# AC Circuit Analysis: RLC & Resonance. Need help!

Discussion in 'Homework Help' started by DTskkaii, Jan 30, 2012.

1. ### DTskkaii Thread Starter New Member

Jan 30, 2012
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0
Hi, I am currently struggling with this AC question that is required for my first year electrical engineering coursework. I know sort of how to go about it all, but am having a really difficult time implementing the correct calculations. Any help would be appreciated!

Context
You are given a simple AC LCR circuit (setup in that order, in series, depicted graphically as a square). The AC Voltage source in the circuit replicates the input energy from the sound waves hitting the glass. The inductor and capacitor replicate the resonance characteristics of the glass, and the resistor replicates the energy being converted by the glass as it vibrates. When the energy converted in the resistor is over 200J/s the glass can no longer support vibration and shatters.

Given Values
Vs = ?
L = 4H
C = 3 x 10^(-10)F
R = 50Ohms

Questions
(1) Find the resonant frequency (rad/sec and Hz) of the sheet of glass.
(2)Find the magnitude of the voltage across the resistor in the equivalent circuit, required to break the glass, at the resonant frequency.
(3) Find the magnitude of the voltage source in the equivalent circuit, required to break the glass at resonant frequency. Compare the last two values found.
(4) Using the voltages found above, find the gain (in dB, gain = 20log(Vout/Vin), where Vout is across the resistor) vs the frequency for the following points: 0.01 x w0, 0.1 x w0, w0, 10 x w0, 100 x w0 (where w0 is the resonant frequency.
(5) The clients plan to use that is only accurate to within +-20% of the desired frequency.
- How much sound power do they need to be sure to break the glass?
- At the worst case scenario (20% away from resonant frequency ) find the voltage magnitude and phase shift across each component; source, inductor, capacitor and resistor.

I am currently trying to determine the V(r) without being given the source voltage.. I haven't done it before without being given that variable.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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At resonance in a series LCR circuit what is the relationship between the source voltage and the resistor voltage drop?

3. ### DTskkaii Thread Starter New Member

Jan 30, 2012
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At resonance, XL and XC phase eachother out, so that VS and VR are equal.
I get that, I'm just having a brain block on how to take the step from having my total impedance (50Ohms as given) to finding I or V across the resistor.
I think it's really obvious and I'm just trying to overthink it or something silly :/

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
You are given the resistor power as 200J/s. Simple!

5. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
Alright so here's the approach i'm thinking is correct.. if you could check it that would be great thankyou!

Q2
Since it at resonant frequency in q 2 and 3, Zt = r = 50.
change V(t) to phasor = M<0 , where M = magnitude.
p = v^2/r
therefore 200 = (M<0)^2/50
10 000 = (M<0)^2
m=100
V(r) = 100cos(rez freq*t) --- since in resonance theres no phase shift.
Rez is [4594.407] from Q1
so V(r) = 17.14V

Q3
I(r) = V(r)/r
from part 1 Vr = 100<0
so Ir = 100<0 / 50 <0
= 2<0
since its a series circuit, currecnt is constant so Is = Ir = 2<0
so Vs = Is * Zt
= 2<0 * 50<0
= 100 < 0
= 100 cos (rez freq *t)
Rez is [4594.407] from Q1
so V(r) = 17.14V

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
As you have correctly determined - the answer in both cases is simply 100V rms.

Hence the peak voltage value or magnitude would be 100*√2 =141.4V.

The time domain sinusoidal voltage expression for the source would then be

Vs(t)=141.4sin(28,867.5*t) Volts

7. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
Thankyou so much!

We actually only learned about RMS this week, with the assignment halfway through and due at the end of the week, so weird :/

One last thing, how do I express Vs(t) as a cos function rather than sin,
I know for some reason the tutor wants it to be in cos.

Appreciate it so much.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Re: sine or cosine

In this case it's arbitrary - simply replace sin with cos

9. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
Sorry t_n_k, i'm just having a problem with that last part.

I understand that in resonance, the phase shift is zero, and hence like you said, the sin or cos is arbitrary.

However, when put into practice with the 2 equations
1) Vs(t)=141.4sin(28,867.5*t) Volts gives 79.24V
2) Vs(t)=141.4cos(28,867.5*t) Volts gives -117.11V

I made sure my calculator was in radians, but I still don't see why those equations should yield different results if that is the resonant frequency :/

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Unfortunately you appear to be missing the point about the idea of this sinusoidally excited circuit.

Firstly. You are confusing instantaneous values in the time domain with steady-state values in the frequency domain. This problem is solved in the steady-state frequency domain - unless you require a transient solution - which you don't.

Secondly. You are making a curious assumption about the time domain value of the source by evaluating the function at a time t=1second. Why are you doing that?

To solve parts (a) to (c) you really only need to know the steady-state rms values.

11. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
I was quite confused, thankyou for clarifying that.

If you could nudge me in the right direction with just one more thing I would be very grateful.

Going on to problem 4, starting say with w=0.01xw0;

finding the impedances:
ZL=jwl=j(288.675x[3x10^-10])
How exactly do I get the answer to that above equation in phasor form?
I believe i've been taught it, but I can't find an example in my workbook.

If you could run me through or maybe point me to a problem like this one, going from a frequency-impendance-power-current-voltage, it would be mindblowingly helpful, but if not thats okay

Thankyou again

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
What do you have in mind with respect to solving part (4)?

Outline your steps without concerning yourself with actual values and then we can discuss the problem further.

13. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
Alright so...
i need to do repeat calculations for 0.01xw0, 0.1xw0, w0 etc...
in order to find the voltages over the resistor for each different frequency, in order to then use the equation for gain = 20log(Vout/Vin) where Vin=Vs and Vout=voltage over resistor.

So, the part that has actual lengthy calculations is going from
-> get new frequency (0.01xw0)
-> use frequency to find impedances of parts in circuit in order to find total impendance ::: ZL=jwl=j(freq x L)=? , ZR=given , ZC=1/(-jwl)=1/(-j(freq x c)=?
:::ZT=Zr+Zl+Zc
From here, I need to find a formula to use to find the new current through the resistor incorporating the new impedance, correct?

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Consider the case of

We have

L = 4H
C = 3 x 10^(-10)F
R = 50 Ohms

XL=ωL=2886.75*4=11,547Ω
XC=1/(ωC)=1,154,701.1Ω

So the total impedance is

$Z_T=R+jX_L-jX_C=50+j11,547-j1,154,701.1 \ \Omega=50-j1,143,154.1\ \Omega$

This can be converted to polar form as

$Z_T=50-j1,143,154.1=1,143,154\angle{-1.57^o}$

From there the current is found as

$I_T=\frac{V_s}{Z_T}=\frac{141.4\angle{0^o}}{1,143,154\angle{-1.57^o}}=1.237E^{-4}\angle{+1.57^o}$

giving VR as

$V_R=50\ast I_T=50\ast 1.237E^{-4}\angle{+1.57^o}=6.185E^{-3}\angle{+1.57^o}$

and the output-to-input ratio of

$\frac{V_R}{V_s}=\frac{6.185E^{-3}\angle{+1.57^o}}{141.4\angle{0^o}}=4.374E^{-5}\angle{+1.57^o}$

and finally converting the absolute magnitude ratio to dB

$20log(\| \frac{V_{R}}{V_{s}} \|)=-87.2 \ dB$

There's a shortcut method which might be of interest if you can grasp the idea. It reduces the calculation workload significantly.

One can treat the output-to-input ratio as a complex voltage divider defined by the relationship

$\frac{V_{R}}{V_{s}}=\frac{R}{(Z_T)}=\frac{R}{(R+jX_L-jX_C)}$

$\frac{V_{R}}{V_{s}}=\frac{50}{1,143,154\angle{-1.57^o}}=4.374E^{-5}\angle{+1.57^o}$

and again

$20log(\| \frac{V_{R}}{V_{s}} \|)=-87.2 \ dB$

Last edited: Feb 2, 2012
15. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
Holy ****, thankyou!

It all makes sense now, even the shortcut version.. god it's nice to actually have it explained bare..

Do you happen to know of an online calculator that can do the final calculation of 20log "number and angle"?
I have been looking and trying to get it to work on my Casio CFX9850 for around 35 mins now.
Have done the full calculations for all 5 instances, just dont know how to do it on my calculator thats stopping me from hitting the home stretch haha.

Last edited: Feb 2, 2012
16. ### DTskkaii Thread Starter New Member

Jan 30, 2012
9
0
Don;t worry about that last post, I realised that gain is magnitude only.

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
DTskkaii,

Re-check post #14 - there was an error in the polar notation argument sign for the impedance ZT. I've made the necessary changes.