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Help with circuits-reactive networks question

  1. Jan 31, 2013 #1
    Hi guys,

    I have this homework question that i have attached. I've done a couple questions but I'm not sure if they are right.




    For the first question i arrived at 15.49 by rearranging the formula; energy stored=1/2.C.V^2. when rearranged you get the square root of 240 for voltage.

    For the second question b I don't understand how you find V. Is this the same as Vs? OR do you have to find the equivalent circuit by having all the inductors behave as a short circuit and capacitors as an open circuit since the circuit has reached a steady state.

    For the third question I think i need the answer from b to continue.

    For the fourth question i used the formula of power=V^2/Resistance. So 15.49^2/92 = 2.608mW

    Not sure if this is right though.

    Any help would be appreciated.

    Thanks

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 1, 2013 #2
    Here's the attachment. The first question should be 21.91 using the formula from the last post.
     

    Attached Files:

  4. Feb 1, 2013 #3

    gneill

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    Staff: Mentor

    Can you past the actual questions? Can't tell the players without a program.
     
  5. Feb 1, 2013 #4
    Sorry i uploaded the wrong file.

    For the second question i got 109.54 by breaking the circuit down, replacing the capacitors with an open circuit and inductors with a short circuit then using the voltage divider equation.
    v0 x (R1+R2/R2) = V.
     

    Attached Files:

  6. Feb 1, 2013 #5

    gneill

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    Staff: Mentor

    Yes, that looks reasonable. The result looks okay to me.
     
  7. Feb 2, 2013 #6
    For the 3rd question i used the formula V=Vo e^(-t/RC) and i got 0.00957 seconds for t. Does this seem correct as its very small.

    For the fourth question i used the formula P=V^2/R, 21.91^2/92=5.21739, then i multiplied it by 1000 to give the answer in mW. Did i use the right numbers for this question?
     
  8. Feb 2, 2013 #7

    gneill

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    Staff: Mentor

    Show your work. How did you determine the time constant ##\tau = RC##?
    Nope. How did you determine the voltage across R9?
     
  9. Feb 2, 2013 #8
    For R i found the equivalent resistant. R8 and R9 in parallel so the resistance when they are combined is 46. Then this new resistor is in series with R7 so they are added together to give 138. C is 0.0001 farads as it says in question 1.

    so time constant is 138 x 0.0001=0.0138
    0.0138 x ln0.5 = -t = -0.00957.

    For question 4 i managed to find the voltage of R9 using the voltage divider formula.

    21.91 x 46/92+46 = 7.3 V
    46 is the equivalent resistance of R8 and R9.

    so for the power i do 7.3^2/92 = 579.71mW

    Is that correct?
     
  10. Feb 2, 2013 #9

    gneill

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    Staff: Mentor

    Ah. The resistors are specified in kΩ (kilo ohms), but you've treated them as just ohms. The equivalent resistance should be 1000X what you've used :wink:
    [/quote]

    For question 4 i managed to find the voltage of R9 using the voltage divider formula.

    21.91 x 46/92+46 = 7.3 V
    46 is the equivalent resistance of R8 and R9.

    so for the power i do 7.3^2/92 = 579.71mW

    Is that correct?[/QUOTE]

    The power calculation looks okay.
     
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