Solving AC Homework: e(t),i(t),Ul(t),Ur(t),Uc(t),Urc(t),Url(t)

In summary, In the frequency domain, the time dependent angle disappears and you can write the voltage as: Ulc = 17.89V*(cos(φ) + j*sin(φ))
  • #1
builder_user
196
0
Hello.This is my new task.

Homework Statement


Find e(t),i(t),Ul(t),Ur(t),Uc(t),Urc(t),Url(t).
Find reactive power and complex power.

Homework Equations


What to do with Ulc(t)?

The Attempt at a Solution


C=200microfarad
L=3millihenry
r=4 Ohms
Ulc(t)=17.89sin(1000t-64)

I've done this
Zc=-j/wC
Zl=jwL
I've deleted L and C and added Zc and Zl to scheme.What is next?
I can't find w because I do not have e(t).

Here my original scheme.
 

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  • #2
There's no Ulc(t) shown in the first figure. Are the diagrams supposed to represent the same circuit redrawn, or are there two circuits to solve?

What are the units for the given Ulc(t)? Is the amplitude in volts? Is t in seconds? Is '64' an angle in degrees or in radians? How about the '1000'? Is it in radians per second? Degrees per second? Something else?
 
  • #3
gneill said:
There's no Ulc(t) shown in the first figure. Are the diagrams supposed to represent the same circuit redrawn, or are there two circuits to solve??

It's one circuit.

gneill said:
What are the units for the given Ulc(t)?
V
gneill said:
Is the amplitude in volts?Is t in seconds?Is '64' an angle in degrees or in radians? How about the '1000'? Is it in radians per second? Degrees per second? Something else?
17.89 - amplitude
t - seconds
64- degrees
1000 -w(sec-1)
 
  • #4
Well, you can pick the value of ω out of the given expression for Ulc(t). Presuming that the 1000 is 1000 degrees/second, you can convert that to radians per second for ω.

That will allow you to calculate the impedance for the series LC. If you then express Ulc as a complex phasor, you should have no difficulty computing the complex current phasor for the current through LC. Since the whole circuit is series connected, the same current flows through the R as well, and you can determine e as a complex phasor.
Convert back to sin(ω*t + θ) form if desired.

Given the current and voltages, you should have no trouble calculating the power.
 
  • #5
64*=0.35pi?
1000 = 5.5pi?

w will be the same for all u(t) and i(t) and e(t)?
 
  • #6
Yes, ω will be the same across the board.

You might find it to be a good idea to carry a few more decimal places for intermediate results. Round the final results at the end. So

64° --> 1.117 radians
1000°/sec --> 17.453 radians/sec
 
  • #7
Is I(t)=Ulc/zl+zc=
17.89sin(5.5pi*t-0.35pi)/(j*5.5pi*L+(-j/5.5pi*L)=17.89sin(17.5t-1.12)/(17.5jL-j/(17.5L)?
 
  • #8
I think a couple of your L's should be C's in the above.

Convert the Ulc(t) to phasor form so you're working entirely in the frequency domain. But you're on the right approach.
 
  • #9
gneill said:
Convert the Ulc(t) to phasor form .

exp?
17.89*(exp^j(17.5t-1.12)-exp^-j(17.5-1.12))/(2*j(17.5j-j/17.5C))

To get u for every element of the circuit i(t) must be multiplyed by every resistance?
 
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  • #10
In the frequency domain the time dependent angle disappears. In complex form you can write the voltage as:

Ulc = 17.89V*(cos(φ) + j*sin(φ))

where φ is your phase angle, -64° = -1.117 radians

Solve for the current by dividing this voltage by the impedance ZL+ZC as you indicated earlier. Then current x individual impedances for the individual voltages.

You didn't say whether the given voltage Ulc was peak or rms. Did the original problem statement mention it?
 
  • #11
gneill said:
Then current x individual impedances for the individual voltages?
So,current is the same for all elems.
For Urc = I*(Zr+Zc)?

And How to find e(t)?

gneill said:
You didn't say whether the given voltage Ulc was peak or rms. Did the original problem statement mention it?

I don't know what is rms.All I know is Ulc - instanteous voltage
 
  • #12
The current is the same for all series connected components. So the overall voltage is given by I*Ztotal. That will equal your e (in complex phasor form).

It seems that your two circuit diagrams have the order of the components changed, so that makes it hard to understand what is meant by Urc or Ulc.
 
  • #13
gneill said:
It seems that your two circuit diagrams have the order of the components changed, so that makes it hard to understand what is meant by Urc or Ulc.

It's made to find LC.I(t) will not change because all elems are the same.

How to find complex power and reactive power?Amplitude e(t)/Ze,
where Ze=Zl+Zr+Zc?

Amplitude for every U&e -> x *(cos...)?
 
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  • #14
How to find active and reactive power?
 
  • #15
builder_user said:
How to find active and reactive power?

If you have the voltage source, e, in complex phasor form, and the current, I, also in complex phasor form, then the apparent power is p = eI*, where I* is the complex conjugate of I.

The components of p are the real and reactive components of the power. Do a search on "power triangle" if you need more information about apparent, real, and reactive power.
 
  • #16
gneill said:
where I* is the complex conjugate of I.
It's means that if I have
e(t)=B*sin(wt-q)
and I(t)=A*sin(wt-q) the result will be like this

p=B*sin(wt-q)/A?
 
  • #17
builder_user said:
It's means that if I have
e(t)=B*sin(wt-q)
and I(t)=A*sin(wt-q) the result will be like this

p=B*sin(wt-q)/A?

No. First of all, the voltage and current waveforms will not have the same phase angle, because of the impedance having both real and imaginary parts. Secondly, you want to be dealing with complex values in the frequency domain, not sines and cosines in the time domain. There should be no t's involved.

Perhaps I'm making an assumption that I shouldn't be making. Do you know what phasors are? How about their representation as complex values?
 
  • #18
I=B*(cosq+jsinq)
e=I*Z=B*Z*(cosq+jsinq)
and
p=e*Bjsinq?
 
  • #19
Okay. Why don't you put some numbers to those expressions?

What values do you have for e, I, and Z?
 
  • #20
I'm not sure but from previous posts and the task it's seems to be
Z=-289j+0.05j+4
I=17.89*(0.45+j*0.89)/(0.05j-289j)
e=I*Z
 
  • #21
Your value for Z does not look right. Can you show your work for finding the impedances of the inductor and capacitor?
 
  • #22
С=-j/WC=-j/5.5pi*200*10e-6

L=jwL=j*5.5pi*3*10e-3
 
  • #23
Okay. You should keep a few more decimal places for intermediate values. I see the capacitive reactance as -286.4 Ohms, and for the inductive reactance, 0.0524 Ohms.

This is assuming that the "1000t" in the time domain expression for Ulc implies 1000 degrees per second.

So, with ZL and ZC in hand, what numerical value are you getting for the current (normalized complex form)? How about for e?
 
  • #24
I=0.03j-0.05Is 17.89 the amplitude of I?
 
  • #25
I'm getting a different value for I. In particular, I'm seeing (0.0561 + 0.0274j) Amps, or if you prefer your current in milliamps, (56.1 + 27.4) mA. Perhaps you should review your calculation. If you don't get this value, show your calculation in detail.

The amplitude of the current would be |I| = |56.1 + 27.4| = 62.5 mA
 
  • #26
С=-286.4j
L=0.0524j

I=17.89*(0.45+0.89j)/(0.0524j-286.4j)=-0.0625*(0.45+j0.89)/j=0.0625j*0.45+j*j*0.89*0.0625=0.028j-0.056
 
  • #27
The expression that you're using for Ulc is suspect: 17.89*(0.45 + 0.89j). The phase angle is -64 degrees, making cos(φ) and sin(φ) 0.438 and -0.899 respectively. Where did you get 0.45 and +0.89?
 
  • #28
I forgot that angle is -64 not 64.
 
  • #29
gneill said:
|I| = |56.1 + 27.4| = 62.5 mA
Why?
 
  • #30
[tex] |I| = \sqrt{56.1^2 + 27.4^2} = 62.4 mA [/tex]

(I think when I originally calculated it I was keeping more decimal places, so the ".4" crept up to ".5" on rounding)
 
  • #31
I see.Thanks
I've found
Uc(t)=7.7328-16.0384j
Ul(t)=0.0014+0.003j
Ur(t)=0.224+0.108j
e(t)=7.9554-15.92746j
Are the result correct? 'cause I'm not sure...
 
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  • #32
I'll have to re-work the problem; I've lost track of the calculations that I did previously. It may take a while, as I've got several things on the go at the moment. I hope that's okay.

To summarize, you are given a series RLC circuit and an expression for the voltage waveform that exists across the inductor and capacitor combination. The inductor is L = 3mH, the resistor is R = 4 Ω, and the capacitor is C = 200μF. The order of the components seems to change depending upon what voltages are to be calculated.

The current across the series connected LC pair is given to us as:

Ulc = 17.89V*sin(1000(degrees/s)*t - 64 degrees)

You are looking to find the values for the voltage supply e(t), and the voltages across the individual components as well as several (shuffled) component pairs, and the reactive and complex power used by the circuit. Does that about sum it up?
 
  • #33
yes.

I've found P
p=e(t)*I*=e(t)*0.027j=0.43+0.216j
P-reactive - - - - - - 0.216 var?
P-active - - - - - - - - - - 0.43 W?
 
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  • #34
Values that I'm seeing (To compare with the values you've calculated):

For:
ω = 1000 deg/sec; φ = -64° ; B = 17.89V; Ulc(t) = B*sin(ωt + φ)
R = 4Ω ; L = 3mH ; C = 200μF

Impedances:
ZL = 52.36 mΩ {milli Ohms}
ZC = -286.48Ω
ZLC = -286.48Ω {ZL + ZC}
Z = 4 - 286.48Ω {Total impedance of series RLC}

Phasors:
Ulc = B(cos(φ) + jsin(φ)) = 7.842 - j16.079 V ; |Ulc| = 17.89V ; Angle: -64°

I = Ulc/ZLC = 0.056 + j0.027 A ; |I| = 62.46 mA ; Angle: 26°

e = I*Z = 8.067 - j15.97 V; |e| = 17.89 V ; Angle: -63.2°

Ul = I*ZL = -1.434 + j2.939 mV ; |Il| = 3.27 mV ; Angle: 116°

Uc = I*ZC = 7.844 - j16.082 V; |Uc| = 17.89 V ; Angle: -64°

Ur = I*R = 0.225 + j0.11 V ; |Ur| = 250 mV ; Angle: 26°

Power:
P = e * conjugate(I) = 0.0156 - j1.117 W ;
|P| = 1.12 VA {Apparent power}
Re(P) = 0.0156 W = ; 15.6 mW {Real power dissipated}
Im(P) = -1.12 VAR {Reactive power -- negative means it's "capacitive" looking - current is leading voltage}
 
  • #35
Some of your results are similar to mine but some of them are completely different.
I think your results are correct so I'll use them.
gneill said:
P = e * conjugate(I) = 0.0156 - j1.117 W ;

I can't get this result
 
<h2>1. What are the main components in an AC circuit?</h2><p>The main components in an AC circuit are a voltage source, a resistor, an inductor, and a capacitor.</p><h2>2. How do I calculate the voltage and current in an AC circuit?</h2><p>To calculate the voltage and current in an AC circuit, you will need to use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In addition, you will need to use the equations that relate voltage and current to inductance (L) and capacitance (C).</p><h2>3. What is the difference between instantaneous and RMS values in an AC circuit?</h2><p>Instantaneous values in an AC circuit refer to the voltage or current at a specific moment in time. RMS (Root Mean Square) values, on the other hand, refer to the effective or average value of the voltage or current over a period of time. RMS values are typically used in calculations and analysis of AC circuits.</p><h2>4. How do I calculate the reactance of an inductor or capacitor in an AC circuit?</h2><p>The reactance of an inductor (XL) can be calculated using the equation XL = 2πfL, where f is the frequency of the AC source and L is the inductance of the inductor. The reactance of a capacitor (XC) can be calculated using the equation XC = 1/(2πfC), where C is the capacitance of the capacitor.</p><h2>5. What is the phase difference between voltage and current in an AC circuit?</h2><p>The phase difference between voltage and current in an AC circuit is determined by the ratio of the reactance (XL or XC) to the resistance (R). When the reactance is greater than the resistance, the current will lag behind the voltage and the phase difference will be positive. When the reactance is less than the resistance, the current will lead the voltage and the phase difference will be negative.</p>

1. What are the main components in an AC circuit?

The main components in an AC circuit are a voltage source, a resistor, an inductor, and a capacitor.

2. How do I calculate the voltage and current in an AC circuit?

To calculate the voltage and current in an AC circuit, you will need to use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In addition, you will need to use the equations that relate voltage and current to inductance (L) and capacitance (C).

3. What is the difference between instantaneous and RMS values in an AC circuit?

Instantaneous values in an AC circuit refer to the voltage or current at a specific moment in time. RMS (Root Mean Square) values, on the other hand, refer to the effective or average value of the voltage or current over a period of time. RMS values are typically used in calculations and analysis of AC circuits.

4. How do I calculate the reactance of an inductor or capacitor in an AC circuit?

The reactance of an inductor (XL) can be calculated using the equation XL = 2πfL, where f is the frequency of the AC source and L is the inductance of the inductor. The reactance of a capacitor (XC) can be calculated using the equation XC = 1/(2πfC), where C is the capacitance of the capacitor.

5. What is the phase difference between voltage and current in an AC circuit?

The phase difference between voltage and current in an AC circuit is determined by the ratio of the reactance (XL or XC) to the resistance (R). When the reactance is greater than the resistance, the current will lag behind the voltage and the phase difference will be positive. When the reactance is less than the resistance, the current will lead the voltage and the phase difference will be negative.

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