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Kirchoff's Loop Law

  1. Apr 3, 2006 #1

    G01

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    I'm having some trouble understanding Kirchoff's loop law. I understand that it is a statement of conservation of energy, and I understand this. My problem comes when we start talking about circuit elements. If we had a shorted battery, the electrons moving through the circuit would loss there potential as they went around. Now lets add a resistor. The law states that the voltage lost in the resistor is equal to the emf of the battery. Why is this? To me it seems that the electrons lose their potential whether there is a circuit element there or not. How can we be sure that all the potential is lost in the resistor? Thank you for you help.
     
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  3. Apr 4, 2006 #2

    Andrew Mason

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    In a circuit containing a battery one has to take into account the internal resistance of the battery. If you short a battery, the current is not infinite. You can think of the battery as a voltage source with a small resistor in series. As you increase the current, there is a drop in the voltage across the battery terminal and the battery will get warm.

    So Kirchoff's voltage law applies. You just have to take into account the internal resistance. In other words, the voltage across the external resistor is equal to the emf of the battery minus the voltage drop across the internal resistance of the battery. This voltage drop depends on the current (V=IR) so for a high applied resistance (low current), the voltage drop across the battery will be small.

    AM
     
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