Stored Magnetic Energy - Inductor Circuit

  1. 1. The problem statement, all variables and given/known data
    In the above circuit, the EMF from the battery is 12 V and the resistor has a resistance is 6.6 Ω. The inductor consists of a long, thin cylindrical coil of wire with 30000 turns, a radius of 5 cm and a length of 61 cm.

    Answer the following questions for a time 1.4 seconds after the battery has been connected.

    (c) How much energy has been delivered by the battery up to this point?


    2. Relevant equations
    U = 0.5LI^2

    ∫Udt from 0 to 1.4s

    3. The attempt at a solution

    I found the inductance and the current for the 1st two parts of the problem:

    (a) What is the inductance of the solenoid? 14.56171141 H
    (b) What is the current through the battery? 0.8542201 A

    I plugged in the numbers to find the energy: U =0.5*14.56171141*(0.8542201^2) = 10.62556402 J

    Then I think that you would have to integrate over time from 0 to 1.4s which gave me a result of 7.4378948 J.

    It did not like the answer. Any suggestions?
     
  2. jcsd
  3. you know the eqn of current in circuit at some time t

    use it to find the charge flown through battery ... q = ∫i dt
    where t changes from 0 to 1.4

    now work done by battery = Q(EMF) = energy delivered
     
  4. Do I just inetrate (V/R)*(1-e^(Rt/L))? Its the only thing i can think of now that will be of any help. I tried to figure this one out from the hints you gave me but no luck.
     
  5. gneill

    Staff: Mentor

    You have the expression for the current with respect to time. That's the current that is being delivered by the battery to the circuit. Do you know what the instantaneous power (watts) delivered by the battery is?
     
  6. Yes, it is P=V*I(t). So, when I find the total current over time I can just plug it in to that!
     
  7. gneill

    Staff: Mentor

    Yup. Integrate the power to find the total energy delivered.
     
  8. But same is integrate current and then multiply by V

    both methods seems different but are same, right?
     
  9. Thanks for the help! I see if i can get it in a bit!
     
  10. gneill

    Staff: Mentor

    Yes; in this case the voltage is constant, so it can be "pulled out of" the integral.
     
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