Time-function,one reactive

Discussion in 'Homework Help' started by Alice10, Jan 6, 2015.

  1. Alice10

    Thread Starter New Member

    Oct 3, 2014
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  2. WBahn

    Moderator

    Mar 31, 2012
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    Which side of Rs is the more positive?
     
  3. Alice10

    Thread Starter New Member

    Oct 3, 2014
    2
    0
    I would say the "south" side on Rs would be positive because the current goes in to the positive side of a resistor?

    But then the next question gets wrong, when I want to replace everything except the capacitor with a "two terminal equivalent".
    (Doing KVL at t=0^+ , starting at bottom through U,Rs,Rc(no current so=0),Uab
    Since then my KVL around is Uab=-U-(-IRs)

    Uab=UT
    answer in book: UT=-U-IRs
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    Okay, so now just track the math.

    Let's call the bottom node 'g', the node at the top of the capacitor 'b' and the top node 'a'.

    As you noted, the voltage on the "south side" of the resistor (node 'g') is higher than the voltage on the north side (node 'a') because the current is entering from the bottom. So

    Vga = Vg - Va = I·Rs = the voltage across the resistor.

    The voltage across the capacitor, Vc, is defined such that it is equal to

    Vc = Vb - Vg

    Because there is no current flowing in Rc, the voltage across it is 0V, hence

    Va - Vb = 0; hence Va = Vb

    Combining these together, we have:

    Vc = Vb - Vg = Va - Vg = -(Vg - Va) = -(I·Rs) = -I·Rs

    For the next part, please show your work -- I'm not sure I'm following your verbal description well enough.
     
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