional approach properly and get the calculations right. They really hardly help at all.
     
  7. Jan 19, 2015 #6
    Ok, thanks. I think I'm getting stuck on differences (or my own invented differences) in the math before and after the resistor. Why, then, would resistance after the resistor be close to zero? Before resistor: 1 amp x 10 ohms = 10V After resistor: 1 amp x 0.0001 ohms = near zero volts. (Based on Russ's explanation that resistance after the resistor returns close to nada.)

    I hope I am being articulate. I am stuck thinking that the initial leg feels 10 ohms resistance before the resistor and close-to-0 ohms after the resistor -- two very different values in the same circuit.
     
    Last edited: Jan 19, 2015
  8. Jan 19, 2015 #7

    sophiecentaur

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    What counts is the Potential Difference across a component. Put the negative lead of a meter on the Earth terminal of the battery. Put the + (red) lead on the Battery + terminal: 12V, connection to Resistor 11.95V (PD across wire = 0.05V). Other resistor terminal 0.05V (PD across resistor = 11.9V). PD across lower bit of wire 0.05V. The PDs add up to 12V. Whatever the Voltmeter reads, it's the difference that counts.

    Take a brick to the top of a mountain and drop it on your foot. It will hurt the same amount as if you do the same thing in a valley. It's the difference in height that counts and not how high you are.
     
  9. Jan 19, 2015 #8

    russ_watters

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    You misread what I said: the resistance before the resistor is also close to nada: 1 amp x .0001 ohms = near zero volts. Just like after.

    You asked:
    And I answered no. No, the resistance before the resistor is not the same value as the resistor, it is much, much lower.
     
  10. Jan 19, 2015 #9
    Thanks. I just realized that all of this stemmed from my incorrect application of Ohm's law. After reading more I see that Ohm's law should only be applied here to the resistor itself, and I am trying to use it to calculate the voltage of the wire before and after a resistor (which is why the math made no $%^& sense). Before the resistor, we have very small resistance and thus no voltage drop. After the resistor, we have very little resistance and very little voltage (again, no drop though).
     
  11. Jan 20, 2015 #10

    Svein

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    Actually, you can calculate in the wire resistances if you are finicky (and I know some applications where you really need to do that). If you do, you have an equivalent resistor string of, say, 0.001Ω + 10Ω + 0.001Ω. Calculate the resulting resistance and you see that the wire resistance is less than 0.02% of the total. Do you really want to bother with that when you know that a standard resistor has an accuracy of 1% or worse?
     
  12. Jan 20, 2015 #11
    Possible not just theoretically but practically as well. Key word is superconductivity.
     
  13. Jan 20, 2015 #12

    jim hardy

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    Nice, zoki...

    jsf :

    Remember limits from calc-1 ?
    Zero volts is what voltage approaches as resistance approaches zero

    if resistance gets to actual zero you have a superconductor which still follows ohm's law but with a little help from Lenz to stabilize what would otherwise entail division by zero......
     
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