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Homework Help: Finals Study Time! A Two Source Circuit w/Inductor, Cap, and Resistor

  1. Dec 8, 2018 #1
    1. The problem statement, all variables and given/known data
    upload_2018-12-8_0-50-22.png

    2. Relevant equations
    Z = √(XC2 + XL2)
    XL = 2πƒL
    XC = 1 / 2πƒC
    I = V/Z

    3. The attempt at a solution
    First off, thank you for all of the help this semester. I'm sure you'll be seeing questions from me in the spring also. Here is how I'm thinking about this problem:

    1. Using the above equations, I find the reactances of the inductor and capacitor to be 3141.59Ω and 1.59Ω, respectively.

    2. Add: XL + (-XC) = 3140Ω = Z

    Note: I used -XC because on a graph the capacitor's reactance would be a downward pointing vector. However, no matter if I add or subtract these reactances the final current would be equivalent since 1.59Ω is much, much smaller than 3141.59Ω.
    upload_2018-12-8_1-5-44.png


    3. I find the current by using the equation of V/Z; however, the issue is that I have two voltage sources-- so which V do I use?

    I decided that I could use superposition. So first equation: I = 1.2V / 3140 = .382mA. Second equation: 0.8V / 3140 = .255mA

    4. Now I superimpose them. Since the 1.2V source moves rightward and the 0.8V source moves leftward, I did the following:
    → ← →
    .382 - .255 = .127mA


    5. My final result seems to be off. Relative to the voltage through the 1.2V source, here is the graph:

    upload_2018-12-8_1-12-0.png


    As you can see, the difference is 56μA, or about .056mA at the voltage peak of the 1.2V source.

    One thing I noticed I did wrong is that the current of .127mA would be present when at the voltage peak of a single source circuit. Since I have 2 voltage sources the way I compared my current to my voltage peak doesn't make sense.

    If someone could highlight where I am going wrong I would be very appreciative. The finals are on Wednesday and I've been studying for 6hrs so I'm gonna head to bed and check on this in the AM. Thanks :)
     

    Attached Files:

  2. jcsd
  3. Dec 8, 2018 #2

    gneill

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    Staff: Mentor

    Check your capacitor impedance. On the circuit diagram the capacitor's value is specified as 200p. Note the p.

    Edit:
    Also note that VR is shown to have a (very) small phase difference to VL:

    upload_2018-12-8_10-47-58.png

    You might want to show that this makes negligible difference to the results.
     
  4. Dec 8, 2018 #3

    gneill

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    You really can't compare voltages and currents in that fashion. They have different units. What you can do though is find the phase difference. What's the phase angle of the current with respect to the VL voltage source?
     
  5. Dec 8, 2018 #4
    Ahh okay... so I had the wrong impedance for the capacitor.

    So Z = 1975.62Ω

    "What's the phase angle of the current with respect to the VL voltage source?"

    Φ = tan-1 (.000607A/1.2V) = .029°

    That amplitude comes from dividing 1.2 volts by 1975.62Ω

    This angle doesn't seem right. Whats my next step?
     
  6. Dec 9, 2018 #5

    gneill

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    That would be its reactance, 1/(ωC). Its impedance would be 1/(jωC) = -j 1975.65 Ω. Note that impedances are complex values, and that for ideal inductors and capacitors they are entirely imaginary with no real component.
    It is not right. You need to first find the current in its complex form. Voltage divided by total impedance (complex values!). The angle will be extracted from that.
     
  7. Dec 9, 2018 #6
    I guess I'm just not understanding what to do with the imaginary number.

    So the reactance of the capacitor is = 1975.62Ω, whereas its impedance is -j1975.62Ω... got it.

    Now what? Would the next step be to find the phase angle? I could divide the above reactance by the resistance and use tan-1 to get 57.86°, but I'm not sure what this angle is really representing. I know that a phase shift is the degree to which one sine wave is leading or lagging another sine wave.

    Is phase angle the same as phase shift? Probably not, since they use two different symbols: Φ vs θ.
     
  8. Dec 9, 2018 #7

    gneill

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    You want to compute the current using the voltage and impedances. The result will be a complex value. Convert to polar form (magnitude and angle). The angle is the phase angle referenced to the voltage supply.
    I've seen people use them interchangeably, along with the symbols.
     
  9. Dec 9, 2018 #8
    Which voltage? I have two voltage sources. I would have already solved this problem if it were only one.
     
  10. Dec 9, 2018 #9

    gneill

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    If you write KVL around the loop you will find that the two sources sum to a single equivalent voltage source (again, counting the minuscule phase angle they've given VR as irrelevant). Or do as you did before and use superposition (more work though).

    The question asked to sketch the current relative to VL, which means letting VL set the reference for phase angles throughout the circuit.
     
  11. Dec 9, 2018 #10
    Here's where I'm at so far:

    Z = √(1k)2 + (1550)2 = 1845Ω

    KVL: 1.2V - 1845ΩI -0.8V = 0 ; I = 217μA

    You said it should be a complex answer, so should the current actually be j217μA?

    If this is correct, what would be my next step? You said to convert it to polar form. This is where my understanding of j is shaky. How do I convert from complex to polar?
     
  12. Dec 9, 2018 #11

    gneill

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    Staff: Mentor

    No, the current will have a phase angle that depends upon the combined impedance of the reactive components (Inductor and capacitor).

    You can work with reactances as you've done, but figuring out the phase angle becomes tricky; You need to look at the net resistance and net reactance and determine what the division does to the resulting phase angle. My old brain doesn't like to wrestle with that sort of puzzle when a more "mechanical" solution is available.

    If you keep all the impedances in complex form for your calculations then the resulting current will be in complex form also. From that you can extract the magnitude and phase of the current. So the impedance of the capacitor is ##Z_C=-j\;1.592\;k\Omega##, the impedance of the inductor is ##Z_Lj\;3.142\;k\Omega##. The resistance impedance is purely real, so ##Z_R = 1\;k\Omega##. The current is then:
    $$I = \frac{VL-VR}{Z_L + Z_R+Z_C}$$
    This will result in the current in complex form (a so-called phasor value).

    Converting to polar form (magnitude and angle) is a matter of a bit of Pythagoras and trigonometry. It's the same as finding the magnitude and angle of a vector. The magnitude is the square root of the sum of squares of the component magnitudes, while the angle is the arctan of the ratio of the components (be sure to check the quadrant that the "vector" lies in).

    If you have some arbitrary complex number ##z = (a + bj)## then

    Magnitude: ##|z| = \sqrt{a^2 + b^2}##
    Angle: ##\phi = \arctan(b/a)## but check the quadrant! Look at the signs associated with the "x" and "y" values.

    Edit: alternatively, if your calculator supports it, use the atan2() function which automatically handles the quadrant placement when working out the angle.
     
  13. Dec 9, 2018 #12
    Here goes:

    VL - VR = 0.4V
    ZL + ZR + ZC = j3.142kΩ + 1kΩ + -j1.592kΩ = j2550

    i = 0.4 / j2250 ←
    a current in complex form

    I'm not sure how to get the above equation into the form z = a + bj

    I can find the magnitude and arctan after that.
     
  14. Dec 9, 2018 #13

    gneill

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    What happened to the real 1kΩ? You cannot just add real and imaginary components. Your total impedance should be a value with both real and imaginary parts. The resistance will appear as the real part. The sum of the impedances of the inductor and capacitor will appear as the imaginary part.
     
  15. Dec 9, 2018 #14
    Oh whoops left that out: i = 0.4V / (1kΩ + j1.55kΩ)
     
  16. Dec 9, 2018 #15

    gneill

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    Okay! What's the resulting current?
     
  17. Dec 9, 2018 #16
    Well, I'm thinking the denominator is: √1k2 + 1.55k2 = 1844.59
    arctan 1.55/1k = 57.17°

    Thus... i = 0.4 / 1.845k∠57.17°

    I'm not sure if there should be some angle in the numerator to be divided. 0.4 / 1.845 = .217mA

    So is the answer i = .217mA∠57.17° ?
     
  18. Dec 10, 2018 #17

    The Electrician

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    Gold Member

    Except that your angle should be the negative of what you have. Moving the angle from the denominator to the numerator changes the sign.
     
  19. Dec 10, 2018 #18
    Ahh, which means it leads the voltage, correct?

    So, what does 217uA represent with regards to a graph? Because when I plot in LTSpice, the peak amplitude is about 450uA.

    I'm thinking it is when the voltage is at 0.4V.
     
  20. Dec 10, 2018 #19

    gneill

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    At 450 μA the driving voltage is more likely to be VR alone (0.8 V).
     
  21. Dec 10, 2018 #20
    Right, but what is the 217uA representing since this is my answer?
     
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