Transient Network Analysis

Discussion in 'Homework Help' started by dalam, Jan 3, 2015.

  1. dalam

    Thread Starter Member

    Aug 9, 2014
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    Untitled.png
    I found out the voltage Vc(t) = 80-40e^(-1000t).
    Voltage at t=1 msec ; Vc=80-40/e =65.28 Volts.
    Using phasor I can find out Vr=√(80^2-65.28^2)=46.23 Volts.
    Now I use P=Vr^2/R to find the power dissipation in resistor.
    I end up getting P=21.378Watts. But this is not correct answer.
    What am I doing wrong? Answer is 2.1Watts.
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    It is not obvious to me what is meant by the function u(t)?
     
  3. dalam

    Thread Starter Member

    Aug 9, 2014
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  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Ok. Here is what I get: V(R) = V(P) because the red trace falls perfectly on top of the green trace. V(p) is the correct equation for the voltage across the resistor.

    The blue trace is the power in the resistor.

    263.gif
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    It's not that the u(t) is unusual, it's the whole 40+40 u(t) statement for the source. This implies that there was 40 volts present before t=0, and after t=0 there is another 40 volts present because of the step 40 u(t). The power at t=0.001 seconds therefore comes out to 2.1 watts, but that's not exact, more exact would be 2.17 watts but that's not exact either, so if you have a more exact result you can post that and we can compare (i'll post the much more exact result).

    So your time expression is wrong to begin with. Perhaps you can show how you approached this problem.
     
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    That is what I concluded, too...
     
  7. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi Mike,

    Wow we posted at the exact same time previously :)

    But anyway, the source originally 40+40 u(t) could be interpreted as a simpler source of 80 u(t) and apply the t<0 existing voltage as the capacitor initial voltage. This leads to a single term equation that is an exponential.
     
  8. dalam

    Thread Starter Member

    Aug 9, 2014
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    The exact answer is 16e^-2.
    I used equation Vc(t)=[v(0+)-v(∞)]*e^(-t/Γ)+v(∞)
    Vc(0+)=40V
    Vc(∞)=80V
    Γ(time constant)=10^-3 sec.
     
  9. dalam

    Thread Starter Member

    Aug 9, 2014
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    Thank you for the simulation. But will you please elaborate on what am I missing in my solution?
     
  10. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Your mistake, is that the source voltage was at 40V at t<0, so after t>= 0, only 40 u[t] was applied across the series res/cap network.
    ie: at t<0 the the circuit had a existing steady state voltage of 40V.

    E
     
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  11. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Your expression for voltage across the resistor vs time was incorrect. I was showing you the correct one V = 40e^(-t/R1C1) at the behavioral voltage source B1, which produces the same voltage at V(p) that the actual circuit produced at V(r).
     
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  12. dalam

    Thread Starter Member

    Aug 9, 2014
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    Thanks a lot. I got the correct answer now.
    Vr(1ms)=(40*e^-1)^2/100= 2.165 Watts
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    I don't understand where phasors are coming into play at all. Phasors are for sinusoidal signals, which this is not.
     
  14. dalam

    Thread Starter Member

    Aug 9, 2014
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    Yes, you are correct. I was doing network analysis after some time so it got messed up.
     
  15. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Your result looks much better now.
     
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