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Finding current across a capacitor in a circuit with two AC voltage sources

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=17817&stc=1&d=1236037718.jpg


    2. Relevant equations
    My answer:
    [tex]I_C=(\frac{1}{6}-\frac{1}{2j})(cos(2t)+\sqrt{2}cos(2t-0.25\pi))[/tex]
    THIS IS WRONG.


    3. The attempt at a solution
    The way I went about doing this was:

    1. Ignored V2 and worked out the total impedance in the circuit for V1.
    2. Calculated the voltage dropped across the resistor and inductor not in parallel.
    3. Subtracted the result in step 2 from V1 to get the voltage across the capacitor as a result of V1.
    4. Repeated the first 3 steps but with V1 and V2 swapped.
    5. Added the results in step 4 and 3 together to get the total voltage across the capacitor.
    6. Worked out I = V/Z

    I could be completely wrong, in which case can someone put me back on the right track.

    Thanks,
    Jimmeh
     

    Attached Files:

    Last edited: Mar 2, 2009
  2. jcsd
  3. Mar 2, 2009 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. I didn't understand your explanation (I only skimmed it to be honest), but do you get the same answer if you just write the KCL for the summing node (top of the capacitor)? Voltage sources are short circuits, so I'm not sure you can "Ignore V2" as part of solving the whole simultaneous circuit...
     
  4. Mar 2, 2009 #3
    My answer is wrong, I've just realised I made a big mistake.

    But by "ignoring V2", I meant I calculate the voltage across the capacitor as a result of V1 and then ignore V1 and calculate the voltage across the capacitor as a result of V2, and then add them both together to get the total voltage across it. Is this the correct approach?
     
  5. Mar 2, 2009 #4

    berkeman

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    Staff: Mentor

    It doesn't seem right, but as long as you also do the problem with the KCL sum to check your answer, you'll be able to see if your method works.

    The problem seems like a pretty straightforward sum, since the V1 and V2 functions are given to you. Since the two voltage sources are out of phase and different amplitudes, there will be current flowing in the V2 leg due to V1, and visa-versa. Open-circuiting each voltage source in turn would seem to miss this simultaneous interaction portion of the circuit...
     
  6. Mar 2, 2009 #5
    What steps would you take exactly to do it as a "straightforward sum"?

    I'm unsure what you mean by doing the question with KCL, as in just verify that the currents I get are consistent with one another?
     
  7. Mar 2, 2009 #6

    berkeman

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    Well in this case, it's a pretty simple KCL, I believe. One equation and one unknown -- the voltage at the summing node at the top of the capacitor. If you solve for that V(t), then you just use the capacitor's impedance to give you the current. Would that work for you?
     
  8. Mar 2, 2009 #7
    That's what I thought I was doing, but how do you get the voltage at the node above the capacitor? What I thought you had to do was get the voltage at the node due to V1 (V1 - voltage drop across the resistor and 0.5H inductor due to V1) and then get the voltage at the node due to V2(V2 - voltage drop across the resistor and 1.5H inductor due to V2) and add them together, but then you seemingly said I couldn't do that...
     
  9. Mar 3, 2009 #8
    Are you talking about the Superposition method? As long as you are dealing in the frequency domain this should work. Just make sure that when you consider the voltage sources separately, that you short circuit and NOT open circuit the one not being considered. Then VC = V'C + V''C

    Is this what you were talking about?
     
  10. Mar 3, 2009 #9
    Yup, I had to google superposition, but yeah, that's what I'm doing.

    By short circuit, do you mean redraw the circuit with a wire in place of the voltage source you're ignoring at the time? If so, that's what I was doing, yeah.
     
  11. Mar 3, 2009 #10
    Can you perhaps show your calculations, so we can check if we find any mistakes? Your method looks ok to me too, though.
     
  12. Mar 9, 2009 #11
    Hey, I'll post my calculations soon, but one question first: when I have the total impedance for the circuit, Z, and I want to get the total current in the circuit, I, by using Ohm's Law, I=V/Z, what form should V be in? Do I leave it as cos(2t) or [itex]\sqrt{2}\cos(2t-0.25\pi)[/itex], or do I have to convert them into complex numbers? (Which would be 1 and [itex]\sqrt{2} + \sqrt{2}j[/itex]?)
     
  13. Mar 9, 2009 #12

    berkeman

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    I think in this problem with only two sources and one node, you could do the addition either way. If you were solving a more complex circuit with a many-node KCL set of equations, it would probably be best to convert it all into complex form, solve the equations, then conmvert back to get the real v(t) answers.
     
  14. Mar 9, 2009 #13
    Ok, here're my calculations:

    Frequencies:

    [itex]cos(2t)[/itex] is 1 when t=0, and is next 1 when [itex]t=\pi [/itex], therefore [itex]T_{V1}=\pi[/itex], and thus [itex]f_{V1}=\frac{1}{\pi}[/itex].

    [itex]\sqrt{2}cos(2t=0.25\p)[/itex] is also 1 when t=0 and next 1 when [itex]t=\pi[/itex], therefore [itex]T_{V2}=\pi [/itex], and thus [itex]f_{V2}=\frac{1}{\pi}[/itex] and [itex]f_{V1}=f_{V2}[/itex]

    General Impedances:

    Let [itex]Z_{L1}[/itex] be the impedance of the 0.5H inductor(L1), [itex]Z_{L2}[/itex] be the impedance of the 1.5H one(L2), and let [itex]Z_C[/itex] be the impedance of the capacitor(C).

    [itex]Z_{L1} = j2\pi fL = j2\pi(\frac{1}{\pi})(0.5) = j[/itex]

    [itex]Z_{L2} = j2\pi fL = j2\pi(\frac{1}{\pi})(1.5) = 3j[/itex]

    [itex]Z_{C} = \frac{-j}{2\pi fC} = \frac{-j}{2\pi(\frac{1}{\pi})(0.25)} = -2j[/itex]

    Let [itex]Z_1[/itex] and [itex]Z_2[/itex] be the impedances of L1 and L2 and the resistors to their immediate right.

    [itex]Z_1 = 1 + j[/itex]

    [itex]Z_2 = 1 + 3j[/itex]

    Now let [itex]Z_3[/itex] be the impedances [itex]Z_{2}[/itex] and [itex]Z_{C}[/itex] in parallel and [itex]Z_4[/itex] be the impedances [itex]Z_{1}[/itex] and [itex]Z_C[/itex] in parallel.

    [itex]Z_3 = \frac{Z_{2} Z_C}{Z_C+Z_2}=\frac{(1+3j)(-2j)}{1+3j-2j)}=\frac{6-2j}{1+j} = 2 - 4j[/itex]

    [itex]Z_4 = \frac{(-2j)(1+j)}{1+j-2j} = \frac{2-2j}{1-j} = 2[/itex]

    Calculating voltage across C due to V1:

    Let [itex]Z_{T1}[/itex] be the total impedance in the circuit as "seen" by V1:

    [itex]Z_{T1} = Z_3 + Z_1 = 2-4j + 1 + j = 3 - 3j[/itex]

    Let [itex]I_{T1}[/itex] be the total current in the circuit as due to V1:

    [itex]I_{T1} = \frac{V1}{Z_{T1}} = \frac{cos(2t)}{3-3j} = \frac{cos(2t)(1+j)}{6}[/itex]

    Let [itex]V_{C1}[/itex] be the voltage across C due to V1:

    [itex]V_{C1} = I_{T1}Z_{3} = \frac{cos(2t)(1+j)}{6}(2-4j) = \frac{cos(2t)(6-2j)}{6}[/itex]

    Calculating voltage across C due to V1:

    Let [itex]Z_{T2}[/itex] be the total impedance in the circuit as "seen" by V2:

    [itex]Z_{T2} = Z_4 + Z_2 = 2 + 1 + 3j = 3 + 3j[/itex]

    Let [itex]I_{T2}[/itex] be the total current in the circuit as due to V1:

    [itex]I_{T2} = \frac{V2}{Z_{T2}} = \frac{\sqrt{2}cos(2t-0.25\pi)}{3+3j} = \frac{\sqrt{2}cos(2t-0.25\pi)(1-j)}{6}[/itex]

    Let [itex]V_{C2}[/itex] be the voltage across C due to V1:

    [itex]V_{C2} = I_{T2}Z_{4} = \frac{\sqrt{2}cos(2t-0.25\pi)(1-j)}{6}(2) = \frac{\sqrt{2}cos(2t-0.25\pi)(2-2j)}{6}[/itex]

    Total current through C

    Let [itex]V_{C}[/itex] be the total voltage across C:

    [itex]V_C = V_{C1}+V_{C2} = \frac{cos(2t)(3-j)+\sqrt{2}cos(2t-0.25\pi)(1-j)}{3}[/itex]

    Let [itex]I_{C}[/itex] be the total current through C:

    [itex]I_C = \frac{V_C}{Z_C} = \frac{\frac{cos(2t)(3-j)+\sqrt{2}cos(2t-0.25\pi)(1-j)}{3}}{-2j} = \frac{cos(2t)(1+3j)+cos(2t-0.25\pi)(\sqrt{2}+\sqrt{2}j)}{6}[/itex]


    Look right? Would I be generally expected to get the magnitude of the current at this point?
     
  15. Mar 10, 2009 #14
    All I can suggest is to give I using only sine or cosine.
     
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