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Galvanometer, voltage drop readings

  1. Sep 9, 2011 #1
    i have problem to conceptualize this old galvanometer readings for voltage drop. i plot a diagram of circuit of interest for bether explanation of whats bothering me.

    so, in DC circuit with some resistor in series, electric current is constant and is of some known value. and if i want to use galvanometer to determine voltage drop on resistors, than it is enough (is it?) to plug it in parallel with resistor. than fraction (lesser the better) of main circuit current will loop in to galvanometer and deflect scale. once we have deflection, by knowing galvanometer resistance, we know interpret that as voltage drop.
    if we repeat measuring on R2, again litle fraction of main current enters galvanometer and again deflection is there.

    my question is why is deflection on galvanometer diferent on diferent resistors if the series current is constant and equal trough all resistors?
    or, why is fraction of main circuit current wich enters galvanometer, diferent on diferent resistor if the electric filed is constant trough wire?
     

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  3. Sep 9, 2011 #2

    Drakkith

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    Staff: Mentor

    I believe it is because you are turning one of the resistors into a parallel circuit and changing the resistance of that part of the circuit. Since the current through the circuit is equal to the voltage divided by the total resistance, changing the resistance of the circuit will alter the current. I'm not experienced with circuits, but wouldn't it be detrimental to the circuit to plug the meter in parallel and alter the current?
     
  4. Sep 9, 2011 #3
    well, the whole point is to make readings without making any changes in circuit( that's the reason for big galvanometer resistance) so, thats not my answer.
    thx on reply.
     
  5. Sep 9, 2011 #4

    Drakkith

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    Staff: Mentor

    Ah, ok. I didn't know a galvanometer had a very large resistance. Anyways, I think my answer still applies. You will affect the circuit, but with a very large resistance the current will only change a very small amount.
     
  6. Sep 9, 2011 #5
    trust me, you can get nothing better from altering initial situation. there is some obvious explanation, but my brain is tilted right know and is of no use.

    i am looking for someone who know explain to me galvanometer in terms of electric field , something like this:
    http://galaxy.cofc.edu/rcircuits.html"
     
    Last edited by a moderator: Apr 26, 2017
  7. Sep 9, 2011 #6

    Drakkith

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    Staff: Mentor

    Alright, sorry I couldn't help.
     
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