Electrical oscillations of an RLC circuit

[JHans]

Electrical oscillations are initiated in a series circuit containing a capacitance C, inductance L, and resistance R.

a) If

R << \sqrt{\frac{4L}{C}}

what time interval elapses before the amplitude of the current oscillation falls to 50.0% of its initial value?

b)Over what time interval does the energy decrease to 50.0% of its initial value?For a), I recognize that current is the first derivative of the charge, which for an RLC circuit with small R is:

Q = Q_i e^\frac{-Rt}{2L} cos(\omega_d t)

Which would make the current:

I = \frac{-I_i e^(\frac{-R t}{2L})}{2L} (2Lw_d sin(w_d t) + cos(w_d t)

From here, I know that I'll set

I = 0.500 I_max

But then, how do I solve for t? My hunch is that the sine/cosine part of the equation just controls the oscillations, not the amplitude. In that case, I would disregard that part of the equation and solve for t using the exponential part of the equation. Is that a correct assumption, or am I doing something horribly wrong?

 

[ehild]

You are right, use only the exponential part.

ehild

 

[JHans]

Alright. I'm a bit confused, though. I've stated that the current is:

<br /> <br /> I = I_i \frac{e^(\frac{-R t}{2L})}{2L} (2L \omega_d sin(\omega_d t) + cos(\omega_d t))<br /> <br />

I could also have stated it as:

<br /> <br /> I = I_i e^(\frac{-R t}{2L}) (\omega_d sin(\omega_d t) + \frac{R}{2L} cos(\omega_d t))<br /> <br />

Which version of the exponential would I be worried? Specifically, I'm wondering if it's necessary to consider (1 / 2L) as part of the exponential, or if I can keep it with the sine/cosine terms and disregard it as only affecting the oscillations.

My thought is that I would follow through as:

<br /> <br /> I = 0.500 I_i<br /> <br />

<br /> <br /> 0.500 I_i = I_i e^(\frac{-R t}{2L})<br /> <br />

<br /> <br /> 0.500 = e^(\frac{-R t}{2L})<br /> <br />

<br /> <br /> ln(\frac{1}{2}) = \frac{-R t}{2L}<br /> <br />

<br /> <br /> ln(2) = \frac{R t}{2L}<br /> <br />

<br /> <br /> t = \frac{ln(4) L}{R}<br /> <br />

Is that correct?

 

[ehild]

The cosine and sine terms can be replaced with a single Acos(ωt+φ) term, and including A into the amplitude (I₀) you have the equation

I=I₀exp(-Rt/(2L))

for the time dependence of the amplitude of the current. So your derivation is correct.

ehild

.

 

[JHans]

I'm sorry, can you explain a bit further? The equation that I have is:

<br /> <br /> I = I_i e^(\frac{-R t}{2L}) (\omega_d sin(\omega_d t) + \frac{R}{2L} cos(\omega_d t))<br /> <br />

How is it that I can convert the sine and cosine terms to a single cosine term? Is there a trigonometric identity that I'm missing? And how would one establish A? Furthermore, would A be a necessary part of the solution to finding the time dependence of the current? Or is it irrelevant, just like the trigonometric term? It would be helpful if you explained a bit more. Thank you!

 

[JHans]

Also, I apologize. The term that comes after e should be an exponent. That is, it should be e^((-Rt)/(2L)).

 

[ehild]

Yes, it comes from the summation lows of trigonometry that the sum or difference of a sine and a cosine function of the same argument is equivalent with a single sine or cosine.

a sin(ωt) + b cos(ωt) = A sin (ωt +φ )

You can find A and φ first by applying the sum rule to the right-hand side then comparing the sinωt and cosωt terms:

A sin(ωt +φ )=A sinωt cosφ + A cosωt sinφ

The two sides must be equal at any time:

a sin(ωt) + b cos(ωt) = A sin(ωt) cosφ + A cos(ωt) sinφ

If sin(ωt)=0, cos(ωt) =1 and b=A sinφ,

if cos(ωt)=0 and sin(ωt)=1 : a=A cosφ.

so

A²=a²+b² and tanφ=b/a.

ehild