# Where is the RLC circuit time constant"2L/R" coming from?

1. Nov 12, 2016

### garylau

Sorry
i dont understand where is this "2L/R"coming from
and also i dont know what is the function of this" time constant"(it is meaningless in physical sens?e)

in the note it states"that the time constant for reaching the steady state here...."

how did he knows this is the time constant that lead to the steady state?

thank you​

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2. Nov 12, 2016

### tech99

This is a resonant circuit consisting of C and L together with some losses represented by R.
The oscillations gradually diminish, in an exponential manner, similar to the discharge of a capacitor. Eventually the amplitude falls almost to zero. The time constant is the period of time after which the amplitude has fallen to 1/e.

3. Nov 12, 2016

### LvW

Unfortunately, you didnt show us neither the circuit diagram nor the corresponding transfer function.
Nevertheless, from the transient response one can deduce how the circuit will look like.
It is a simple RLC- lowpass.

The transfer function is H(s)=(1/sC)/(R+sL+1/sC)=1/(1+sRC+s²LC)
The poles of this function can be found by settimg the denominator to zero: 1+sRC+s²LC=0.
Result (two real poles):
s1,2=-R/2L(+-)SQRT(R²/4L²-1/LC).

Knowing that
(1) the denominator of the transfer function (frequency domain) is identical to the characteristic polynominal (time domain) and
(2) that, therefore, the the real part of the pole (-R/2L) appears in the solution of the differential equation as exp(-R/2L)t,

we can derive that the damping coefficient for the decaying oscillation is σ=R/2L.
If we compare this result with the general expression exp(-t/τ) we get τ=2L/R.

4. Nov 12, 2016

### garylau

Oh Sorry

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5. Nov 12, 2016

### LvW

Also for your circuit arrangement, the result of my calculation example still applies.
The reason is as follows: The current through the whole circuit (your task) has the the same form (in the time domain) as the voltage across the capacitor (my example). That means, the damping properties are identical - and hence the time constant.

6. Nov 12, 2016

### garylau

Oh i See

but when i try to put "2L/R" into the equation
i didn't see anything special?
there is no well-defined "amplitude" in this case?
because there are two terms
De^(Rt/2L) * Sin(omega*t)

if i try to put t=2L/R then it will become (63.4%)*D*sin(omega* 2L/R)

it seems (63.4%)* D = Amplitude of that D which the power is dissipated 50%?

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7. Nov 12, 2016

### garylau

However i got confused here
there is no well-defined definition of amplitude in this case as the sin function terms never allow the function reaches "the new amplitude" value?

8. Nov 12, 2016

### LvW

garylau - what now is your problem? Please, try a clear question or problem description.

9. Nov 12, 2016

### garylau

Does RLC circuit has well-defined time constant?
In the case of RL and Rc circuit,it is clearly that there is time constant as there is only "e" terms in the equation.
but in the case of RLC circuit,the whole equation include both "e" and "sin" terms and it behaves strange.
the sin terms never allow the the function reaches its "new Amplitude" value except time=0
So it looks different from the case of LC and RC cases,is it meaningless to understand what is "time constant in RLC circuit"?
from a website
somebody said "I think that in a RLC circuit one doesn't define time constant. There has two expressions with dimensional unit: time, but that is not time constant.
T=1/((-R/2L)±SQRT((R/2L)^2-1/(LC)))"

10. Nov 12, 2016

### cnh1995

RC and RL circuits are called 'first order circuits' and their behaviour is analysed using a defined 'time constant'.
An RLC circuit is a second order circuit and its behaviour can be analysed using parameters like rise time, peak time, delay time, damping factor, natural frequency of oscillation etc. (look up step response of a second order circuit). I have not seen the concept of time constant being used directly in case of 2nd order circuits.

Last edited: Nov 12, 2016
11. Nov 13, 2016

### LvW

Cnh1995 - didnt you read my contribution in post#3 ?
In a 2nd-order sysytem the time constant defines the slope of the envelope of the decaying oscillation.
Hence, it has a direct relation to the settling time.