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RLC circuits and phase angle

  • Thread starter brollysan
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  • #1
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Homework Statement


I am given a RLC circuit in series with a fully charged C. There is no applied voltage at this part of the problem.

Using kirchhoff I arrive at a second order diff.eq which leads me to the (correct) solution for the charge in the capacitor: q(t) = Q0exp(-t/T)cos(wt + Ø)

Now I am to derive an expression for the current in the circuit.

I am also given I(0) = 0, and q(0) = 1. T and W are known. The problem here is finding I(t) for the circuit, I have no clue what to do with the phase angle Ø.


Homework Equations



Unknowns are Q0 and Ø

The Attempt at a Solution


I know that since all elements are in series the current is equal everywhere, and that I(t) = dq/dt.

I know that if the circuit is driven the current/voltage is out of phase over the RLC. But this is not driven, what happens to the phase? If someone could clarify the physics of the phase in a driven and undriven RLC I would greatly appreciate it.

I can make the math work but the expression for I(t) is looking very with a cos(wt + atan(ugly expression)).
 

Answers and Replies

  • #2
berkeman
Mentor
57,451
7,469

Homework Statement


I am given a RLC circuit in series with a fully charged C. There is no applied voltage at this part of the problem.

Using kirchhoff I arrive at a second order diff.eq which leads me to the (correct) solution for the charge in the capacitor: q(t) = Q0exp(-t/T)cos(wt + Ø)

Now I am to derive an expression for the current in the circuit.

I am also given I(0) = 0, and q(0) = 1. T and W are known. The problem here is finding I(t) for the circuit, I have no clue what to do with the phase angle Ø.


Homework Equations



Unknowns are Q0 and Ø

The Attempt at a Solution


I know that since all elements are in series the current is equal everywhere, and that I(t) = dq/dt.

I know that if the circuit is driven the current/voltage is out of phase over the RLC. But this is not driven, what happens to the phase? If someone could clarify the physics of the phase in a driven and undriven RLC I would greatly appreciate it.

I can make the math work but the expression for I(t) is looking very with a cos(wt + atan(ugly expression)).
I think you need one or more of the following equations:

[tex]I(t) = C \frac{dV_C(t)}{dt}[/tex] current through a capacitor

[tex]V_L(t) = L \frac{dI(t)}{dt}[/tex] voltage across an inductor
 
  • #3
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0
Well

With T = 2L/R and W = sqrt [(R/2L)^2 - 1/LC]

Since q(t) = Q0exp(-t/T)cos(wt + Ø), and q(0) = 1 then =>

a: 1 = Q0cos(Ø)

then dq/dt = I(t) and I(t=0) = 0 =>

b: dq/dt = Q0[(-1/T)*exp(-t/T)*cos(wt +Ø) - wexp(-t/T)sin(wt +Ø)] which implies:

0 = (-1/T)*cos(Ø) - w*sin(Ø) => Ø = atan(-1/wT)

No we have a solution for Q0 and Ø, so we should have a known expression for I(t) but the problem is that if you insert for Q0 and Ø in the equation for I(t) you get something very ugly and something dependant on the phase Ø.

Now my problem is that although this makes mathematic sense is this correct in a physical since? I am to later in the assigment compare this with a numerical solution with L = 1, C = 1 og R = 0.2 (dimensionless) but you can already see that the phase will become imaginery for the analytic solution.

Edit: Actually atan doesn't make sense for an imaginery number..

What does that mean? If the current/voltage is out of phase over a component in the circuit, is it also out of phase in the wiring in between? I(t) is the same everywhere in a series so if they are out of phase in the wiring this implies that the voltage is also constant?

There is no added voltage to the circuit btw, it is just a charged capacitor.
 
Last edited:
  • #4
berkeman
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Current and voltage are out of phase at the reactive components L & C, per the differential equations in my previous post. I and V are in phase at the resistor R.
 
  • #5
ehild
Homework Helper
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a: 1 = Q0cos(Ø)

Ø = atan(-1/wT)
You got these two equations for Q0 and Ø, and the expression for I=dq/dt

I =Q0[(-1/T)*exp(-t/T)*cos(wt +Ø) - wexp(-t/T)sin(wt +Ø)].

You know that Ø=-atan(1/(wT)), that is tan(Ø)=-1/(wT). Use the relation between cosine and tangent to get cos(Ø):

cos(Ø)=√[1/(1+tan2(Ø))]

ehild
 

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