# Solving a series LCR circuit via two methods

1. May 6, 2014

### 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.

2. May 6, 2014

### Staff: Mentor

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.

3. May 6, 2014

### fayled

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.

4. May 7, 2014

### fayled

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=-ωV0eiωt
and then take the imaginary part of the solution.

Try I=Aeiωt, then -Lω2A+ARωi+A/C=-ωV0. This means that A=-ωV0/(1/C-Lω2+Rωi)=ωV0/Be 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.
ZTOT=R+iωL+1/iωC=√[R2+(ωL-1/ωC)2]e where tanθ=(ωL-1/ωC)/R. Then I=V0eiωt/ZTOT 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.

5. May 7, 2014

### fayled

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=-ωV0eiωt
and then take the imaginary part of the solution.

Try I=Aeiωt, then -Lω2A+ARωi+A/C=-ωV0. This means that A=-ωV0/(1/C-Lω2+Rωi)=ωV0/Be 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.
ZTOT=R+iωL+1/iωC=√[R2+(ωL-1/ωC)2]e where tanθ=(ωL-1/ωC)/R. Then I=V0eiωt/ZTOT 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.

6. May 7, 2014

### ehild

I=Aeiωt 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 Voeiωt on the right side, or find the steady-state solution in the form of A(sinωt+θ).

ehild

7. May 7, 2014

### fayled

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

8. May 7, 2014

### ehild

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

ehild

9. May 7, 2014

### fayled

Asinωt. Why do I need to equate them?

10. May 7, 2014

### ehild

No, it is not the imaginary part of Aeiωt. Note that A is complex.

ehild

11. May 7, 2014

### fayled

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...

12. May 7, 2014

### ehild

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

ehild

13. May 8, 2014

### fayled

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

14. May 8, 2014

### 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=Voeiωt, I=Io ei(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