# LCR circuit.

1. May 24, 2014

### albega

1. The problem statement, all variables and given/known data
I have the LCR circuit attached below.

At time t=0 the capacitor is uncharged and the switch is closed. By solving an appropriate
differential equation, show that the current through the resistor is oscillatory provided
L<4CR2. By considering the boundary conditions at t=0 and as t→∞, sketch the
form of this current as a function of time.

2. Relevant equations
V=IR, V=LdI/dt, V=Q/C.

3. The attempt at a solution
So the first bit is pretty simple, giving a DE of
LCRd2I/dt2+LdI/dt+RI=V0.

Solving for the transient complentary functions using the quadratic formula gives the required condition for an oscillatory current.

I can solve the DE to obtain
I=exp(-t/2CR)(Asinβt+Bcosβt)+(V0/R), where β=[√(4LCR2-L2)]/2LCR and we are assuming oscillatory solutions do exist as the question wants.

I have the initial condition Q=0 for the capacitor when t=0, so then I=0. Then B=-V0/R. However the condition as t→∞ is problematic. I expect I→Vo/R (the steady-state solution) as t→∞. If I let t tend to infinity in my solution, the CF vanishes, so it doesn't allow me to implement any sort of condition. I can only think of this meaning I can let A be zero but I don't think that would be ok.

Any clues would be great, thanks!

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2. May 24, 2014

### SammyS

Staff Emeritus
What is $\displaystyle\ \lim_{t\,\to\,\infty}e^{-t/(2RC)} \ ?$

3. May 24, 2014

### albega

Zero... As I said, that causes the CF, the transient response, to vanish.

However this isn't useful in finding the arbitrary constant A because it all just disappears.

4. May 24, 2014

### SammyS

Staff Emeritus
I = 0 initially because of the nature of inductors, not because Q = 0.

Q = 0 implies that at t = 0, the voltage drop across the capacitor is zero.

Take the derivative of your solution and apply a boundary condition to that at t = 0. That should give what A is.

5. May 25, 2014

### albega

I don't see how Q=0 at t=0 implies anything about dI/dt at t=0...

6. May 25, 2014

### ehild

The voltage across the resistor and capacitor is U=Vo-LdI/dt. At t=0 it is zero.
Note that the problem asks the current through the resistor, which is U/R.
Oscillatory means also damped oscillation.

ehild

7. May 25, 2014

### SammyS

Staff Emeritus
It wouldn't except that the inductor is involved.

Initially, t = 0, the voltage drop across the resistor/capacitor is zero, so you know that the voltage drop across the inductor is V0. But you also know how the voltage drop across the inductor is related to the current. Right?

Last edited: May 25, 2014
8. May 26, 2014

### albega

Yes, that gives dItotal/dt=V0/L at t=0. How can I relate that to I through the resistor? Also I notice we're dealing with t=0 here. What about t→∞ as the question suggests?

9. May 26, 2014

### SammyS

Staff Emeritus
Last question first:
Presumably the initial solution to the differential equation contained an additive constant, which was set to V0/R because of the behavior of I(t) as t→∞ .

How to find A:
You have an expression for dI/dt at t = 0: dI/dt=V0/L .

That's why I suggested evaluating the first derivative of your solution for I(t) at t = 0.

Take the derivative of $\displaystyle \ e^{-t/(2RC)}\left(A\sin(\beta t)+B\cos(\beta t) \right) +\frac{V_0}{R} \,,\$ evaluate that at t = 0 & set it equal to V0/L

10. May 26, 2014

### albega

Right, I'm fine with all that, but surely I have an expression for dItot/dt at t=0 and not dI/dt, because the current through the resistor is not the same as that through the inductor.

11. May 26, 2014

### SammyS

Staff Emeritus
What is I(t) ? Current through the source? ... Current through the inductor ? ... Current through the resistor/capacitor combination ?

Yes to all three.

Although it doesn't matter for the issue at hand: At t = 0, the voltage drop across the resistor is zero. Therefore, no current flows through the resistor.

12. May 26, 2014

### albega

Is there a mathematical way to show dItot/dt=dI/dt though? I can't find a way which is a little annoying.

13. May 26, 2014

### ehild

It is not true. You solved the differential equation for the total current I.

The current through the resistor is i = U/R where U is the voltage across the resistor.

The voltages add along the loop (Kirchhoff's loop rule) so the voltage across the inductor + voltage across the capacitor= voltage of the source. UL+U=U0.
The voltage across the inductor is proportional to the time derivative of the current. UL=LdI/dt.

So LdI/dt + U = U0 →U=U0-LdI/dt and the current through the resistor is i=U/R.

ehild

14. May 26, 2014

### albega

I solved it for the current through the resistor only.

Here:
Let the current through the inductor be I1+I2, that through the resistor I2 and that through the capacitor I1.

Going around the resistor capacitor loop,
Q1/C-I2R=0 and differentiating this gives I1=CRdI2/dt.
Going around the cell, inductor and capacitor loop,
V0-Ld(I1+I2)/dt-Q1/C=0.
V0-LdI1/dt-LdI2/dt-Q1/C=0

Substituting for I1 we have

LCRd2I2/dt2+LdI2/dt2+RI2=V0.

Then solve for the complementary function, assuming L<4CR2 for underdamped solutions. Then solve for the PI.

I obtain the general solution I2=exp(-t/2CR)(Asinβt+Bcosβt)+(V0/R), where β=[√(4LCR2-L2)]/2LCR.

15. May 26, 2014

### ehild

Sorry, I did not recognize it was the equation for the resistor current in the OP.
It is still valid that LdI(total)/dt+RI2=Vo.
To find the other constant, A, use that the current through the inductor (the total current) is zero at t=0.
I(total)=I1+I2, and I1=dQ/dt=d(UC)/dt=RCdI2/dt.

ehild

Last edited: May 26, 2014
16. May 26, 2014

### SammyS

Staff Emeritus
Thanks ehild for clearing up this issue!

@albega,
I apologize for misreading the question. You're in excellent hands now, with my friend ehild.

And of course, you were correct in stating that Q=0 @ t=0 → I2 = 0 at t=0 !

Placing subscripts on the currents sure does help clarify things!

Last edited: May 26, 2014
17. May 26, 2014

### albega

So my initial conditions are I2=dI2/dt=0 at t=0.

If so, thanks for clearing that up!

18. May 26, 2014

### albega

Ah sorry, I should have made it clearer by using subscripts from the start! That will teach me not to take shortcuts in the future...

19. May 26, 2014

### ehild

Yes, we figured it out at the end The problem text was confusing: t→∞ helped to find the particular solution.

ehild