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Electromagnetics Question Circuits

  1. Sep 4, 2012 #1
    1. The problem statement, all variables and given/known data
    If [itex]V_{s}(t) = 12 sin(4*10^{4}t-\frac{∏}{4})[/itex], R = 2 Ω, L = .1 mH and C = .3 mF, obtain an expression for the current though the capacitor.

    See attached file for circuit.

    I'm really confused. I applied KVL around the loop with the coil and got that

    [itex]\tilde{I} = \frac{e^{-\frac{j3∏}{4}}}{R+4*10^{4}Lj}[/itex]

    I applied the KVL around the loop with the capacitor and got

    [itex]\tilde{I} = \frac{4*10^{4}e^{-\frac{j3∏}{4}}}{4*10^{4}R-j}[/itex]

    I don't really know where to go from here or if what I'm doing will produce the answer I'm looking for. Any help at all solving this problem would be greatly appreciated.

    The first thing I did was adopt a cosine reference
    [itex]V_{s}(t) = sin(4*10^{4}t-\frac{∏}{4})[/itex] = [itex]cos(4*10^{4}t-\frac{∏}{4} - \frac{∏}{2}) = cos(4*10^{4}t-\frac{3∏}{4})[/itex]

    For the loop around the coil and applying KVL

    [itex]cos(4*10^{4}t-\frac{3∏}{4}) = R i(t) + L \frac{di}{dt}[/itex]

    Applying phasor analysis

    [itex]V_{s}(t) = Re[e^{j(4*10^{4}t-\frac{3∏}{4})}] = Re[e^{j*4*10^{4}t}e^{-\frac{3∏j}{4}}] = Re[ e^{4*10^{4}tj}*\tilde{V_{s}(t)}][/itex]

    And so on.
     

    Attached Files:

    Last edited: Sep 4, 2012
  2. jcsd
  3. Sep 4, 2012 #2

    gneill

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    Staff: Mentor

    Have you ever worked with impedances (the complex version of 'resistance')?
     
  4. Sep 4, 2012 #3
    I have not but I'm willing to learn and am pretty much desperate to understand this stuff at this point. I have studied complex numbers before and have not used phasors but am understanding it. It's the beginning of the semester and this stuff is already pretty confusing. Any help you can provide me would be appreciated. I don't even know were to go from where I started.

    I want to my uni's tutor center and this graduate student couldn't really help me. He was showing me some stuff with impediances and apparently it's a lot easier way to solve this problem and was able to get a solution. Apparently I'm suppose to use phasors though in the class and should learn how to solve this problem in this way because I'll need it later in the class. I would know how to solve this problem if the inductor and capacitor weren't in parallel but sense they are in parallel I'm really confused as to what to do.

    Every day in the class I'm always just sitting there with a blank stare like what is that symbol?? lol but I think I understand the information just don't know how to solve this particular problem with the parallel inductor and capcitor.

    The person at the tutor center said that people with knowledge on how to solve this problem isn't to common. The fact that this was moved to introductory physics however gives me hope I'll learn how to solve this problem. Thanks for any help.
     
  5. Sep 4, 2012 #4

    gneill

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    Impedances work hand-in-hand with phasors. For the steady-state solution to circuits, it's really just phasor analysis in a slightly different mathematical guise. It relies on the fact that
    $$\epsilon^{j \theta} = cos(\theta) + j\;sin(\theta)$$
    The impedances for R, L, and C become: ZR = R, ZL = jωL, and ZC = 1/(jωC). You get the ω from the radial frequency of the applied source (a sinewave in this case). Given numerical values for ω, R, L, and C, then you have simple complex numbers to work with. They are used in the same way as plain old resistance values are manipulated in the circuit formulas for DC circuits.

    The source is converted to a phasor value and is then treated (mathematically) as you would a normal voltage or current during circuit analysis. Only the magnitude and phase values are important, and you don't even have to worry about whether the original is a sine or cosine provided that in the end you convert the solution back to sine or cosine respectively using the magnitude and phase of the result. (If there happens to be multiple sources, some sine and others cosines, then you should convert them all to one or the other in order to keep all the relative phases consistent). If A is the source magnitude and θ the given phase angle, then for your voltage source let

    $$Vs = A\;(cos(\theta) + j\;sin(\theta))$$
    be the phasor version for Vs. Note that this yields a complex number that incorporates magnitude and angle information of the source (i.e., a phasor). You probably won't need to worry about this complex number until the end when you want to generate actual current or voltage phasors from your solution equations.

    You can perform all the basic analysis symbolically just as though it were a DC circuit, plugging in the appropriate impedance expressions (R, jωL, and 1/(jωC)) for the circuit components, or just substitute R, ZL, ZC for them for purposes of mathematical manipulation; After initial simplification you can then substitute the complex expressions or the complex numerical values. Then it's just plug and chug the complex arithmetic.

    Once you obtain a solution in phasor form you can convert back to time-domain sine or cosine by determining the magnitude and phase of the resulting phasor; Just apply the magnitude and phase values into a sine or cosine expression respectively. No phase shifting to convert from sines to cosines required.
     
  6. Sep 4, 2012 #5
    Thanks for your help. So what I don't understand is this whole thing

    [itex]\tilde{V}[/itex] = [itex]Z_{Tot}[/itex] [itex]\tilde{I}[/itex]
    [itex]\tilde{I}[/itex] = [itex]\frac{\tilde{V}}{Z_{Tot}}[/itex]
    [itex]Z_{||} = {(\frac{1}{jωL}+jωc)}^{-1} = {(\frac{1}{jωL}-\frac{{ω}^{2}cL}{jωL})}^{-1} = {(\frac{1-{ω}^{2}cL}{jωL})}^{-1}[/itex] = [itex]\frac{jωL}{1-{ω}^{2}cL}[/itex]
    [itex]Z_{Tot} = R + \frac{jωL}{1-{ω}^{2}cL}[/itex]
    I'm confused about what to put in for [itex]\tilde{V}[/itex] in the equation.

    Maybe I'm doing it wrong,

    [itex]V_{s}(t) = 12 sin(4*10^{4}t-\frac{∏}{4}) = -j Im[12 e^{j(4*10^{4}t-\frac{∏}{4})}] = -j Im[12 e^{j*4*10^{4}t}e^{-\frac{∏j}{4}}] = -j Im[ e^{4*10^{4}tj}*\tilde{V_{s}(t)}][/itex]

    So

    For [itex]\tilde{V_{s}(t)}[/itex] I plug in 12 [itex]e^{-\frac{∏j}{4}}[/itex]?

    So

    [itex]\tilde{I}[/itex] = [itex]\frac{\tilde{V}}{Z_{Tot}}[/itex] = [itex]\frac{ 12 e^{-\frac{∏j}{4}}}{R + \frac{jωL}{1-{ω}^{2}cL}}[/itex]

    Am I doing something wrong? How do I get the phasor current back into the time domain when it's all like this?

    Thanks for your help!
     
    Last edited: Sep 4, 2012
  7. Sep 4, 2012 #6

    gneill

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    Suppose this was a DC circuit with just resistors, say, R, RL, RC, and DC voltage supply Vs. Can you solve (symbolically) for the current through RC? You can employ any circuit analysis methods that you are familiar with. I can see opportunities to use loop, mesh, nodal, or even Thevenin/Norton transformations, your choice.

    Once you have an expression for the current through RC in terms of those variables you can switch them to impedances, R, ZL, ZC. Then calculate (complex) numerical values for them as appropriate:

    R = R {unchanged}
    ZL = jωL
    ZC = 1/(jωC)
    ##Vs = 12(cos(-\pi/4) + j\;sin(-\pi/4)\;V##

    Note: you have component values for R, L, and C, and you have the angular frequency ω.

    Calculate the complex value for the current by plugging these values into your expression for the current. Chug through the complex arithmetic. Then convert the resulting complex value (which is in fact the phasor for the current) back to the time domain. Do this by finding the magnitude and angle of the complex value and constructing the corresponding sine function from them.
     
  8. Sep 4, 2012 #7
    done editing

    So is this expression correct?

    [itex]\tilde{I}[/itex] = [itex]\frac{\tilde{V}}{Z_{Tot}}[/itex] = [itex]\frac{ 12 e^{-\frac{∏j}{4}}}{R + \frac{jωL}{1-{ω}^{2}cL}}[/itex]

    All I have to do is plug in the numbers at this point?

    [itex]\tilde{I}[/itex] = [itex]\frac{\tilde{V}}{Z_{Tot}}[/itex] = [itex]\frac{ 12 e^{-\frac{∏j}{4}}}{R + \frac{jωL}{1-{ω}^{2}cL}} = \frac{12cos(\frac{∏}{4})-12jsin(\frac{∏}{4})}{2+j\frac{4*10^{4}.1}{10^{3}(1-4*10^{4}.3*\frac{1}{10^{3}}.1*\frac{1}{10^{3}})}} = \frac{8.485-8.485j}{2+j\frac{4000}{998.8}}=\frac{8.485-8.485j}{2+4.00j} = \frac{8.485-8.485j}{2+4.00j}*\frac{2-4.00j}{2-4.00j} = \frac{16.97 - 33.94 j -16.97 j - 33.94}{4+16} = \frac{-16.97 - 50.91 j}{20} = .849 - 30.91 j[/itex]

    Not sure what to do from here...
     
    Last edited: Sep 4, 2012
  9. Sep 4, 2012 #8

    gneill

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    The question as posed states that you want the current through the capacitor, not the total current. Your ##Z_{Tot}## would appear to be the total impedance presented to the source voltage, so it will yield the total current; You need to solve for the current through the capacitor only.
    Once you have the appropriate expression for ##i_c## in SYMBOLIC terms of Vs, R, ZL, ZC, then you can plug in the numerical values for Vs, R, ZL, ZC to yield the phasor (in complex form) for ##i_c##. Then you can convert that to magnitude and angle form, and from that write the time domain version for ##i_c##. It will be of the form:
    $$i_c = A\cdot sin(\omega t + \theta)$$
    where A is the magnitude and ##\theta## the angle.
     
  10. Sep 4, 2012 #9
    hmm thanks lets see...

    All I know is KCL and am not sure how to apply this here to this situation. The total current is entering the node. I found it above. It gets split at the node down the capcitor and down the inductor. Not sure how to find it here. I thought in order to apply this I had to first find the total current. According to KCL

    [itex]\tilde{I_{Tot}} = \tilde{I_{Capacitor}}+ \tilde{I_{Inductor}}[/itex]

    I found [itex]\tilde{I_{Tot}}[/itex] above but it dosen't seem to help me solve for the current across the capictor, i thought I needed it to solver for it in this way, but sense i don't the current across the inductor it dose me no good

    I thought of applying KVL along the loop starting at the voltage source and going around to the inductor and then back to the voltage source.
    The other option is starting at the voltage source and going to the capcitor and then back to the voltage source.
    I don't see how KVL will work here.
     
    Last edited: Sep 4, 2012
  11. Sep 4, 2012 #10

    gneill

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    Staff: Mentor

    I'm surprised that you would be studying reactive components (inductors, capacitors) without first having covered loop and nodal analysis at least for purely resistive circuits.

    However. If you find a numerical value for i (as it is depicted in the diagram), then the rule for current division through two parallel paths is that it splits in inverse proportion to the impedances. That is, suppose the two paths have impedances Z1 and Z2 and that the total current is I. Then the current through Z1 will be given by I*Z2/(Z1 + Z2).

    Personally I would probably apply nodal analysis to solve for the node voltage Vn where the three components meet, then find the capacitor current as Vn/ZC.
     
  12. Sep 4, 2012 #11
    X_X More stuff I don't know lol lets see thanks um hmm... ok thanks i'll try and see that given this rule
     
  13. Sep 4, 2012 #12

    gneill

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    Note: it would appear that somethings gone wrong in your hammering out the value for I in post #7. You should recheck your math.
     
  14. Sep 4, 2012 #13
    Do I even need I total? And in this case you mean node analysis by applying KCL as I put above? The sum of the two currents leaving the node is equal to the total current?
     
  15. Sep 4, 2012 #14

    gneill

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    It's possible to write an expression for the capacitor current that doesn't require first solving for the total current. But solving for the total current first is also valid if you know how to apply the current division trick.

    Nodal analysis applies KCL at the nodes of a circuit in order to solve for the node voltages. Once you have the node voltages, finding current through any component is just a matter of using Ohm's law with the component impedance and potential difference. Writing the KCL for the currents at a given node is the starting point.
     
  16. Sep 4, 2012 #15
    [itex]V_{n} = \frac{V_{s}}{1+\frac{R}{jωc}+jRωc}[/itex]

    [itex]\tilde{V_{n}} = \tilde{Z}\tilde{I_{c}}[/itex]

    IS the voltage in the first equation phasor voltage at the node and should have a telda above it?
     
    Last edited: Sep 4, 2012
  17. Sep 4, 2012 #16

    gneill

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    That should be L, not C.
    Yes, good.
    Yes it's a phasor. If your notation convention is to use tildes to specify phasors, then yes.
     
    Last edited: Sep 4, 2012
  18. Sep 4, 2012 #17
    [itex]\tilde{V_{n}} = \frac{V_{s}}{1+\frac{R}{jωL}+jRωC} = \tilde{Z_{c}}\tilde{I_{c}}[/itex]
    [itex]\tilde{I_{c}} = \frac{V_{s}}{\tilde{Z_{c}}(1+\frac{R}{jωL}+jRωC)} = \frac{V_{s}}{\frac{1}{jωC}(1+\frac{R}{jωL}+jRωC)} = \frac{jωC V_{s}}{1+\frac{R}{jωL}+jRωC} = \frac{jωC V_{0}sin(ωt + ∅)}{1+\frac{R}{jωL}+jRωC} = -\frac{ωC V_{0}sin(ωt + ∅)}{j+\frac{R}{ωL}-RωC} = -\frac{ωC * Im[ V_{0} e^{j(ωt+∅)}]}{j+\frac{R}{ωL}-RωC} = -\frac{ωC * Im[ V_{0} e^{jωt}e^{j∅}]}{j+\frac{R}{ωL}-RωC} = -\frac{ωC * Im[\tilde{V_{s}}e^{jωt}]}{j+\frac{R}{ωL}-RωC}[/itex]

    I think I'm actually getting some where but now what? I'm not sure how to proceed from here.
     
    Last edited: Sep 4, 2012
  19. Sep 5, 2012 #18

    gneill

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    Staff: Mentor

    You don't have to translate back to the time domain until you have a final complex number result; It just makes for more work if you convert too soon. Also, unless you particularly need a symbolic expression for the phasor result, it suffices to work with impedances as variables since you can assign numerical values to them.

    You can plug in the numerical values for each of the variables, simplify to a complex number, take its magnitude and angle, and then write the time domain result from them.

    What are the complex values associated with the following (numerical, not symbols)?

    Vs = ?
    ω = ?
    ZL = ?
    ZC = ?

    Using these variables the capacitor current will take the form

    $$i_c = \frac{ZL}{ZL \cdot ZC + R \cdot ZC + R \cdot ZL} Vs$$

    Just plug in the values to obtain the complex result, then determine the magnitude and angle. Call them A and θ. Then the time domain result is then Ic = A*sin(ωt + θ), since the original source had the form sin(...).
     
  20. Sep 5, 2012 #19
    Editing

    [itex]-\frac{ωC * Im[\tilde{V_{s}}e^{jωt}]}{j+\frac{R}{ωL}-RωC} = - \frac{4*10^{4}*.3 mF*\frac{F}{1000 mF}*Im[\tilde{V_{s}}e^{j*4*10^{4}t}]}{j+ \frac{2}{4*10^{4}.1 mH*\frac{H}{1000 mH}}-2*4*10^{4}.3 mF*\frac{F}{1000 mF}} = -\frac{12*Im[\tilde{V_{s}}e^{40,000t j}]}{j+ \frac{1}{2}-24} = -\frac{12*Im[\tilde{V_{s}}e^{40,000t j}]}{- \frac{47}{2}+j} = -\frac{12*Im[\tilde{V_{s}}e^{40,000t j}]}{- \frac{47}{2}+j}*\frac{{- \frac{47}{2}-j}}{{- \frac{47}{2}-j}} = \frac{282*Im[\tilde{V_{s}}e^{40,000t j}]}{\frac{2213}{4}}+\frac{48*Im[\tilde{V_{s}}e^{40,000t j}]}{2213}j = \frac{1128*Im[\tilde{V_{s}}e^{40,000t j}]}{2213}+\frac{48*Im[\tilde{V_{s}}e^{40,000t j}]}{2213}j = \sqrt{.260+4.705*10^{-4}}<arctan(\frac{1128}{48}) = .510<1.528[/itex]

    Is this
    [itex]I_{c} = .510 sin(4*10^{4}+1.528)[/itex]
    correct?

    I'm not sure how to get around this
    [itex]Im[\tilde{V_{s}}e^{40,000t j}][/itex]
    When finding the magnitude do I need to consider this as part of the magnitude of each component? If so it would just factor out of the square root and I would be left with
    [itex].510*Im[\tilde{V_{s}}e^{40,000t j}]<1.528[/itex]

    and my solution would be
    [itex]I_{c} = .510*Im[\tilde{V_{s}}e^{40,000t j}]* sin(4*10^{4}+1.528)[/itex]

    I think I'm almost done but I'm not sure how to handle this part.
     
    Last edited: Sep 5, 2012
  21. Sep 5, 2012 #20

    gneill

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    Nope. Why don't you try following my suggestions above?
     
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