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Frequency of circuit with purely real impedance

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Two resistors, a capacitor, and an inductor are connected as shown. What is the frequency at which the impedance is purely real?

    Diagram below

    2. Relevant equations

    Inductor: Z = jwL
    Capacitor: Z = [tex]\frac{1}{jwL}=\frac{-j}{wC}[/tex]
    Resistor Z = R

    3. The attempt at a solution

    I added all the components, but I don't see how I can get very far without any values of them.


    I'm not sure if the capacitor was added correctly, but that's just taken from my notes.

    So is that equation I found equal to Z?

    I need to set Im[Z] = 0, which is the imaginary part of the impedance, and then solve for w. Did I find Z correctly? And how do I know the imaginary part of Z?

    Attached Files:

  2. jcsd
  3. Feb 10, 2013 #2


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    Staff: Mentor

    The impedance of the inductor should involve jω. Your equation shows only L. Otherwise it looks okay.
    You will need to separate the expression into real and imaginary parts (algebraically).
  4. Feb 10, 2013 #3
    Ok so I have [tex]\frac{1}{\frac{1}{R}+\frac{1}{jwL}}+\frac{1}{\frac{1}{R}+jwC}[/tex]

    So now I need to separate the imaginary parts, which are j. I need to have the equation set equal to something for me to be able to do that, right? Since that equation is the [tex]Z_{eq}[/tex], then I set it equal to that? And then solve for j?
  5. Feb 10, 2013 #4


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    Staff: Mentor

    Yes, that's an expression for ##Z_{eq}##. It's not j that you want to solve for. You want to rearrange the expression so that it is put into the form ##Z_{eq} = A + jB##, where A and B are real expressions. In other words, separate the impedance expression into separate real and imaginary parts. Then work out what value of ω will make the imaginary part zero.
  6. Feb 10, 2013 #5
    So the real parts would be the resistance.

    So [tex]Z_{eq}=\frac{2}{R}+\frac{jwL}{1+j^2w^2LC}[/tex]

    Well a value of 0 would make the imaginary part 0, wouldn't it?
  7. Feb 10, 2013 #6


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    Staff: Mentor

    Can you expand upon how you arrived at that expression for the impedance?
  8. Feb 10, 2013 #7
    Yeah, I'm not even sure if I can do that. I inverted both sides of the equation to get rid of the numerator on the right, then I combined the resistance terms, and then flipped it again to get rid of the numerator on the left.

    After thinking about it, there's no way that's right, is it?
  9. Feb 10, 2013 #8


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    Staff: Mentor

    It looks like the way you performed the inversion was not correct.

    Invert the two principal terms separately and massage them into canonical form (a + jb) before attempting to add them (or extracting their imaginary parts for addition).
  10. Feb 10, 2013 #9
    Ok what I'm doing right now on paper is turning into a giant mess.
    I multiplied both sides by the denominators, then I tried getting rid of some of the R's by multiplying both sides by R, but now this looks like I have no where else to go. I have Z's, j's, L's, and w's everywhere.

    This isn't the way this problem is supposed to be solved, is it?

    I'm not even sure what this thing would look like if it was in A + jB form. The A term is supposed to be just the resistance? What is the B term? Everything stuck to the j's? I'm so confused. I wish books actually explained stuff instead of just mentioning stuff.
  11. Feb 10, 2013 #10


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    Staff: Mentor

    Yeah, it can get a bit messy and tedious working with complex algebra. But it's what has to be done here.

    Start with just one of the main terms, say the inductor in parallel with the resistor and see if you can simplify it into the normalized form. You can use the "two resistor expression" for parallel resistors to begin if you wish:
    $$z1 = \frac{R \cdot jωL}{R + jωL}$$
    Clear the imaginary from the denominator.

    Then do the same thing for the other, capacitor term.
  12. Feb 10, 2013 #11
    Just so I'm clear, the imaginary part is the j, right? Well in my notes I found something that might be useful. This is from a parallel RLC circuit:


    Then the next line


    How did that j get factored out? In one term it's in the numerator and the other it's in the denominator.

    But then the absolute value of [tex]Z_{eff}[/tex] is given:


    Where did the j go in this last one? It just disappeared.

    But to do what you asked, I multiplied both sides by the denominator and then did a little rearranging to get only the j's on the right side and I got:


    I'm kinda confused on which letters are considered imaginary and which ones aren't, because if it's just the j that's considered imaginary, then I can't isolate it from the rest of the variables.
    Last edited: Feb 10, 2013
  13. Feb 10, 2013 #12


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    Staff: Mentor

    Yes, the terms with j multiplying them are imaginary. j is the variable name associated with ##\sqrt{-1}##. j can "disappear" if it is multiplied by j (so it's squared) as j2 = -1.

    It sounds like you should take a few minutes to review the properties and basic manipulations of imaginary numbers.
  14. Feb 10, 2013 #13
    But in that last expression, everything was raised to the second power? Well, the reason I said it disappeared was because the + sign in front of it stayed +. If j just turned into a -1, then that + would turn into a -, right?

    I also don't understand how that j was factored out in the 1st one.
  15. Feb 10, 2013 #14


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    Staff: Mentor

    The last expression is finding the magnitude of the complex expression. The real and imaginary parts add in quadrature (square root of sum of squares). This is a property of imaginary numbers that you need to know.
    Multiply the numerator and denomination of the 1/(jωL) term by j.
  16. Feb 10, 2013 #15


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    Homework Helper

    The calculation is simpler if you make the denominators real, by multiplying both the numerator and the denominator by the complex conjugate of the denominator.
    The RL part becomes
    [tex]Z_{RL}= \frac {1}{\frac{1}{R}+ \frac{1}{jwL} }=\frac{\frac{1}{R}-\frac{1}{jwL}}{(\frac{1}{R}+\frac{1}{jwL})(\frac{1}{R}-\frac{1}{jwL})}=\frac{\frac{1}{R}-\frac{1}{jwL}}{(\frac{1}{R})^2+(\frac{1}{wL})^2}[/tex]

    Decomposing into real and imaginary terms, the RL term of the impedance is

    [tex]Z_{RL}=\frac{\frac{1}{R}}{(\frac{1}{R})^2+(\frac{1}{wL})^2}+j\frac{\frac{1}{wL}}{(\frac{1}{R})^2 + ( \frac{1}{wL} )^2 }[/tex]

    Do the same with the capacitor term. Simplify, add the imaginary parts and make the sum zero.

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