# How Long Does It Take for a High-Resistance RLC Circuit to Lose Half Its Energy?

• Nivlac2425
In summary, The solution for a simple RLC circuit with a large resistance is found to be t= (tdiss/2) ln(2) where tdiss is the energy dissipation time, equal to L/R. This is a shorter amount of time compared to the solution for small resistance, indicating that the energy drains faster in a circuit with a large resistance.
Nivlac2425

## Homework Statement

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)

## 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!

Nivlac2425 said:

## 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) , :

You lost a factor 2 from the denominator of ω

ehild

ehild said:
You lost a factor 2 from the denominator of ω

ehild

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!

Nivlac2425 said:
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.

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

ehild said:
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

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

## 1. What is a resistance in an RLC circuit?

Resistance in an RLC circuit refers to the opposition that an electrical component, such as a resistor, presents to the flow of current. It is measured in ohms and can affect the flow and magnitude of current in the circuit.

## 2. How does resistance affect an RLC circuit?

Resistance can affect an RLC circuit in several ways. It can limit the flow of current, causing a voltage drop across the circuit. It can also affect the resonant frequency of the circuit, as well as the amplitude and phase shift of the current.

## 3. How is resistance calculated in an RLC circuit?

Resistance in an RLC circuit can be calculated using Ohm's law, which states that resistance (R) is equal to the ratio of voltage (V) to current (I) in the circuit, or R = V/I. You can also calculate the equivalent resistance in a series or parallel RLC circuit using the appropriate formulas.

## 4. What are the effects of high and low resistance in an RLC circuit?

High resistance in an RLC circuit can result in a decrease in current flow and a voltage drop across the circuit. This can cause the circuit to operate at a lower frequency and with a smaller amplitude. Low resistance, on the other hand, can lead to a larger current flow and a smaller voltage drop, resulting in a higher frequency and larger amplitude in the circuit.

## 5. How can resistance be controlled in an RLC circuit?

Resistance in an RLC circuit can be controlled by using different components, such as resistors, in the circuit. By changing the value of the resistor, you can increase or decrease the overall resistance in the circuit. You can also use techniques like voltage division or current division to control the resistance in a circuit.

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