# Math Help - Resistance and attenuation

1. ## 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.

2. Originally Posted by totalnewbie
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

3. $
T_0=2*Pi*\sqrt{(LC)}
$

$
T=(2*Pi)/\sqrt{((1/LC)-(R^2/4L))}
$

These T gives different values and dividing both, for example $
T_0/T
$
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: $logaritmic decrement=attenuation*T$

4. Originally Posted by totalnewbie
$
T_0=2*Pi*\sqrt{(LC)}
$

$
T=(2*Pi)/\sqrt{((1/LC)-(R^2/4L))}
$

These T gives different values and dividing both, for example $
T_0/T
$
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

5. Originally Posted by topsquark
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:

6. Originally Posted by totalnewbie
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

7. 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!

8. Originally Posted by totalnewbie
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.