How can current disappear in a bridge?

Discussion in 'Homework Help' started by neolines101, Feb 18, 2013.

  1. neolines101

    Thread Starter New Member

    Feb 17, 2013
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    My Basic Electronics text states: Current is the same amount in all the series components. It baffles me that matched resistances can render current and voltage to zero as in the Wheatstone Bridge. If one constructed a series-parallel circuit using four matched 100 watt light bulbs for the resistances in the same configuration as the Wheatstone Bridge, does that mean you could grab the bare wires and not be electrocuted? How can there be current in some parts of the circuit (enough to light up the bulbs) and yet zero current where the galvanometer is connected? What am I missing?
     
  2. Bryon

    New Member

    Mar 13, 2008
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    The current is still present in the circuit. It has to do with how the galvonometer is connected. For example, if you are trying to measure a voltage with respect to a potential that is equal to the point you are measuring, it will be zero.
     
  3. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    A circuit can be balanced in such a way that there are currents in some portions and no currents in others. The portion of the circuit that has no currnt can definitely have voltage associated with it, that is, it does not have to be 0 volts. That is the case for the balanced bridge circuit. The voltage is the same at both sides of the galvanometers connection, thus no current. No voltage difference , therefore no current. Don't forget though, that if you touch a circuit you can change the flow of currents through that circuit because you yourself also have electrical characteristics.

    So, no guarantee that you won't feel it if you touch the bare wire.
     
  4. neolines101

    Thread Starter New Member

    Feb 17, 2013
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    Thank-you for explaining the circuit characteristics to me! Electronics is, today, slightly less mystifying than yesterday.
     
  5. neolines101

    Thread Starter New Member

    Feb 17, 2013
    3
    0
    I appreciate your explanation; thank-you for helping me understand this circuit!
     
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