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A Few Inductor/Capacitor Problems - Am I going in the right direction?

  1. Oct 9, 2006 #1
    Hey,

    I've got a couple of problems here that seem pretty simple... And I've worked through them and everything, I just have no basis of comparison, no way to even remotley guess if I'm doing the right/wrong thing. At all... What I'm doing makes sense, but I was wondering if someone could possibly glance through these and tell me if what I'm doing is correct??


    First Problem:
    The voltlage across a 2F capacitor is given by the waveform (on the left). Find the waveform for the current in the capacitor.

    (Work shown with my waveform to right)
    [​IMG]


    Second Problem:
    The voltage across a 2H inductor is given by the waveform shown below. Find the wave form for the curren in the inductor.
    [​IMG]

    My answer:
    [​IMG]


    Third Problem:
    Find the value of C if the energy stored in the capacitor below equals the energy stored in the inductor.
    [​IMG]

    My steps:
    [​IMG]
    [​IMG]


    Fourth Problem:
    Given the network shown below, find the (a) equivalent resistance at terminals A-B with terminals C-D short circuited, and (b) the equivlaent inductance at terminals C-D with terminals A-B open circuited.
    [​IMG]

    Now, I'm a bit confused about this one... I've never really encountered anything like this... In part a, I figure I'd treat this circuit like it were a bunch of resistors being shorted, so I end up with 20mH+6mH, which gives me 26mH...

    And then for part b (this is where I'm lost), would I just treat this like a ladder of resistors? I know inductors act like resistors in that you add them up when they're in serial and add their inverses when in parallel... So would I have 20mH+12mH+6mH=38mH in parallel with the 5mH... And I get a nuts answer like 190/43=4.4186mH

    How would I approach this? I'm a bit lost.



    Thanks much for your time!
     
  2. jcsd
  3. Oct 10, 2006 #2

    SGT

    User Avatar

    The first and third problems are correct. For the second problem, the answer is correct in the interval 0-2s. In the interval 2-4s the voltage is zero, so the current must remain constant at 5A (you are adding a null area). From 4s to 6s you have a negative voltage, so the current will drop to zero and remain there afterwards, since the voltage is zero.
    In the fourth problem, if you short-circuit terminals C and D, the 20mH and 5mH inductors will be in parallel (one common terminal at A and another common terminal at CD). In the same way the 6mH and 12mH inductors will also be in parallel.
    For part b, untwist the circuit, putting A and D at the left and B and C at the right.
     
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