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Emfs, work and heat

  1. Jan 19, 2009 #1
    When we say heat released in a circuit is i^2 RT , who/what is the cause of this energy loss; is it the battery mechanism? Since work done on charge in the circuit by E field is zero, the only other force seems to be the battery mechanism, right? But in text books there is so much implication that the resistor is the root cause of heat, I am confused.
  2. jcsd
  3. Jan 19, 2009 #2
    The work done by the battery = emf times charge.The electrons pick up energy in the battery and convert this to other forms of energy in the circuit.If the circuit is resistive only heat is generated.In terms of free electron theory this can be explained in terms of the drifting current electrons continually colliding with the vibrating metal lattice atoms and transferring some of their energy with each collision.
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