Amplitude of currents

Discussion in 'Homework Help' started by Niles, Dec 10, 2008.

  1. Niles

    Thread Starter Active Member

    Nov 23, 2008
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    Hi all.

    Please take a look at the attached circuit. I've have found the amplitude of the current through the resistance to be:


    <br />
\left| {I_0 } \right| = \frac{{\varepsilon _0 }}{{\left| {R - \frac{R}{{\omega ^2 LC}} + \frac{i}{{\omega C}}}\right|}},<br />

    where ε_0 is the amplitude of the EMF, and the EMF is given by ε_0 cos(ωt).

    This is all good (and correct too!), but in my book it says that in general, the amplitude of the current is given by:

    <br />
\left| {I_0 } \right| = \frac{{\varepsilon _0 }}{{\left| Z \right|}},<br />

    where Z is the impedance. So according to my book, the amplitude of the current through the resistance must be:

    <br />
\left| {I_0 } \right| = \frac{{\varepsilon _0 }}{{\left| {R - i\omega L + \frac{i}{{\omega C}}} \right|}}.<br />

    What's wrong here? I mean, I know my result is correct, but it is obviously not the same as the one my book wants. What impedance is it I have in my denominator then?

    Thanks in advance.

    Sincerely,
    Niles.

     
    Last edited: Dec 10, 2008
  2. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,040
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    " alt="
    \left| {I_0 } \right| = \frac{{\varepsilon _0 }}{{\left| Z \right|}},​
    " />

    where Z is the impedance. So according to my book, the amplitude of the current through the resistance must be:​

    <br />
<br />
\left| {I_0 } \right| = \frac{{\varepsilon _0 }}{{\left| {R - i\omega L + \frac{i}{{\omega C}}} \right|}}.<br />



    What's wrong here? I mean, I know my result is correct, but it is obviously not the same as the one my book wants. What impedance is it I have in my denominator then?

    Thanks in advance.

    Sincerely,
    Niles.
    Click to expand...
    I'll give you a partial answer....but will let you figure out the rest. You have the complete circuit impedance. You also have the "AC Ohms Law" e=i/z. What you need to do now is figure out the VOLTAGE across R1. The revelation should now snap into place. :)

    Good luck!

    eric
     
  • Niles

    Thread Starter Active Member

    Nov 23, 2008
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    Thanks for replying so fast.

    So the expression in my book assumes that I am dividing the EMF across the resistance, and not the total EMF of the circuit, with the total impedance?
     
  • KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,040
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    Bingo!

    By the way....I'm not sure how they teach this any more....but it greatly simplifies these sorts of problems is you use the ADMITTANCE, SUSCEPTANCE, and CONDUCTANCE concepts. Complex parallel circuits become simpler when using the inverse functions....if you're facile with both impedances AND admittances, it's a powerful weapon in your arsenal.

    Eric
     
  • Niles

    Thread Starter Active Member

    Nov 23, 2008
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    Hmm, but I think that there is something fundementally I have not understood.

    Ohms law (let us just look at DC) says that V = RI, and this "law" is valid for resistors. It says that "the voltage across the resistor is equal to the resistance of the resistor times the current through the resistor". This makes sense.

    Now we look at AC currents, and Ohms "law" for AC-circuits is e = ZI. As in the DC-example, we have that "the voltage across the component is equal to the impedance of the component times the current through the component".

    But in our case (my first post), you said I should divide the EMF across the resistor with the total impedance, and not just the impedance of the resistor?
     
  • steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    You probably are just confusing the notation in the book with your notation. Your formula is correct and your concepts are correct. Basically, the book is saying that the magnitude of current (of any circuit or component) is the magnitude of voltgage across it divided by the magnitude of impedance. In your case, the  I_0 and the  \epsilon_0 are both different than the book, but the notation is the same so it is confusing.

    In other words you can calculate the voltage on the resistor and just divide by R to get the current. You would get the same answer you already have.
     
  • Niles

    Thread Starter Active Member

    Nov 23, 2008
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    That is just what I wanted to hear! Thanks.

    By the way, is it valid to talk about "replacement impedances" just like we talk about "replacement resistances"? E.g. in the circuit I attached, can all 3 components be gathered in 1 single component with a specific impedance?
     
  • steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    You can do that for sinusoidal signals in linear circuits. I prefer to do this with Laplace transforms using "s" rather than directly with jw, but both are valid. The laplace transform formulas are a little more versitile for both theory and Matlab, and you can always substitute s=jw when you need to get the actual frequency response and magnitude/phase relations.

    Xc=1/jwC or Xc=1/sC
    Xl = jwL or Xl=sL
    R=R

    Then just use your parallel and series resistor combinations as you suspected.
     
  • Niles

    Thread Starter Active Member

    Nov 23, 2008
    56
    0
    Thanks to both of you; I'm glad that I understand it now.
     
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