# Homework Help: Resistance in RLC circuit

1. Apr 22, 2012

### Nivlac2425

1. The problem statement, all variables and given/known data
Consider a simple RLC circuit. If R >> (4L)/C , compute approximately how long it takes for the circuit to lose half of the initial energy stored in the system.

(There is no circuit drawing given; the problem assumes a general RLC circuit)
2. Relevant equations

3. The attempt at a solution
I have already worked out the differential equation that is associated with RLC circuits, and got an expression for

ω=[iR±sqrt(4L/C - R2)]/(2L)

and since R >> (4L)/C ,

ω= 0, (iR)/L

We consider the non-trivial case, ω= (iR)/L
Then it turns out that

Q=Aoexp(-Rt/L) , and

dQ/dt = I = -R/L Aoexp(-Rt/L)

From these, I solved for the total energy stored in the RLC system (capacitor energy + inductor energy) and got:

Etot= Ao2 [(L+CR2)/2LC] exp(-2Rt/L)

Then at this point, I'm sort of stuck and not sure where to go. Although I can derive the average power in a cycle by:

P= Etot/tdiss = Ao2(R/L)[(L+CR2)/2LC]exp(-2Rt/L)
where tdiss is the energy dissipation time, equal to L/R

From here, I can integrate the power expression to come up with a relation between the energy and initial energy,

E= (Eo/2)exp(-2t/tdiss)

And then, finally attempting to answer the question, I set E=Eo/2 and get t=0.

Somehow this doesn't make much sense to me and makes me believe that I did something very wrong in my work above. In my class lecture notes, a similar derivation was done assuming that the resistance was small (R << (4L)/C), but now the resistance is huge.

From inspection, I can see that the current through the circuit must be very small and that perhaps the capacitor is discharging very slowly as well. But what does this mean about the energy dissipation? Assuming that my derivation above is correct (it is probably not), the circuit can only store half of the total energy predicted from the Etot expression.

Can anyone make some sense of this?
Thanks!!

2. Apr 23, 2012

### ehild

You lost a factor 2 from the denominator of ω

ehild

3. Apr 23, 2012

### Nivlac2425

I believe it was

ω=(iR±iR)/2L, (after using R>>(4L)/C

ω=2iR/2L

ω=iR/L

Correct me if I'm wrong with this

Thanks!

4. Apr 23, 2012

### ehild

You are right, Q(t)=Aoexp(-Rt/L)
The formula you got for the total energy at time t is also correct:

Etot= Ao2 [(L+CR2)/2LC] exp(-2Rt/L). The initial stored energy is Etot(0)=Ao2 isn't it? The question is at what time becomes the energy half of the initial value.

ehild

5. Apr 23, 2012

### Nivlac2425

At t=0, Etot(0)= Ao2 [(L+CR2)/2LC] ?
I thought this would be true since exp(-2Rt/L) is the exponential decay term and the "coefficient" would be the initial value. Setting t=0 also makes the exp term become 1.

Assuming this is true, I could set Etot= Eo/2 where Eo is the intial energy, and solve for time t.

I would get

Ao2/2 [(L+CR2)/2LC] = Ao2 [(L+CR2)/2LC] exp(-2Rt/L)

and t= (L/2R)ln(2) , or t= (tdiss/2) ln(2) when tdiss=L/R

Comparing this to the solution for small resistance, it takes a shorter amount of time to drain the energy to half when we have a large resistance. This somehow makes more sense than my previous solution