I LRC Series Circuit with an AC Source

[rtareen]

TL;DR Summary

The voltages across all components at any instant is equal to the voltage of the source at that instant. But we vectorially add the amplitudes of each component to find the source amplitude. Also, the source voltage is usually out of phase with the source current. We get the phase angle from the phasor diagram, but it doesn't make intuitive sense that the source voltage and source current are out of phase.

Attached is the section from the book. I am doing section 31.3

We know that an AC source gives a sinusoidal varying current, and as far as I know its always given by $i(t) = Icos(wt)$. Its like we take the current to be the base of all other quantities, so we use it to derive everything else. But nothing was explained as to how this current is generated by the AC source. I know one way where we can alternate the current by the rotating a conducting loop with a dc source in a magnetic field, but I doubt that's what this is. Just out of curiosity, would we be able to configure the exact same kind of LRC circuit described in this section using the rotating loop configuration?

Next we try to find a function for the ac source voltage as a function of time. But to do this we need the amplitude. Can somebody explain why we have to use phasors to find the amplitude?

Also, for equation 31.21, will this work to find the impedance if we have more than one resistor, or more than one inductor? I don't even want to think about what will happen if we have more than one capacitor.

Anyways, once we know the current amplitude and the impedance we have the ac source voltage amplitude $V$. So now we need to find the function for it. I assume we are using the cosine function because earlier in the chapter we defined the instantaneous current, resistor voltage, inductor voltage, and capaitor voltage to the projection of their phasors onto the horizontal axis. We can use the phasor diagram to solve for the phase angle. But why are the current, and voltage, which come from the same source, not in phase? What is the explanation?

 

[BvU]

Why not consider a simpler circuit with only a capacitor ?
Can you understand $i$ and $V$ are not in phase ?

 

[anorlunda]

I know one way where we can alternate the current by the rotating a conducting loop with a dc source in a magnetic field, but I doubt that's what this is.

That's exactly how almost all the AC power you buy from the power company is generated, so don't doubt it. We call it the synchronous generator.

But why are the current, and voltage, which come from the same source, not in phase?

That's easiest to understand if you know how to write the transient equations for inductors and capacitors.

The voltage across an inductor is $V=L\frac{dI}{dt}$ let's say that $I=\cos\omega t$, then $\frac{dI}{dt}=\sin\omega t$, which is equal to I shifted 90 degrees.

The current in a capacitor is $I=C\frac{dV}{dt}$. let's say that $V=\sin\omega t$ then $\frac{dV}{dt}=-\cos\omega t$, which is equal to V shifted 90 degrees.

So the answer to your question is that differentiation and integration of sinusoidal functions are the sources of phase shifts that you see in circuits.

You may learn a few more basics from this PF Insights article.
https://www.physicsforums.com/insights/ac-power-analysis-part-1-basics/

 

[rtareen]

Why not consider a simpler circuit with only a capacitor ?
Can you understand $i$ and $V$ are not in phase ?

We have $i(t) = Icos(wt)$
and we have $v_c(t) = \frac{I}{wC}cos(wt - \pi/2)$
So $v_c$ is behind i by 90 degrees.
Since the total voltages add to zero, we get $v_s(t) = -v_c(t) = -\frac{I}{wC}cos(wt - \pi/2)= \frac{I}{wC}cos(wt + \pi/2)$.

So the source voltage is not in phase with the current when there is a capacitor only.
Is this ok?

 

[rtareen]

I just realized that I attached the completely wrong pdf! I deleted the old one and attached the correct one.

I have a new question regarding figure 31.13. It looks like they are vectorially adding $V_L$ and $V_C$ instead of subtracting them in parts (b) and (c). Since they are in opposite directions when you subtract one from the other, the resultant should be longer. But in (b) the resultant is shorter than what it should be. There is a similar problem in (c).

 

[BvU]

They add the vectors . The magnitude of the result is $|V_L| - |V_C|$ .
The direction is that of $\vec V_L$ in 31.13(b).
The direction is that of $\vec V_C$ in 31.13(c).

 

[rtareen]

They add the vectors . The magnitude of the result is $|V_L| - |V_C|$ .
The direction is that of $\vec V_L$ in 31.13(b).
The direction is that of $\vec V_C$ in 31.13(c).

Thanks for clearing that up!

 

[vanhees71]

That looks very complicated. Are no complex numbers allowed or used?

 

[BvU]

That looks very complicated. Are no complex numbers allowed or used?

Book is at page 1030 and no complex numbers yet.

 

[vanhees71]

Sigh

 

[alan123hk]

I have a new question regarding figure 31.13. It looks like they are vectorially adding VL and VC instead of subtracting them in parts (b) and (c). Since they are in opposite directions when you subtract one from the other, the resultant should be longer. But in (b) the resultant is shorter than what it should be. There is a similar problem in (c).

There should be no problem with those expressions and diagrams in the attachment.

Since V_L and V_c are in opposite directions in the phase diagram, we can simply assign V_L as the positive direction and V_c as the negative direction, that is, +V_L and -V_c , so the potential difference between the series LC circuit, V_LC , is equal to sum of them (+V_L - V_c).

If their sum is positive, then the direction of V_LC will of course follow the V_L, otherwise it will of course follow V_C.

The phasor diagrams can help visualize the relationship between different phase voltages and currents, but for more complex situations, it is more convenient to use complex number algebra.