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Inductors in a circuit

  1. Apr 21, 2008 #1
    The problem is: There is a circuit such that a battery (source of constant emf) is connected to an inductor and a resister (in series). After an unspecified time period, a switch is thrown, effectively bypassing the battery.

    I know (becuase I've seen the answer) that a) the initial current is the battery's emf divided by R and b) that once the switch has been thrown, in order to find the current, one would set the emf of the inductor equal to IR (then solved the differential equation). I just want to clarify as to why this is so.

    The initial current: the inductor is acting as "back emf" so it's pushing against the current created by the battery. The battery has to do more work in order to maintain the same emf as it would were the inductor not there, however, it doesn't affect that fact that the current is (eventually) equal to the initial emf divided by the resistance. Is that right?

    After the switch: the inductor is providing the only emf in the circuit, and it is effectively slowing it down. The potential difference across the resistor will then be equal to the only emf present in the circuit (ie. the inductor).

    Does that sound about right?

    Another question: why is the current in a circuit consisting of a capacitor at an initial V and an inductor oscillatory? Does that mean it never diminishes, but simply constantly changes direction?
     
  2. jcsd
  3. Apr 22, 2008 #2
    The inductor capacitor circuit, commonly called a 'tank circuit' is oscillatory because the inductor charges the capacitor then the capacitor discharges and the current flows back to the inductor then the inductor charges the capacitor again until the current finally decays.
     
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