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Voltage problem

  1. Apr 16, 2006 #1
    One application of an RL circuit is the generation of time varying high-voltage from a low-voltage source as shown in the figure.
    1257 ohms

    14 ohms 2 H (switch) b
    a
    12.6 V.
    Sorry, I couldn't get the picture, but heres a rough idea.. it's in a circuit.
    What is the current in the circuit a long time after the switch has been in position "a"?
    I got this part... it was 0.9 A
    Part 2 says: Now the switch is thrown quickly from "a" to "b". Compute the initial voltage across the inductor.
    I got this too.. it was 1143.9 V
    Part 3 says: How much time elapses before the voltage across the inductor drops to 13 V? Answer in units of ms.

    I used the equation [tex] I= I_0 e^ (-R/L)t [/tex]
    and then figured out that [tex] V= I_0 Re^(-R/L)t [/tex]
    so 13.0 V=(0.9)(1143.9)e^(1143.9t/2)
    solving for t gave me 7.64 ms which isnt' right.. can someone please help me?
     
  2. jcsd
  3. Apr 17, 2006 #2

    berkeman

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    Staff: Mentor

    You're putting in 1143.9 for the resistance in the last equation? Are you just a factor of 0.9 low on the correct answer?
     
  4. Apr 17, 2006 #3
    I tried using just 1143, but it still isn't right.
     
  5. Apr 17, 2006 #4

    berkeman

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    Staff: Mentor

    Maybe show in a bit more detail how you got 1143V. A picture would definitely help, if you could just upload some simple DOC or PDF file.
     
  6. Apr 17, 2006 #5
    [​IMG]
    I don't know if this will work. I found the 1143 V by:
    first I calculated the initial current with was 0.9 A.
    then I figured out the voltage from the 14 ohm resistor by using IR.
    V= (14)(0.9)= 12.6 V
    then I did the same thing for the larger resistor
    V= (1257)(0.9)=1131.3 V
    Then to get the total voltage, I added these together..
    1131.3+12.6=1143.9 V.
    so that's why I was using it to find how much time it would take.
     
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