A basic circuit calculation problem

[bittware]

Hello all,
These days, I am thinking about a basic circuit calculation problem which just involves a switch, DC voltage source,C,L,R. I can't figure out how to get the current i vs. time relationship. Could anyone help me solve it or give some hints.
Thanks in advance
Best regards,
bittware

 

[eng_Semi]

1. if switch is connected to 1:
i = 0, and C is charging in a very small time till it reach the voltage equal to the supply voltage.

2. if switch is connected to 2:
The capacitor acts as the source for the other circuit and it will discharge with time, i.e. its voltage will be Vc = Vo e^(-t/τ); where τ is the circuit time constant, which can be calculated easily

By applying KVL u may get the relation u want

 

[VVV]

You need to do a transient analysis.
The easiest way is to use the Laplace transform to get the complex impedances and transient power sources, easily calculate the Laplace image of the current, I9s) and finally apply the inverse Laplace transform to get the current as a function of time.

Replace the inductor by its complex impedance, sL, the capacitor by its complex impedance, 1/sC. The resistor's impedance is still its resistance, R.
Now, the cap has been charged up to the input voltage, so consider a power source in series with the cap, of magnitude Ui/s. There was initially no current flowing through the inductor, so you need not add anything.

With these transformations, the current after flipping the switch can be calculated simply as:
I(s)=Ui/s/(sL+1/sC+R)=Ui/(s^2L+1/C+sR)

A few more algebraic manipulations can be performed, to make the function look closer to a Laplace transform whose inverse can be easily obtained from a Laplace transform table: K*ω^2/(s^2+2ζωs+ω^2)

Using that table inverse, you get the current as a function of time.
It will be, of course, sinusoidal, with a frequency of 1/(2*π*√LC), decaying exponentially to zero with a time constant given by L/R (if I recall correctly).

Hope this helps.
If not , I will post the full solution as I have more time than now.

 

[bittware]

You need to do a transient analysis.
The easiest way is to use the Laplace transform to get the complex impedances and transient power sources, easily calculate the Laplace image of the current, I9s) and finally apply the inverse Laplace transform to get the current as a function of time.

Replace the inductor by its complex impedance, sL, the capacitor by its complex impedance, 1/sC. The resistor's impedance is still its resistance, R.
Now, the cap has been charged up to the input voltage, so consider a power source in series with the cap, of magnitude Ui/s. There was initially no current flowing through the inductor, so you need not add anything.

With these transformations, the current after flipping the switch can be calculated simply as:
I(s)=Ui/s/(sL+1/sC+R)=Ui/(s^2L+1/C+sR)

A few more algebraic manipulations can be performed, to make the function look closer to a Laplace transform whose inverse can be easily obtained from a Laplace transform table: K*ω^2/(s^2+2ζωs+ω^2)

Using that table inverse, you get the current as a function of time.
It will be, of course, sinusoidal, with a frequency of 1/(2*π*√LC), decaying exponentially to zero with a time constant given by L/R (if I recall correctly).

Hope this helps.
If not , I will post the full solution as I have more time than now.

Hello VVV,
Nice to see you again.
I can't understand where the "magnitude Ui/s" comes up? Any rules to be applied?
Thanks for your explanation.
Best regards,
bittware

 

[boot_strap]

u should write the kvl:
v +Cdv/dt+Ld2v/dt2=0
v(0-)=Ui

an alternative solution can also be done using Laplace tran.

 

[tara1]

h**p://en.wikipedia.org/wiki/RLC_circuit

 

[bittware]

h**p://en.wikipedia.org/wiki/RLC_circuit

It's a good reference. Thanks, tara1!

 

[ms_talebi]

Hi,
Simply solve this ODE,
v+RCdv/dt+LCd2v/dt2=0
v(0-)=Ui
dv/dt(0-)=Ui/RC\]

 

[bittware]

Hi,
Simply solve this ODE,
v+RCdv/dt+LCd2v/dt2=0
v(0-)=Ui
dv/dt(0-)=Ui/RC\]

How do you get "dv/dt(0-)=Ui/RC"?

 

[VVV]

OK, I got a little bit of time now, so here is my explanation.

When doing a transient analysis, you replace the caps and inductors by their complex impedances: sL for inductors, 1/sC for capacitors. You leave the resistors alone, R. The justification for the above is simple:
Applying the Laplace transform, the differentiation means multiplying by s, integrating means division by s:

1. The voltage across an inductor is uL=L*di/dt. Thus, UL(s)=L*sI(s) Since we define impedance as U/I, it follows that UL(s)/I(s)=sL
2. The voltage across the cap is uC=(1/C)∫i(t)dt. Thus, UC(s)=(1/C)*I(s)/s. Hence, UC(s)/I(s)=1/(sC)

When doing a transient analysis, the voltage sources must be replaced by their Laplace images. For DCsources, they behave just like voltage steps, hence their images are U*(1/s)=U/s. Other sources are replaced with their Laplace image, calculated depending on their behaviour.

Voltages across capacitors are considered as step voltages and you include in the circuit a DC voltage source in series with each charged cap, with the correct polarity, with the magnitude: Uo(s)=Uc(0-)/s
For inductors, the current flowing through them before the transition moment, will generate a n EMF. This is included in the circuit as a voltage source in series with the inductor, with the polarity such that the "+" of the source is corresponds to the tip of the "arrow" that represents the inductor current. The magnitude is Uo=L*IL(0-).

With these modifications, your circuit will be an RLC series circuit, having a Ui/s source in series with the cap, since the cap was previously charged up to Ui.

Now, write Ohm's law (for other circuits, Kirchhoff's laws can be applied, too) and you get:

I(s)=Ui/s/(sL+1/sC+R)=Ui/(s^2*L+1/C+sR)=(Ui/L)*1/(s^2+sR/L+1/LC).
Arrange the second factor to look like: 1/(s^2+2ζωn+ωn^2)

With the notation ωn=1/(√(LC)), we have the last term. ωn is called the natural frequency of the circuit. If we did not have anything else in the circuit, it would resonate at f=1/(2π√(LC)). This should be a familiar formula.

Then if follows that 2ζωn=sR/L, so ζ=R/(2*√(L/C))
The parameter √(L/C) is called the characteristic impedance., probably familiar, too.

The quantity ζ is called the damping factor, or the damping ratio.
With these notations, the equation can be written:
I(s)=(Ui/L)*1/(s^2+2ζωn+ωn^2).

The first factor is a constant, so we only need to worry about the second paranthesis. Its inverse Laplace transform is, from a table:
(1/(ωn)*√(1-ζ^2))*exp(-ζωnt)*sin(ωn*√(1-ζ^2)t)

With the notation, ωd, called the damped frequency, we can write:
i(t)=Ui/L*(1/ωd)*exp(-ζωnt)*sin(ωd)t)

As you can see, the current is sinusoidal, decaying exponentially to zero with the time constant 1/(ζωn)

For ζ<1, you can plot the function directly.
I generally use Excel for this type of job. Fig. 1 shows a picture of the current, for ζ=0.2. I used ζωn=1 for simplicity only. The sine wave has a frequency lower than ωn and its amplitude always stays within the exponentially decaying curves.

Most textbooks stop here and they do not tell you HOW to get the rest of the plots.

For ζ=1, if you try to plot the function directly you run into trouble, since the denominator becomes zero. But, consider sin(ωd)t)/(ωd). In the limit, when ωd->0 (for ζ->1), you will find that the limit is t. Just apply l'Hopital's rule. Then the equation becomes i(t)=(Ui/L)*t*exp(-ζωnt)
This is called critically dampled, when ζ=1, that is, when R=2*√(L/C).
A picture of the response is in Fig.2. The peak is 1/e=0.37.

The last case is ζ>1.
In this case, the quantities √(1-ζ^2) become imaginary. To plot the response, rewrite: √(1-ζ^2)=j*√(ζ^2-1).
Then, remember that sinx=(exp(jx)-exp(-jx))/2j.
With that, the equation becomes:
i(t)=Ui/L*(1/j*√(ζ^2-1)*ωn)*exp(-ζωnt)*((exp(j*j*√(ζ^2-1)*ωn)-exp(-j*j*√(ζ^2-1)*ωn))/(2j)

i(t)=(Ui/L)*(1/-2(√(ζ^2-1))*exp(-ζωnt)*((exp(-√(ζ^2-1)*ωn)-exp(√(ζ^2-1)*ωn))

This is the function I plotted in Fig. 3 for ζ=2 (unless I made some typing errors).
The major difference between this picture and Fig. 2 is the lower amplitude and lower frequency.

Note that in my plots I considered Ui/L=1 and ωn=1, for simplicity.

The important things to remember are:
1. the current decays exponentially with the time constant ζωn
2. the frequency is lower than ωn by the factor √(1-ζ^2)
3. the critically damped response is obtained when R=2*Zo, where Zo=√(L/C), the characteristic impedance.
4. For overdamped circuits, there are no oscillations.

As you can see, the circuit is simple, but its analysis is not. You will probably have more respect now for the simulation programs. I have a lot of respect for Sir Isaac Newton and Monsieur Laplace.
You can simulate the circuit, to check the validity of my explanation.

Note that if the circuit does not include BOTH capacitors and inductors, the analysis is simpler.

VVV