RLC calculations

Discussion in 'Homework Help' started by Hitesh Modh, Dec 26, 2014.

  1. Hitesh Modh

    Thread Starter New Member

    Dec 26, 2014
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    Please, please, please, please help me. help me


    Questions:

    Three impedance are connected in series across a supply of V=100 V, 1.0 kHz supply:

    # 125 Micro Henry inductor that a 0.4 ohm resistance. (Note: V1= Voltage across inductor)
    # 20 Ohm Resistor, (Note V2= voltage across resistor)
    # 120 pF capacitor in parallel with a 150 ohm resistor, (V3= voltage across these two components.)

    a. Sketch the circuit diagram

    b. Calculate the magnitude of the voltage V1 across the first impedance

    c. Calculate the magnitude of the voltage V3 across the third impedance
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Please don't hijack someone else's thread. Not only is it rude, it causes nothing but chaos and confusion.

    Start a thread of your own, state your problem, and then show YOUR attempt to solve YOUR homework problem so that we have a starting point for discussion.
     
  3. bertus

    Administrator

    Apr 5, 2008
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  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Makes no sense.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    It appears that the inductor has a series resistance of it's own, the ESR. This it typical.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Do you know how to calculate complex impedances and use them in an analysis?
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    So the total impedance of the inductor is: jwL+ESR? j(1 kHz)(125 uH)+0.4 Ohm?
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    The dc resistance of the coil wire is 0.4 ohms in this example. It is typically represented by adding a resistor to the circuit, unless of course, you can just add that parameter to your simulation software. Adding the resistor would demonstrate clarity in your thinking.
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Yes. You can actually think of this in at least two ways:
    1. The inductor has a series resistor of Rs so the impedance is j*w*L+Rs (Rs is the ESR as you noted)
    2. If there is an already present resistor in series with the inductor then just add it to that resistance. So if the resistor was 100 ohms just add 0.4 to that and get 100.4 ohms. This is only valid if the resistor is in series with the inductor and there is nothing else connected to that lead of the inductor.
    .
    Inductors always have at least some series resistance so we end up doing this for almost every inductor. This is more important in power circuits because it affects efficiency and even whether or not the circuit works as all sometimes. Some boost circuits wont work at all if the series resistance of the inductor is above a certain value because the inductor can never absorb enough energy from the power supply to allow the output to be boosted to a higher level with the required current demand.
     
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Thank you. I don't think I ever had this in my circuits class.
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Yes, this is correct because you properly used units. The unit Hz·H IS a unit of resistance, but just not ohms. There are 6.28Ω per hertz-henry.
     
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