Resistor, Resonance Frequency And Damping Factor

[EEE]

Recently I did a project where I studied the Electriacl Resonance frequency. Now, the resonance frequency depends on the value of resistance and inductance only and there is no impact of resistance on it. Whereas the damping factor depends on all the three (R,L and C). Why is it so?

[billvon]

Recently I did a project where I studied the Electriacl Resonance frequency. Now, the resonance frequency depends on the value of resistance and inductance only and there is no impact of resistance on it. Whereas the damping factor depends on all the three (R,L and C). Why is it so?

I believe that frequency depends on CAPACITANCE and inductance.  A perfect resonant circuit would resonate forever, if nothing wasted the circulating energy, since ideal capacitors and inductors are lossless.  The thing that wastes energy (by turning it into heat) is the resistor.

[OceanBreeze]

Recently I did a project where I studied the Electriacl Resonance frequency. Now, the resonance frequency depends on the value of resistance and inductance only and there is no impact of resistance on it. Whereas the damping factor depends on all the three (R,L and C). Why is it so?

 

 

I assume you meant to write capacitance and inductance. . .

 

As Billvon has already pointed out, if there is no resistance at all, the circuit would oscillate forever, but there is always some resistance present as there are no perfect reactive components that have no resistance at all.

 

But your statement that resistance affects only damping and has no effect on frequency is not correct.

 

Most often you will see this equation for the resonant frequency: 

 

[math]{ F }_{ r }=\frac { 1 }{ 2\pi \sqrt { LC }  }[/math]

 

As you can see this equation does not take resistance into account. This does not mean that resistance has no effect on the resonant frequency, only that in many cases the Inductance and Capacitance have much greater effect and resistance can be ignored in those cases. So, this equation is not a general solution, only the particular solution when resistance is zero or at least negligible.

 

The general solution for resonant frequency that does include the resistance follows:

 

[math]{ F }_{ r }=\quad \frac { \sqrt { 4L/C\quad -\quad { R }^{ 2 } }  }{ 4\pi L }[/math]

 

As an exercise, try setting R = 0  in this second equation and see if it becomes the same as the first one.

 

 

[exchemist]

I assume you meant to write capacitance and inductance. . .

 

As Billvon has already pointed out, if there is no resistance at all, the circuit would oscillate forever, but there is always some resistance present as there are no perfect reactive components that have no resistance at all.

 

But your statement that resistance affects only damping and has no effect on frequency is not correct.

 

Most often you will see this equation for the resonant frequency: 

 

[math]{ F }_{ r }=\frac { 1 }{ 2\pi \sqrt { LC }  }[/math]

 

As you can see this equation does not take resistance into account. This does not mean that resistance has no effect on the resonant frequency, only that in many cases the Inductance and Capacitance have much greater effect and resistance can be ignored in those cases. So, this equation is not a general solution, only the particular solution when resistance is zero or at least negligible.

 

The general solution for resonant frequency that does include the resistance follows:

 

[math]{ F }_{ r }=\quad \frac { \sqrt { 4L/C\quad -\quad { R }^{ 2 } }  }{ 4\pi L }[/math]

 

As an exercise, try setting R = 0  in this second equation and see if it becomes the same as the first one.

 

 

Interesting. 

 

I don't remember any AC theory. Can you explain why damping depends on inductance and reactance? I'd naively have expected it to depend only on resistance.

[OceanBreeze]

Interesting. 

 

I don't remember any AC theory. Can you explain why damping depends on inductance and reactance? I'd naively have expected it to depend only on resistance.

 

 

I am glad you find it interesting, as do I! The physical laws that govern RLC circuits have precise equivalents in mechanical engineering, and I deal with this sort of problem on a continuous basis.

 

Ship vibrations are something that I am very alert to, and I have found the simplest way to analyze them is by using an equivalent electric circuit, a type of analog computer (although I admit to using digital computer programs as well but don’t trust them programmers dagnabbit)

 

OK, first of all, I didn’t actually say that L and C contribute to damping; I did say that R contributes to the determination of the resonance frequency. I am guessing that (with your alert mind) you read into that and realized there must be an inverse relationship? That is, if R contributes to the resonance, then L and C must contribute to the damping, right?

 

Well, that is true but it can do with a bit of clarification. The resistance does all of the damping, but the amount of damping also depends on the values of L and C.

 

That may sound like confusing, so maybe it is best to go to an algebraic example.

 

The differential equation for the usual RLC circuit is:

 

[math] L\frac { di }{ dt } +Ri+\frac { q }{ C } =E[/math]

 

Where E is some constant voltage

 

Differentiating this yields a second-order DE that is easier to solve:

 

[math]L\frac { { d }^{ 2 }i }{ d{ t }^{ 2 } } +R\frac { di }{ dt } +\frac { i }{ C } =0[/math]

 

Now we can use the auxiliary equation and make use of the Quadratic Formula:

 

[math]L{ m }^{ 2 }+Rm+\frac { i }{ C } =0[/math]

 

The roots are:

 

[math]m\quad =\quad \frac { -R\quad \pm \quad \sqrt { { R }^{ 2 }-4\frac { L }{ C }  }  }{ 2L }[/math]

 

And that brings us to the root of the matter!

 

If R² is less than 4L/C the roots are complex and the circuit will oscillate with some damping.

 

If R² = 4 L/C that is critical damping that will cause the current to drop to zero with no overshoot in minimum time. (This is the ideal I try to obtain to suppress ship vibrations)

 

If R²  is greater than the critical value there will be two real roots and the current will be in an exponential decay that in theory can take forever, but in practice is considered to be zero after 5 time constants.

 

As you can see, there is an interdependence between the value of R and L and C that determines the damping, and not just the value of R by itself.

Edited June 5, 2017 by OceanBreeze

[exchemist]

I am glad you find it interesting, as do I! The physical laws that govern RLC circuits have precise equivalents in mechanical engineering, and I deal with this sort of problem on a continuous basis.

 

Ship vibrations are something that I am very alert to, and I have found the simplest way to analyze them is by using an equivalent electric circuit, a type of analog computer (although I admit to using digital computer programs as well but don’t trust them programmers dagnabbit)

 

OK, first of all, I didn’t actually say that L and C contribute to damping; I did say that R contributes to the determination of the resonance frequency. I am guessing that (with your alert mind) you read into that and realized there must be an inverse relationship? That is, if R contributes to the resonance, then L and C must contribute to the damping, right?

 

Well, that is true but it can do with a bit of clarification. The resistance does all of the damping, but the amount of damping also depends on the values of L and C.

 

That may sound like confusing, so maybe it is best to go to an algebraic example.

 

The differential equation for the usual RLC circuit is:

 

[math] L\frac { di }{ dt } +Ri+\frac { q }{ C } =E[/math]

 

Where E is some constant voltage

 

Differentiating this yields a second-order DE that is easier to solve:

 

[math]L\frac { { d }^{ 2 }i }{ d{ t }^{ 2 } } +R\frac { di }{ dt } +\frac { i }{ C } =0[/math]

 

Now we can use the auxiliary equation and make use of the Quadratic Formula:

 

[math]L{ m }^{ 2 }+Rm+\frac { i }{ C } =0[/math]

 

The roots are:

 

[math]m\quad =\quad \frac { -R\quad \pm \quad \sqrt { { R }^{ 2 }-4\frac { L }{ C }  }  }{ 2L }[/math]

 

And that brings us to the root of the matter!

 

If R² is less than 4L/C the roots are complex and the circuit will oscillate with some damping.

 

If R² = 4 L/C that is critical damping that will cause the current to drop to zero with no overshoot in minimum time. (This is the ideal I try to obtain to suppress ship vibrations)

 

If R²  is greater than the critical value there will be two real roots and the current will be in an exponential decay that in theory can take forever, but in practice is considered to be zero after 5 time constants.

 

As you can see, there is an interdependence between the value of R and L and C that determines the damping, and not just the value of R by itself.

It was the OP who said it depended on all 3. 

 

But your explanation is quite clear. Thinking of it in terms (if I may try to paraphrase you)  of L and C determining the intrinsic vigour of the oscillation tendency that R tries to quench helps me understand why all 3 are important. And yes the 3 scenarios for the roots of the quadratic make sense.  

[connor090]

I am not the expert on this field. But I have learnt a lot form your conversation.

Edited October 2, 2017 by connor090