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Simple question about batteries and directions

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Imagine a simple circuit loop with a battery and two bulbs connected in series.
    I'm having trouble with the voltage of them, where are the + and - poles on the bulbs?
    At first I simply told myself that the current flows (using the passive convention) from plus to minus, and the + poles of the two bulbs are such that the current flows from + to -.
    In a problem I had there's a bulb in parallell with one of the bulbs as well.
    I used mesh analysis (2 meshes) and got that both currents are zero, so obviously something is wrong. Did a whole other stuff as well but have been keep getting different results, it's kinda frustrating since this is very basic.
    Anyway, am I wrong to think that the current flows through the battery from + to - (passive convention)? or does it go from the + pole, through the circuit (not through the battery) and then on to the - pole of the battery?

    2. Relevant equations

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 8, 2011 #2


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    You realize that bulbs are not actually polarized devices and that the voltage across them depends on the direction of the current. As current goes through the device it experiences a voltage drop. One side--the side the current is entering--will be at a higher voltage (call it the + side) than the other side (call it the - side).

    By the way, sometimes the direction of current is unknown, like when doing mesh analysis. In this case, assume a direction, draw the current arrow, and continue the analysis. If you guessed correctly the current will be a positive number.

    Probably overkill for this circuit, but show what you did. Let's see what went wrong.

    If the battery is not a passive device why would it follow the "passive convention"?

    Yes, this would be the "active convention"
  4. Sep 8, 2011 #3
    I was a bit tired, somewhat new to circuit analysis and was confusing the battery with the ideal voltage source. Anyway.

    So what I did was I named the two meshes a and b, with the currents: i(a), i(b).
    a: 3 + i(a)*R + R(i(a) - i(b) = 0 <=> 3 + 2R*i(a) = i(b)
    b: R*i(b) + R(i(a) - i(b)) = 0 <=> R*i(a) = 0
    => i(b) = 3 & i(a) = 0, which doesn't seem right.
    All bulbs have the same resistance, very simple indeed.
  5. Sep 8, 2011 #4


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    Yes, some of your choices for voltage polarity seem a bit random. That will get better with practice. Check out as many tutorials on the subject as you can. Here's one: http://www.allaboutcircuits.com/vol_1/chpt_6/2.html

    It helps to include a picture of your circuit with your voltage polarities and current directions indicated.
  6. Sep 9, 2011 #5
    [PLAIN]http://img109.imageshack.us/img109/2564/namnlswq.jpg [Broken]

    There you go, I still think the polarities are right?
    Last edited by a moderator: May 5, 2017
  7. Sep 9, 2011 #6


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    Thanks for the drawing. Your polarities, as indicated, are fine. By "random", I meant that the polarities in your mesh equations did not make sense.
    As you go around the mesh, you need to be consistent in how you write down the voltages. The rules I use which force me to be consistent are:
    1) Start at some node.
    2) Go clockwise (in the direction of the indicated mesh current)
    3) If the polarity of the first encountered element is "+" then the voltage of that element is positive (or if "-", then negative).
    4) Currents from adjacent meshes are positive if going in same direction as the mesh current, negative if otherwise.

    For example, starting at the bottom of the battery, the equation becomes:

    a: -3 + RiA + R(iA - iB) = 0

    Sometimes it is more intuitive to equate the voltages of the mesh elements to a particular element, like the battery, for example:

    a: 3 = RiA + R(iA - iB)

    You get the same equation.

    Try applying these "rules" for the second mesh. [EDIT1: standby, something doesn't seem right...]

    [EDIT 2: "rule" #3 only applies to active elements. For resistors, just use the voltage drop with respect to the loop current. Apologies.]
    Last edited: Sep 9, 2011
  8. Sep 9, 2011 #7
    Alright, no problem. What do you mean by the voltage drop with respect to the loop current?
    Another thing I don't understand is, why is the voltage from the battery negative?
    It goes from + to -? through the circuit though.
    Does it still count as if the current goes "through" the battery direction-wise, and therefor it goes from - to +? That wouldn't make sense.
  9. Sep 9, 2011 #8


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    "voltage drop with respect to the loop current" = the voltage drops/decreases as you travel through the resistor in the same direction as the current.

    You can look at it this way--take loop A. You want to sum the "voltage drops" across all the elements as you travel in the direction of IA. Starting at the bottom of the battery, you travel through the battery and experience a voltage drop of -3V. Why? because the voltage increased (same as saying it decreased, or dropped by a negative amount).

    Continuing to the next resistor the voltage drop is +RIA.

    Continuing to the resistor shared by IA--since IA is the loop current we care about for the moment, the current through the resistor is IA -IB. So the voltage drop is +R(IA - IB).
  10. Sep 9, 2011 #9
    Oh, right. But that's pretty much "rule" #3?
    I did that, except for inverting the voltage value of the battery. But what about loop B?
    R*iB - R(iB - iA) = 0 <=> iA = 0?
    Because iB passes the right-most resistor from higher to lower voltage (positive) and then the left-most resistor from lower to higher voltage (thus the minus) plus iA which passes that same resistor from higher to lower voltage (thus positive).
  11. Sep 9, 2011 #10


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    For loop B, there is no battery so ignore "rule 3". Start somewhere. How about the top of the rightmost R. In the direction of IB, there is a voltage drop of RIB. For the other R, in the direction of IB, and since IB is the loop current we care about for now, there is a voltage drop of R(IB - IA).

    so you get equation b: RIB + R(IB - IA) = 0

    I think what is confusing is polarity markings on the shared R. They appear when the direction of current is known, in this case the direction of the loop B current is known so the "+" would be on the bottom of the shared R, for the loop B equation. For the loop A equation the "+" would appear on the top. All clear now? :smile:

    [EDIT: here are some more examples: http://www.calvin.edu/~svleest/circuitExamples/NodeVoltageMeshCurrent/ ]
    Last edited: Sep 9, 2011
  12. Sep 11, 2011 #11
    Thanks a lot, I understand now :).
    Another thing I didn't know about/think about until recently is the fact that you can move around components in series or parallell. Anything I should be careful about when doing this?
    For example when using superposition. Say that I have to find a current or voltage through/over a resistor that is in parallell with other resistors. If it's a voltage, then I can merge them, right?
    And if it's a current I can't, since the current is different for each branch?
  13. Sep 11, 2011 #12


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    Nothing comes to mind...maybe keep things like diodes (LEDs or regular) pointing in the same direction.


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