How to Simplify and Sketch RLC Circuit Analysis?

In summary, when w is low, the circuit appears as a short-circuit, so the load is basically R1 to ground. When w is high, the capacitor appears as a short-circuit, so again the load is basically R1 to ground. However, when w is high, the inductor appears as a short-circuit, so again the load is basically R1 to ground. Your expression for current magnitude is correct, but when it comes to the phase I just couldn't sketch the whole thing. I simulated the circuit and the phase is just too complicated! Unfortunately, the first equations are incorrect.
  • #1
Lunat1c
66
0
Consider the circuit show below, where [tex] V1 = V_{peak}sin(\omega t - 36^\circ) [/tex]

http://img195.imageshack.us/img195/9687/rlc.png

I wish to find an expression for I in the time domain and in phasor form and then sketch the magnitude/phase vs frequency plots. I already managed to get a solution but it seems way too complex to be able to sketch the above mentioned plots from it. So I'm just going to list my solution and perhaps someone will be kind enough to suggest other ways to tackle this problem (if there are any).

to find i(t):
[tex] V1(t) = i(t)R_1 + i(t)\bigg[2 \pi fL || \frac{1}{2 \pi fC} || R_2\bigg] [/tex]

But [itex] f = \frac{\omega}{2 \pi} [/itex]

Hence, [itex] V1(t) = i(t)R_1 + i(t)\bigg[\omega L || \frac{1}{\omega C} || R_2\bigg] = i(t)\bigg[R_1 + \frac{\omega LR_2}{R_2 + \omega^2 LCR_2 + \omega L}\bigg] [/itex]

After substituting for V1 and simplifying, I ended up with:

[tex] i(t) = \frac{\omega^2 LCR_2 + \omega L + R_2}{\omega^2 LCR_1R_2 + \omega L(R_1+R_2) + R_1R_2}V_{peak} sin(\omega t - 36^\circ) [/tex]

Which seems ok.

Now, for the phasor part:
[tex] v(j \omega) = i(j \omega)R_1 + i(j \omega)\bigg[j\omega L || \frac{1}{j\omega C} || R_2\bigg] = i(j \omega)R_1 + i(j\omega)\bigg[\frac{1}{\frac{1}{j \omega L} + j \omega C + \frac{1}{R_2}}\bigg] = i(j\omega)\bigg(R_1 + \frac{j\omega LR_2}{R_2 - \omega^2 LCR_2 + j\omega L}\bigg) = i(j \omega)\bigg(\frac{R_1R_2 + j \omega LR_1 - \omega^2 LCR_1R_2 + j\omega LR_2}{R_2 + j \omega L - \omega^2 LCR_2}\bigg)[/tex]

Hence, [tex] i(j \omega) = \frac{R_2(1 - \omega^2 LC) + j \omega L}{R_1R_2 + j \omega L (R_1+R_2) - \omega^2 LCR_1R_2}v(j \omega) [/tex]

And now I proceded to finding [itex] |\bar{I}| [/itex] and [itex] \phi [/itex]

Note that [itex] V1 = V_{peak}sin(\omega t - 36^\circ)= V_{peak} \angle{-36^\circ} [/itex]

[tex] \bar{I} = \frac{\sqrt{R_2^2(1 - \omega^2 LC)^2 + (\omega L)^2}}{\sqrt{R_1^2R_2^2(1 - LC \omega^2)^2 + (\omega L)^2(R_1 + R_2)^2}} V_{peak} [/tex]
and
[tex] \phi = tan^{-1}\frac{\omega L}{R_2(1 - \omega^2 LC)} - 36^\circ - tan^{-1}\frac{\omega L(R_1+R_2)}{R_1R_2 - \omega^2 LCR_1R_2} [/tex]

How would one go about sketching this without the help of any software package? There must be another way to solve this problem which is way simpler. Any ideas? Any help will be greatly appreciated
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
In your expression for current, set w to a low value (e.g., w=0) to allow you to determine the current at low frequencies. Set w to a high value (e.g., w -> infinity) to determine the current at high frequencies. The current will reach a maximum or minimum where the circuit is resonant, i.e., close to w = 1/ sqrt (LC).

Looking at your circuit, when w is low the inductor appears as a short-circuit, so the load is basically R1 to ground. When w is high, the capacitor appears as a short-circuit, so again the load is basically R1 to ground. Your expression for i should support these figures. (It's an easy check.)
 
  • #3
That is honestly what I did for the current magnitude. However, when it came to the phase I just couldn't sketch the whole thing. I simulated the circuit and the phase is just too complicated!
 
  • #4
Unfortunately, the first equations are incorrect. You can't compute the impedance of a resistive and reactive component in parallel by pretending they are both purely resistive or reactive. You have to use complex numbers thruout.

Look at your expression for i(t). The phase of i with respect to V is independent of w! You know that's not right.

BTW when you see parallel components it's much easier to use admittance, susceptance and conductance than their impedance duals. So for example I can readily write
Y = G2 + jwC + 1/jwL for the parallel network, where G2 = 1/R2. Then Z = 1/Y and i = V/(Z + R1)

(BTW giving you V at a lagging angle of 36 degrees makes little sense. I would ignore it until the very end, then with your i(t) expression you can just subtract 36 deg for the phase angle. Which I think you've already done. That number has no meaning in the context of your schematic unless there is in addition some external time reference defining phase = 0.)

When you started using complex numbers that part looks OK though I haven't checked your math. If done right your complex expression for current is what you use for getting i(t): compute i(jw) = i1 + j i2, then
i(t) = √[i1^2 + i2^2]sin(wt + φ), φ = arc tan i2/i1. Minus 36 deg of course.

The frequency response graph is made complicated by the second-order LC network. As another poster suggested, the parallel circuit looks like infinite impedance at w^2 = 1/LC so your graph of |i|will start flat at i = V/R1, droop down to i = V/(R1 + R2) at
w = 1/√LC, then pick up & flatten again to V/R1 at high frequencies. I have to confess I forgot the rules on constructing the asymptotes for a resonant network, but you should be able to dig them up somewhere. They involve the damping coefficient ζ and "Q" of the i/V transfer function, both of which are just functions of your component values. Hopefully some kind and informed soul will help us?

Graph log |i| on the y-axis and log w or log f on the x axis.

For the phase, graph phase in degrees on the y-axis and log w or log f on the x axis. Keep track of the signs of i1 and i2. Example: arc tan (-1/2) is not the same angle as arc tan (1/-2.) All logs are base 10.
 
  • #5
Forget the edit. Should be OK as is.
 
Last edited:
  • #6
rude man said:
Forget the edit. Should be OK as is.
I think you should indicate more clearly exactly what the reader should "forget"? Are you saying the reader should disregard your entire post where you pointed out [apparently] glaring errors in his analysis? If so, perhaps it might be possible to request an administrator delete your posts in this thread.
 
  • #7
NascentOxygen said:
I think you should indicate more clearly exactly what the reader should "forget"? Are you saying the reader should disregard your entire post where you pointed out [apparently] glaring errors in his analysis? If so, perhaps it might be possible to request an administrator delete your posts in this thread.

What I meant was that my post of 4:40 a.m. on 9/21/11 is OK. Sorry.
 
  • #8
rude man said:
What I meant was that my post of 4:40 a.m. on 9/21/11 is OK. Sorry.

So you stand by your post stating
Look at your expression for i(t). The phase of i with respect to V is independent of w!

In that case I'm now confused. The OP's post appears not to have been edited subsequent to your criticism, yet in it he provides a formula for phase that involves both w and w2, just as one would expect.
 
  • #9
nascent - yes, his expression for i(t) contains w and w^2 but that is the amplitude, not the phase. His phase is a constant 36 degrees! Phase will obviously vary with w since there are reactive components.

And if we disregard the 36 degrees, which as I say is a nonsensical spec to begin with, then he has 0 phase shift between V and i thruout 0 < w < ∞ !
 

Related to How to Simplify and Sketch RLC Circuit Analysis?

1. What is an RLC circuit in phasor form?

An RLC circuit is a type of electrical circuit that contains a resistor, inductor, and capacitor. It is represented in phasor form, which is a mathematical technique used to simplify the analysis of circuits that have sinusoidal inputs.

2. How is a phasor diagram used to represent an RLC circuit?

A phasor diagram is used to represent an RLC circuit by converting the sinusoidal voltage and current values into vectors. The length of the vector represents the magnitude of the voltage or current, and the angle of the vector represents the phase difference between the voltage and current. This allows for a visual representation of the circuit's behavior over time.

3. What is the difference between a series and parallel RLC circuit in phasor form?

In a series RLC circuit, the components (resistor, inductor, and capacitor) are connected in a single loop, while in a parallel RLC circuit, the components are connected in multiple branches. This results in different phasor diagrams, as the voltage and current values will be affected by the arrangement of the components.

4. How are impedance and admittance represented in phasor form for an RLC circuit?

Impedance is represented as a complex number in phasor form, with the real part representing the resistance and the imaginary part representing the reactance. Admittance, on the other hand, is represented as the inverse of impedance, with the real part representing the conductance and the imaginary part representing the susceptance.

5. What is the significance of the resonance frequency in an RLC circuit in phasor form?

The resonance frequency is the frequency at which the inductive reactance and capacitive reactance cancel each other out, resulting in a purely resistive circuit. In phasor form, this can be represented by a phasor diagram where the voltage and current vectors are aligned, indicating maximum power transfer and minimum impedance.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
17
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
19
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
0
Views
928
  • Introductory Physics Homework Help
Replies
3
Views
268
  • Introductory Physics Homework Help
Replies
3
Views
383
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
464
Back
Top