Laplace transforms are virtually always covered later, usually in a second-semester circuits course. In the first-semester course, you are just given a technique, usually using terms like "phasor analysis" that magically works. Sometimes you are given some hand-wavy explanation as to why it works.

The situation is actually only a little better when you get to transform methods in the second semester. At least there the curtain is pulled back far enough to reveal that what you are doing is solving differential equations using Laplace transforms, but for most students, they have never seen Laplace transforms and why they work is never covered -- heck, the why isn't even covered in a differential equations class, which is where they are introduced in the math curriculum. To see the why, you need to take a course in math physics, or possibly a complex variables course. Very, very few students ever do that. This has always struck me as an almost unique instance in which a mathematical concept is not developed first prior to being applied, but I have to admit that understanding the 'why' requires such a deeper math grounding than the application does that there's pretty much no alternative, given how powerful and useful transform methods are, whether you understand why they work or not. For most topics, the 'why' and the 'what' are close enough conceptually that the 'why' can be used to push the math grounding along, but this just isn't the case for Laplace transforms.

As for the mistake you caught in my work earlier, you are correct. My eyes are getting bad enough that I missed the negative sign on the -105° when I set that up. That would have resulted in my check not working out, so it would have been caught. To make tracking down the error easier, I should have included the more direct set up in the Latex equations I posted (which I did on the hand-written notes I made). So it should have been:

\(
I_C \; = \; \frac{V_1 - V_2}{-j2 \; \Omega} \; = \; \frac{10 \; \angle 30^\circ \; V \; - \; 10 \sqrt{2} \; \angle 105^\circ \; V}{-j2 \; \Omega} \; = \; \frac{\left( 8.660 + j5.0 \right) \; V \; - \; \left( -3.660 + j13.660 \right) \; V}{2 \; \angle -90^\circ \; \Omega} \; = \; \frac{ \left( 12.320 - j8.660\right) \; V}{2 \; \angle -90^\circ \; \Omega} \; = \; \frac{15.060 \; \angle -35.104^\circ \; V}{2 \; \angle -90^\circ \; \Omega} \; = \; 7.530 \; \angle 54.894^\circ \; A
\)

When you said that both components of V2 should be negative, I looked at my notes and said, "Ah, no. Only the real part is negative." So then I looked at your solution and immediately saw that I was missing that minus sign.

Making that correction should result in (using 'align' to make things better organized):

\(
\begin{align}
I_C \; & = \; \frac{V_1 - V_2}{-j2 \; \Omega} \\
& = \; \frac{10 \; \angle 30^\circ \; V \; - \; 10 \sqrt{2} \; \angle -105^\circ \; V}{-j2 \; \Omega} \\
& = \; \frac{\left( 8.660 + j5.0 \right) \; V \; - \; \left( -3.660 - j13.660 \right) \; V}{2 \; \angle -90^\circ \; \Omega} \\
& = \; \frac{ \left( 12.320 + j18.660\right) \; V}{2 \; \angle -90^\circ \; \Omega} \; \\ & = \; \frac{22.361 \; \angle 56.565^\circ \; V}{2 \; \angle -90^\circ \; \Omega} \\
& = \; 11.180 \; \angle 146.565^\circ \; A
\end{align}
\)

I'm not going to pursue this any further, but it looks like your calculation for Ix works out.

Im so grateful for the effort and energy you put into your detailled answer, concerning the laplace transform i studied it alone to solve PDE/ODE, but i had never ever thought i could apply it to circuit, but after what you said and @MrAI 's insightful answer, a circuit specially in pure AC , embodies a differential equation for it to be solved. outside of this, so my answer/solution is a valid response right?Infinite thanks.

 

Great insight on oscillators but dont worry i will make effort to study them on my own, to not waste your time, well the point of this problem is ONLY finding the voltage V2, so from what i've conclude from your generous replies, the inductor doesnt care about the input's frequency so it will have a frequency of its own (definition of oscillator)? but why is AC analysis is needed though? isnt basically circuit analysis techniques enough?

Oh no problem. Follow WBahn's notes for the AC stuff.

From WBahn's notes on the initial conditions, I found there actually is one case where we can get just the sinusoidal output with the same frequency as the input frequency.
For one example, when w=1 and L=1 and C=1/2, if we set i0 to 10amps and v0 to 10 volts, we get just one frequency output.
i0 is the inductor current, and v0 is the capacitor voltage, both at start time. That's because in the full time solution when we set those values we have amplitude multipliers of the 'other' frequency of:
(10-v0)
and
(i0-10)
and when they both go to zero, the amplitude of that 'other' frequency goes to zero, so no output from that.

Yes, you can use differential equations if you want the solution to include initial conditions, or Laplace Transforms if you don't, or you know how to convert Laplace Transforms to include initial conditions.

 

Hi,

Yes, an oscillator is a device that develops it's own frequency like a sine wave although not limited to that.
Now if you power an oscillator made to be an oscillator with a DC voltage, it produces an output that may be that sine wave, and it will be a frequency dependent on the values like R, C, and L. That's with a DC voltage.
If you power it with an AC voltage of a different frequency, you will get that output frequency and also the frequency of the oscillator itself, so there are two frequencies, and because an AC analysis depends on all the frequencies being the same, an AC analysis will not be able to provide the correct result, because an AC analysis will produce only one frequency at the output.

You may not have learned this yet, but an ideal capacitor and ideal inductor connected together in various ways becomes an oscillator when at least some energy is introduced to the circuit. That means the L and C can produce a frequency dependent on their values, and it is often:
w=sqrt(1/(L*C))
and w=2*pi*f.

Now if you power it with a sine sin(w2*t) that will be a second frequency and so both frequencies may appear at the output.
It depends a lot on the damping of the LC network. If there is enough damping you won't get that second frequency, but without that you do get that second frequency. We could go over damping factors if you like and how that affects a network.

This is why doing a time domain analysis shows two frequencies, one is the frequency of the input sine and the other is the resonate frequency of the LC combination.

R1 does not contribute to the damping because it is always powered directly by the source, and never takes any energy away from the LC part of it. If R1 was in series with C or even in parallel to L, it would take energy from the LC part and that could provide enough damping to stop he LC part from oscillating on it's own.

The output will look like a sine wave modulated by another sine wave of a different frequency, although it may just look erratic when the frequencies are close to each other.
In the Laplace domain analysis, we end up with two parts in the denominator of the form s^2+w^2 and one w is one frequency and the other w is another frequency. In the time domain, we see two parts that are cosine and two parts that are sine, and one sine and one cosine can be combined into one sine or cosine with a phase shift, but then we still end up with two sinusoidal parts with phase shifts. The two parts are of different frequencies.

See attachment for a rough idea what this would look like.

There is one little catch to all this though.
If the instructor or the book this problem came from does not know that the output will be two frequencies then the result may be assumed to be a regular AC output. You'd have to find out if that is the case. If the problem was constructed using an AC analysis to check it, it may have not been checked in the time domain so they think the right result is that AC analysis only.

hey there! i think you might have seen that the resistor is in parallel to the inductor, thus the resistor is contributing to the LC oscillator no?

 

My main point though was that doing a pure AC analysis seems to lead to results that are nothing like what we would really see. That's why I was talking about this. I was surprised because I did not look at the drawing carefully enough the first time. When I saw and R, L, and C, I did not notice that the only R was not really part of an "RLC" circuit.

The word "really" is the adverb form of "real", with "real" referring to the real world. In the real world even if somehow everything could be ideal (it wouldn't be the real world, would it?), if the (ideal) inductor is carrying any AC current, there will be losses via radiation. So even if the wires connecting the components are truly zero ohms, there would be a very small radiation resistance in the loop containing the inductor, say 1 nanoohm, and that will eventually eliminate any transient, leaving only the steady state solution.

There's always something to mess up "ideal", isn't there?

 

hey there! i think you might have seen that the resistor is in parallel to the inductor, thus the resistor is contributing to the LC oscillator no?

Hi,

Well the resistor is not in parallel with the inductor.
The input source has zero impedance, so all the resistor does is draw current and act as a sensor for the current source.
If there was a resistor in parallel to the inductor really, it would dampen the oscillation.
I'll try to get back here later with more kind of busy today.

 

Hi,

Well the resistor is not in parallel with the inductor.
The input source has zero impedance, so all the resistor does is draw current and act as a sensor for the current source.
If there was a resistor in parallel to the inductor really, it would dampen the oscillation.
I'll try to get back here later with more kind of busy today.

hey MrAi thank you so much for your time i appreciate it, well the resistor and capacitor and the inductor is forming a triangle shape so it isnt in parallel right? well why isnt the resistor not contributing to the dampening isnt part of the LC oscillator? tysm!!

 

hey MrAi thank you so much for your time i appreciate it, well the resistor and capacitor and the inductor is forming a triangle shape so it isnt in parallel right? well why isnt the resistor not contributing to the dampening isnt part of the LC oscillator? tysm!!

Here's an easy way to determine if components are in series or parallel.

If whatever current flows in one of them MUST flow in the other, then they are in series.

If whatever voltage appears across one of them MUST appear across the other, then they are in parallel.

In this case, you can see that a current that flows in the resistor doesn't have to flow in the capacitor, so they are not in series. Similarly, whatever voltage appears across the resistor doesn't have to appear across the capacitor or the inductor, so they aren't in parallel.

 

hey MrAi thank you so much for your time i appreciate it, well the resistor and capacitor and the inductor is forming a triangle shape so it isnt in parallel right? well why isnt the resistor not contributing to the dampening isnt part of the LC oscillator? tysm!!

Hi again,

If you want to skip the math that's ok for now the other parts here should help also.

There are a couple hard and fast rules for determining if something is in parallel or in series as follows.

In a series circuit, the current is the SAME for every element in series,
In a parallel circuit, the voltage is the SAME for every element in parallel.

For this circuit you can see that the resistor voltage would be different than the voltage across the inductor so it is not in parallel. In order for it to be in parallel the capacitor would have to be shorted out, and that of course would alter the circuit completely.

In this circuit it's a little trickier though because the resistor is right across the source so you have to think a little and do a complete analysis to figure out what is going on. When you do that, you find that the only energy the resistor gets is from the source alone. That means it never takes any energy away from the LC part of the circuit. A resistor will only take energy from the LC part when it is part of the LC network itself.
It may be hard to see this at first though I also missed it with my first look.
In theory, when there is no damping for the LC network (no resistance to eat up energy) the LC network keeps oscillating, where the energy flows from the cap to the inductor and from the inductor to the cap repeatedly, and that's why we see an oscillation basicially. When there is a resistor involved, some of the energy from the cap and inductor gets dissipated in the resistor so the oscillations will gradually diminish unless more energy is constantly added. This can be explained with a little bit of math.
As you know, the expression:
V=sin(w*t)
is a sinusoidal wave, and never ceases to end. Add a exponential factor and we see this instead:
V=e^(-a*t)*sin(w*t)
and as 't' gets larger and larger, the exponential part gets smaller and smaller, and thus the voltage 'V' gets smaller and smaller. It tends to zero as 't' tends toward infinity.

If we look at the example of an RLC series circuit being driven by a constant voltage (step) source, there are three different solutions and with the wildest of them we see the current is:
I=(2*sqrt(C)*E*e^(-(t*R)/(2*L))*sin((t*sqrt(4*L-C*R^2))/(2*sqrt(C)*L)))/sqrt(4*L-C*R^2)

and there you can see that exponential part. Because of that exponential part, as 't' goes toward infinity that part goes to zero so we get:
I=0

To see this in a simpler way, we set R=1, E=1, C=1, L=1 and the solution reduces to:
I=(2*e^(-t/2)*sin((sqrt(3)*t)/2))/sqrt(3)

and this is the same thing. As 't' gets larger the current 'I' gets smaller, and tends to zero.

Now if we set R=0, it's different because now there is no damping. The solution then becomes:
I=(sqrt(C)*E*sin(t/(sqrt(C)*sqrt(L))))/sqrt(L)

and here we see no exponential part so the oscillation never dies. That's just an L and C in series driven by a voltage source and that is what we see in this circuit (without the current source).
This is a highly theoretical result however, in that ANY small resistance would eat up some of the energy and thus the oscillation would eventually stop if we did not put more energy into the circuit. Even R=0.001 would eat up some of the energy so the trade of energy between the L and C would be less and less until it was all absorbed by the resistor as 't' goes toward infinity.

If you have any more questions I'd be happy to help you understand this better. Keep in mind though it does take a little time to understand these circuit. To understand as much as possible, do an analysis where you find out what all the currents and voltages are doing over time. This could be done partly in a simulation as that will tell you a lot. Plot the currents and voltages and compare them, and it also helps to plot the energy in the inductor and capacitor. You can see the energy being transferred back and forth between the L and C.
BTW the energy in the inductor is (1/2)*L*i^2 and in the cap it is (1/2)*C*v^2. This means you can plot both of these and watch the energy waveforms change as they go to zero and then back up and then back down and back up, etc.

Analysis is the key to all these circuits and the more you do the more you realize that there is more that you can do to understand better and better. That's why the first step is really to learn a general analysis method and learn how to use a simulator that other people here use so everyone can converse easily. In the meantime members here are happy to answer any questions.

 

Hello again,

A question about series and parallel circuits that comes up from time to time on the internet in electronical forums involves a very simple circuit consisting of just two components such as a battery and a resistor, where the battery is providing energy to the resistor and the resistor is dissipating that energy as heat.
In this example, the resistor is connected directly across the battery and there are no other parts in the circuit. The question that comes up is, "Is this a series or parallel circuit?"
Since the voltage is the same across both elements, AND the current is the same through both elements, it may be possible to call it both series and parallel. In this simple case though, it is so very simple that it probably does not matter that much what we call it. It's the more complex circuits that usually need more attention to details like these.

 

Hello again,

A question about series and parallel circuits that comes up from time to time on the internet in electronical forums involves a very simple circuit consisting of just two components such as a battery and a resistor, where the battery is providing energy to the resistor and the resistor is dissipating that energy as heat.
In this example, the resistor is connected directly across the battery and there are no other parts in the circuit. The question that comes up is, "Is this a series or parallel circuit?"
Since the voltage is the same across both elements, AND the current is the same through both elements, it may be possible to call it both series and parallel. In this simple case though, it is so very simple that it probably does not matter that much what we call it. It's the more complex circuits that usually need more attention to details like these.

why isnt it in series? and no the voltage isnt the same across the resistor

 

In this example, the resistor is connected directly across the battery and there are no other parts in the circuit. The question that comes up is, "Is this a series or parallel circuit?"
Since the voltage is the same across both elements, AND the current is the same through both elements, it may be possible to call it both series and parallel.

I cannot agree. The question "Is this a series or parallel circuit?" is not a valid one and it sounds confusing to me.
This is because - I think - there are no "parallel circuits". The terms "parallel" or "series" are applied to electronic parts only which consumes power. The number of parts must >2. Otherwise, the term "parallel" or "in series" is not applicable.

 

why isnt it in series? and no the voltage isnt the same across the resistor View attachment 310468

Why don't you think they are in series?

Does the same current that flows in the supply not also flow in the resistor?

The voltage drop you are showing is a drop across a piece of wire, which is normally taken to be zero. Yes, in the real world, unless it is a superconductor there will be some voltage drop (and in certain situations it might be large enough to be important), but in most situations it is sufficiently negligible to be ignored (i.e., taken to be identically zero).

 

I cannot agree. The question "Is this a series or parallel circuit?" is not a valid one and it sounds confusing to me.
This is because - I think - there are no "parallel circuits". The terms "parallel" or "series" are applied to electronic parts only which consumes power. The numer of parts must >2. Othewise. the term "parallel" or "in series" is not applicable.

So it makes no sense to talk about placing two batteries in parallel or in series?

 

So it makes no sense to talk about placing two batteries in parallel or in series?

OK - I agree. I should reword my comment.
It is better to say that some parts or devices (active and/or passiv) can be combined "in parallel" resp. "in series" instead of saying " Is this a series or parallel circuit ?"

 

why isnt it in series? and no the voltage isnt the same across the resistor View attachment 310468

Hi,

That's a good question because it highlights the difference between theory and practice with circuit drawings.

In circuit drawings, for a purely theoretical analysis we assume that everything is clearly and completely shown. In this case that would mean that the 'wires' that connect the two parts have zero resistance. If we add resistance to the wire, then we no longer have just two elements and so the question of if they are in series or parallel becomes a question of three elements not just two as in the original question.
In practice, we may decided that parasitic elements play an important role and so we have to include them in the circuit either explicitly or just in a more casual perusing of the drawing. In this example, if the current is very low we may realize that adding wire resistance will have little effect, but with a higher current the wire resistance starts to drop some significant voltage so we are forced to add that to the circuit.

For the theoretical question of if two elements are in series or parallel though, we keep it at just two elements, otherwise there is no question about it also as you noted.

There is one more caveat to this.
If we dig very deep into theory, we could say that there is usually no such thing as series or parallel, because there are emissions from each element that are interacting with other things in space. For example, magnetic coupling, electrostatic coupling. With lumped circuit elements we don't do that though, and that means we ignore the size of the components versus the wavelengths involved, as well as any interactions. If there are interactions or the size is comparable to the wavelength, then we are forced to dig deeper into the theory of electromagnetics if we want useable results.

 

I cannot agree. The question "Is this a series or parallel circuit?" is not a valid one and it sounds confusing to me.
This is because - I think - there are no "parallel circuits". The terms "parallel" or "series" are applied to electronic parts only which consumes power. The number of parts must >2. Otherwise, the term "parallel" or "in series" is not applicable.

I am not sure what you mean.
If I connect a battery in 'parallel' to a capacitor, you have a good idea what I mean, and there is no dissipation in theory if we consider, as we must, zero resistance in the connections.

This question comes up from time to time so it's not my wording either. Maybe it's just one of those more playful questions just to poke at pure theory to see what others are thinking.

 

OK - I agree. I should reword my comment.
It is better to say that some parts or devices (active and/or passiv) can be combined "in parallel" resp. "in series" instead of saying " Is this a series or parallel circuit ?"

In general, I agree. The phrase "series circuit" (or "parallel circuit") would only be appropriate for the special cases in which ALL components in the circuit are in series (or parallel).

I think the imprecision comes in because the early circuits that students work with when learning these concepts often fit that description. So they get a misimpression early on that carries forward stubbornly. This isn't the only one, by a long shot. Another related one is that when they first learn about series and parallel connections, they work with circuits that can be reduced by a sequence of combining series and parallel elements (usually just resistors) until they have a single effective resistance. So they get the notion that ANY circuit can be simplified this way, even after they are introduced to circuits in which this can't be done.