Results 1 to 8 of 8

Math Help - Resistance and attenuation

  1. #1
    Member
    Joined
    Jul 2005
    Posts
    187

    Resistance and attenuation

    The topic was actually meant to be 'resistance and electromagnetic oscillation'.

    It is hard to translate specific words. I hope you understand what I meant.

    I have an oscillating circuit. The data I know are:
    Capacitance C=20*10^(-6) F
    Inductance L=180*10^(-6) H
    Attenuation (logaritmic decrement) = 1,4*Pi
    What is resistance and period ?
    There should be two periods, for example T and T_0.
    If T_0=2*Pi*square root of(L*C), then what is T ?

    I believe that it is possible to find the resistance using formula which relates damping factor to resistance and inductance. Damping factor = R/2L

    Resistance should be 3,4 according to book.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by totalnewbie View Post
    The topic was actually meant to be 'resistance and electromagnetic oscillation'.

    It is hard to translate specific words. I hope you understand what I meant.

    I have an oscillating circuit. The data I know are:
    Capacitance C=20*10^(-6) F
    Inductance L=180*10^(-6) H
    Attenuation (logaritmic decrement) = 1,4*Pi
    What is resistance and period ?
    There should be two periods, for example T and T_0.
    If T_0=2*Pi*square root of(L*C), then what is T ?

    I believe that it is possible to find the resistance using formula which relates damping factor to resistance and inductance. Damping factor = R/2L

    Resistance should be 3,4 according to book.
    Why should you have two periods? Perhaps if you gave me what your definition of "logarithmic decrement" I could understand the problem better. However I don't see why an RLC circuit would produce a second period in the signal...

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2005
    Posts
    187
    <br />
T_0=2*Pi*\sqrt{(LC)}<br />
    <br />
T=(2*Pi)/\sqrt{((1/LC)-(R^2/4L))}<br />

    These T gives different values and dividing both, for example <br />
T_0/T<br />
means the relative oscillation change because of attenuation. (I don't know if these words are right in this context in english).

    What I meant with logaritmic decrement was that: logaritmic decrement=attenuation*T
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by totalnewbie View Post
    <br />
T_0=2*Pi*\sqrt{(LC)}<br />
    <br />
T=(2*Pi)/\sqrt{((1/LC)-(R^2/4L))}<br />


    These T gives different values and dividing both, for example <br />
T_0/T<br />
means the relative oscillation change because of attenuation. (I don't know if these words are right in this context in english).

    What I meant with logaritmic decrement was that: logaritmic decrement=attenuation*T
    I see now. Your T0 is not actually a period in the circuit, it's due to the resonant frequency of the circuit. (At least in Intro Physics we don't normally talk about the period of this frequency which is why I didn't understand what you were referring to.)

    I'm afraid I still can't give you an answer though. I've looked through my material (and a Google search) and can't find a formula for the attenuation. It must somehow be related to the damping factor (R/(2L)), but I don't know what the precise relationship is.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2005
    Posts
    187
    Quote Originally Posted by topsquark View Post
    I see now. Your T0 is not actually a period in the circuit, it's due to the resonant frequency of the circuit. (At least in Intro Physics we don't normally talk about the period of this frequency which is why I didn't understand what you were referring to.)

    I'm afraid I still can't give you an answer though. I've looked through my material (and a Google search) and can't find a formula for the attenuation. It must somehow be related to the damping factor (R/(2L)), but I don't know what the precise relationship is.

    -Dan
    Did you look at the following formula where delta is attenuation and beeta is damping factor:
    Attached Thumbnails Attached Thumbnails Resistance and attenuation-dekrement.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by totalnewbie View Post
    Did you look at the following formula where delta is attenuation and beeta is damping factor:
    I can honestly say I've never seen that equation before. (I don't claim any great knowledge about circuits.)

    So it looks to me that you have two equations in two unknowns:
    \delta = \left ( \frac{R}{2L} \right ) T

    T = \frac{2 \pi}{ \sqrt{\frac{1}{LC} - \frac{R^2}{4L}}}

    From the first equation
    T = \frac{2 \delta L}{R}

    So inserting this into the second equation:
    \frac{2 \delta L}{R} = \frac{2 \pi}{ \sqrt{\frac{1}{LC} - \frac{R^2}{4L}}}

    Now solve for R.
    \frac{ \delta L}{R} = \frac{ \pi}{ \sqrt{\frac{1}{LC} - \frac{R^2}{4L}}}

    \frac{R}{ \delta L} = \frac{ \sqrt{\frac{1}{LC} - \frac{R^2}{4L}}}{ \pi}

    \pi R = \delta L \sqrt{\frac{1}{LC} - \frac{R^2}{4L}}

     \pi ^2 R^2 = \delta ^2 L^2 \left ( \frac{1}{LC} - \frac{R^2}{4L} \right )

     \pi ^2 R^2 = \frac{\delta ^2 L^2}{LC} - \frac{\delta ^2 L^2 R^2}{4L}

     \pi ^2 R^2 = \frac{\delta ^2 L}{C} - \frac{\delta ^2 L}{4}R^2

    \pi ^2 R^2 + \frac{\delta ^2 L}{4}R^2 = \frac{\delta ^2 L}{C}

    \left ( \pi ^2 + \frac{\delta ^2 L}{4} \right ) R^2 = \frac{\delta ^2 L}{C}

    R^2 = \frac{\delta ^2 L}{\left ( \pi ^2 + \frac{\delta ^2 L}{4} \right ) C}

    R = \sqrt{ \frac{\delta ^2 L}{\left ( \pi ^2 + \frac{\delta ^2 L}{4} \right ) C}}

    I get R = 313.036 \, \Omega

    Thus
    T = \frac{2 \delta L}{R} = 50.581 \mu s

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2005
    Posts
    187
    The formulas you are using lead to the correct answer which is 3,4 Ooms. You got the resistance over 300. You have a calculation mistake somewhere.
    Anyway, thank you very much for help!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by totalnewbie View Post
    The formulas you are using lead to the correct answer which is 3,4 Ooms. You got the resistance over 300. You have a calculation mistake somewhere.
    Anyway, thank you very much for help!
    Never trust me on the easy stuff.

    Sorry about that!

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Shunt Resistance.
    Posted in the Math Topics Forum
    Replies: 9
    Last Post: March 13th 2011, 03:14 PM
  2. circuit resistance
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 29th 2010, 10:06 PM
  3. air resistance
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 28th 2010, 02:57 AM
  4. electrical resistance
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 13th 2008, 11:40 PM
  5. Resistance at certain lux
    Posted in the Pre-Calculus Forum
    Replies: 14
    Last Post: October 12th 2007, 01:51 PM

Search Tags


/mathhelpforum @mathhelpforum