Solving a series LCR circuit via two methods

[fayled]

Given a series LCR circuit with a driving voltage V=V₀cos(ωt), would it be possible to obtain a solution for I(t) by the two methods listed below?
1. Summing the voltages, i.e LdI/dt+IR+Q/C=V₀cos(ωt) and solving the DE.
2. Using I=V/Z where Z is the total complex impedance and solving for I in terms of known values.
I ask this because this is something I have been trying to do. However I keep obtaining different solutions for the phase (although quite similar) which are
tanø=Rω/(-Lω²+1/C) using method 1. and tanø=(ωL-1/ωC/R) by method 2. with the current lagging the voltage in each case by ø.

I will be able to post further working if it is possible but I thought writing it all down now may be a waste of time if the method is flawed. Thanks.

 

[gneill]

The differential equation approach will give you the transient response and the steady state response. The impedance method (or more properly the phasor approach) will yield the steady state response alone.

Note that the cosine function starts at cos(0) = 1 for time t = 0, so that the circuit is immediately "hit" with an impulse of magnitude Vo which leads to the transient part. This will die away after some amount of time that depends upon the time constants of the circuit (damping). The phasor method ignores the transient part of the solution and yields only the steady state response.

 

[fayled]

The differential equation approach will give you the transient response and the steady state response. The impedance method (or more properly the phasor approach) will yield the steady state response alone.

Note that the cosine function starts at cos(0) = 1 for time t = 0, so that the circuit is immediately "hit" with an impulse of magnitude Vo which leads to the transient part. This will die away after some amount of time that depends upon the time constants of the circuit (damping). The phasor method ignores the transient part of the solution and yields only the steady state response.

This makes sense. However, even ignoring the transient solution via the DE method, the steady state solutions disagree. The amplitudes agree which gives me hope, but not the phases...

It will probably be best if I post my working which I will do tomorrow.

 

[fayled]

The differential equation approach will give you the transient response and the steady state response. The impedance method (or more properly the phasor approach) will yield the steady state response alone.

Note that the cosine function starts at cos(0) = 1 for time t = 0, so that the circuit is immediately "hit" with an impulse of magnitude Vo which leads to the transient part. This will die away after some amount of time that depends upon the time constants of the circuit (damping). The phasor method ignores the transient part of the solution and yields only the steady state response.

So I find the steady state solution via the two methods.

For the differential equation method:
LdI/dt+IR+Q/C=V₀cosωt
Ld²I/dt²+RdI/dt+I/C=-ωV₀sinωt
The steady state solution comes from the particular integral. Instead solve for the PI of
Ld²I/dt²+RdI/dt+I/C=-ωV₀e^iωt
and then take the imaginary part of the solution.

Try I=Ae^iωt, then -Lω²A+ARωi+A/C=-ωV₀. This means that A=-ωV₀/(1/C-Lω²+Rωi)=ωV₀/Be^iø where B=√[(1/C-Lω²)²+(Rω)²] and tanø=Rω/(1/C-Lω²).

Then the solution for the current has amplitude, ωV₀/√[(1/C-Lω²)²+(Rω)²] and lags the driving voltage by a phase ø.

For the complex impedance method.
Z_TOT=R+iωL+1/iωC=√[R²+(ωL-1/ωC)²]e^iθ where tanθ=(ωL-1/ωC)/R. Then I=V₀e^iωt/Z_TOT giving a current I with amplitude V₀/√[R²+(ωL-1/ωC)²] and lagging the driving voltage by a phase θ. The amplitudes match but the phases do not.

 

[fayled]

The differential equation approach will give you the transient response and the steady state response. The impedance method (or more properly the phasor approach) will yield the steady state response alone.

Note that the cosine function starts at cos(0) = 1 for time t = 0, so that the circuit is immediately "hit" with an impulse of magnitude Vo which leads to the transient part. This will die away after some amount of time that depends upon the time constants of the circuit (damping). The phasor method ignores the transient part of the solution and yields only the steady state response.

So I find the steady state solution via the two methods.

For the differential equation method:
LdI/dt+IR+Q/C=V₀cosωt
Ld²I/dt²+RdI/dt+I/C=-ωV₀sinωt
The steady state solution comes from the particular integral. Instead solve for the PI of
Ld²I/dt²+RdI/dt+I/C=-ωV₀e^iωt
and then take the imaginary part of the solution.

Try I=Ae^iωt, then -Lω²A+ARωi+A/C=-ωV₀. This means that A=-ωV₀/(1/C-Lω²+Rωi)=ωV₀/Be^iø where B=√[(1/C-Lω²)²+(Rω)²] and tanø=Rω/(1/C-Lω²).

Then the solution for the current has amplitude, ωV₀/√[(1/C-Lω²)²+(Rω)²] and lags the driving voltage by a phase ø.

For the complex impedance method.
Z_TOT=R+iωL+1/iωC=√[R²+(ωL-1/ωC)²]e^iθ where tanθ=(ωL-1/ωC)/R. Then I=V₀e^iωt/Z_TOT giving a current I with amplitude V₀/√[R²+(ωL-1/ωC)²] and lagging the driving voltage by a phase θ. The amplitudes match but the phases do not although they are similar.

 

[ehild]

So I find the steady state solution via the two methods.

For the differential equation method:
LdI/dt+IR+Q/C=V₀cosωt
Ld²I/dt²+RdI/dt+I/C=-ωV₀sinωt
The steady state solution comes from the particular integral. Instead solve for the PI of
Ld²I/dt²+RdI/dt+I/C=-ωV₀e^iωt
and then take the imaginary part of the solution.

Try I=Ae^iωt,

I=Ae^iωt is not solution of the equation Ld²I/dt²+RdI/dt+I/C=-ωV₀sinωt.

Do not mix real and complex forms. Either solve the equation with V_oe^iωt on the right side, or find the steady-state solution in the form of A(sinωt+θ).

ehild

 

[fayled]

I=Ae^iωt is not solution of the equation Ld²I/dt²+RdI/dt+I/C=-ωV₀sinωt.

Do not mix real and complex forms. Either solve the equation with V_oe^iωt on the right side, or find the steady-state solution in the form of A(sinωt+θ).

ehild

But I'm pretty sure Im(Ae^iωt) is a solution to the differential equation as I said above...

 

[ehild]

It is, but then you have to equate Im(Ae^iωt) with -ωVosin(wt). What is the imaginary part of Ae^iωt?

ehild

 

[fayled]

It is, but then you have to equate Im(Ae^iωt) with -ωVosin(wt). What is the imaginary part of Ae^iωt?

ehild

Asinωt. Why do I need to equate them?

 

[ehild]

Asinωt. Why do I need to equate them?

No, it is not the imaginary part of Ae^iωt. Note that A is complex.

ehild

 

[fayled]

No, it is not the imaginary part of Ae^iωt. Note that A is complex.

ehild

Oh sorry, it is ωV₀sin(ωt-ø)/B where B=√[(1/C-Lω2)2+(Rω)2] and tanø=Rω/(1/C-Lω2). I didn't bother explicitly taking the imaginary part above because I wanted to highlight the different phase solutions and didn't really care about the whole solution...

 

[ehild]

You get the same phase difference between current and voltage if you apply both methods correctly.

ehild

 

[fayled]

You get the same phase difference between current and voltage if you apply both methods correctly.

ehild

Well I can't apply them correctly. I can't see how anything you've suggested affects the outcome :(

 

[ehild]

The problem is that you mix the real and complex formalism.
The equation is real: LdI/dt+IR+Q/C=Vocos(ωt). You want the steady-state current. It has the same frequency as the diving voltage, but it has some phase constant with respect to the voltage.

So you want the solution in form I=Acos(ωt+θ).
You take the derivative of the original equation:

LI'' + RI'+I/C=-ωVosin(ωt)
multiply by C, and substitute the derivatives:

-LCω²Acos(ωt+θ)-RCωAsin(ωt+θ)+Acos(ωt+θ)=-ωCVosin(ωt)--->A[cos(ωt+θ)(1-ω²LC)-RCωsin(ωt+θ)]=-ωCVosin(ωt)

Expand cos(ωt+θ) and sin(ωt+θ):
cos(ωt+θ)=cos(ωt)cos(θ)-sin(ωt)sin(θ)
sin(ωt+θ)=sin(ωt)cos(θ)+cos(ωt)sin(θ).

Substitute:

A[(1-ω²LC)(cos(ωt)cos(θ)-sin(ωt)sin(θ))-RCω(sin(ωt)cos(θ)+cos(ωt)sin(θ))]=-ωCVosin(ωt)

Collect the cos(ωt) and sin(ωt) terms. As cos(ωt) and sin(ωt) are independent, the equation must hold separately for both.

cos(ωt)A{(1-ω²LC)cos(θ)-RCωsin(θ)}=0
sin(ωt)A{-(1-ω²LC)sin(θ)-RCωcos(θ)}=-ωCVosin(ωt)

θ is obtained from the first equation : (1-ω²LC)cos(θ)-RCωsin(θ)=0

$$θ=\arctan\left(\frac{1-ω^2 LC}{RCω}\right)$$

With the impedance method, we work with complex voltages and currents in the form V=Voe^iωt, I=Io e^i(ωt+θ). I=V/Z, the phase constant of I is the negative of the phase ψ of the impedance. Z=i(ωL-1/(ωC))+R, its phase is

$$ψ=\arctan\left(\frac{ωL-\frac{1}{ωC}}{R}\right)=\arctan\left(\frac{ω^2LC-1}{RωC}\right)$$

The phase of I is negative of that

$$θ=\arctan\left(\frac{1-ω^2LC}{RωC}\right)$$

ehild