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AC Electronics Problem (Problem with Peak values)

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data

    This is my first time posting so I'm sorry if the formatting isn't right, just let me know what I need to change. Thanks.

    Note: If it helps, I am using "Electronics with Discrete Components" by Enrique J. Galvez and this problem is #4 on page 166.

    "The AC supply had a peak voltage of 10 V and a frequency of 1 kHz, and the capacitor had a capacitance of 0.1 μF.

    a: Find the peak value of the current through the capacitor.

    b: Explain how the current flows through the capacitor.

    c: What is the peak voltage between points A and B?

    d: We change the frequency of the source so that the peak voltage through the capacitor is the same as the peak voltage through the resistor.
    i: What is the value of the frequency?
    ii: What are the peak voltages across the resistor and capacitor?
    iii: Explain why they are not 5 V."


    2. Relevant equations

    ZR = R
    ZC = [itex]\frac{1}{iωc}[/itex]
    ω = 2πf
    V0 = |Zeq|I0

    3. The attempt at a solution

    In this problem, for part a, I found that Zeq = R + [itex]\frac{1}{iωc}[/itex]
    and so |Zeq| = [itex]\sqrt{R^{2} - (\frac{1}{ωc})^{2}}[/itex]

    From there, I entered what I knew into V0 = |Zeq|I0 in order to solve for I0.

    I0 = [itex]\frac{V_{0}}{\sqrt{R^{2} - (\frac{1}{ωc})^{2}}}[/itex]

    Plug in with:
    V0 = 10 V
    R2 = 1000 Ω
    ω = 2000π rad/s
    c = 10-7 F

    I0 = [itex]\frac{10}{\sqrt{(1000)^{2} - (\frac{1}{(2000π)(10)^{-7}})^{2}}}[/itex]

    I0 = 0.0081 A

    I believe this is the correct solution for part a but I am not quite confident with it.

    From what I have read, I am pretty sure that the current is 90° out of phase with the voltage when it flows through the capacitor but I would appreciate it if someone could explain to me how to understand phase shifts. I understand at a high level but mathematically I don't really know what to do.

    The current is the same everywhere since the circuit is in series.
    Vab = I0ZC
    Vab = I0 [itex]\frac{1}{ωc}[/itex]
    Vab = (0.0081)([itex]\frac{1}{(2000π)(10^{-7})}[/itex]
    Vab = 12.89 V

    This does not seem like it should be possible to me so do I treat it like a DC circuit instead and:
    VR = I0ZR
    VR = I0R
    VR = (0.0081 A)(1000 Ω)
    VR = 8.1 V
    and then subtract from the peak voltage?
    10 V - 8.1 V =1.9 V ?
    or from the 12.89 V calculated before?
    I don't really understand how to approach this aspect of AC.

    VR = I0ZR and VC = I0ZC

    VR = VC so ZR = ZC

    R = [itex]\frac{1}{ωc}[/itex]

    ω = [itex]\frac{1}{Rc}[/itex]

    ω = [itex]\frac{1}{(1000)(10^{-7})}[/itex]

    ω = 10000 rad/s

    f ≈ 1591.55 Hz

    The peak voltage will be the same for both (since that is the state of the problem) but I first need to find the peak current again.

    I0 = [itex]\frac{V_{0}}{\sqrt{R^{2} - (\frac{1}{ωc})^{2}}}[/itex]

    I0 = [itex]\frac{10}{\sqrt{(1000)^{2} - (\frac{1}{(10000)(10)^{-7}})^{2}}}[/itex]

    I0 = [itex]\frac{10}{0}[/itex]

    Here I run into this issue and I don't know where I made a mistake.

    I can't really answer this yet.

    Thank you for any help. I really want to understand this.
    Last edited: Mar 15, 2013
  2. jcsd
  3. Mar 15, 2013 #2


    User Avatar

    Staff: Mentor

    Think of the real and imaginary components of the impedance like the components of a vector, and here the "j" is the unit vector that's at 90° to the "real" axis. In that case, to find the magnitude of this "vector", drop the unit vectors and take the square root of the sum of the squares of the component magnitudes. In other words, your expression should read

    [itex]\sqrt{R^{2} + (\frac{1}{ωc})^{2}}[/itex]

    Redo your current calculation accordingly.
    It takes a change in charge on the capacitor in order for the potential across the capacitor to change. That means the current must flow to change the charge to change the potential. Thus the current leads the voltage change for a capacitor.

    But this may not be quite what the question is asking. It may be that they are referring to the fact that the two plates of a capacitor are separated by an insulator through which charge carriers cannot pass. The question then becomes, how is it we have a current flowing in a circuit with a capacitor?
    Redo with your recalculated current.
  4. Mar 15, 2013 #3
    Okay, I think I understand. At the very least it makes sense that I want to take the sum of the component magnitudes (since they should each be a magnitude and not subtracted from each other.

    So if I have [itex]|Z_{eq}| = \sqrt{R^{2}+(\frac{1}{ωc})^{2}}[/itex]

    Then for [itex]V_{0} = |Z_{eq}|I_{0}[/itex] I obtain:

    [itex]I_{0} = \frac{V_{0}}{\sqrt{R^{2}+(\frac{1}{ωc})^{2}}}[/itex]

    [itex]I_{0} = \frac{10}{\sqrt{(1000)^{2}+(\frac{1}{(2000π)(10^{-7})})^{2}}}[/itex]

    [itex]I_{0} ≈ 0.0053 A[/itex]

    If I correctly understand then a complete answer might say that there is current flowing through the capacitor because there is a potential difference across the two plates (which causes movement of electrons despite the fact that they are not physically leaping across the capacitor.) In addition, the potential across the capacitor occurs out of phase from the source because there must be a current through the capacitor before it has a potential difference across itself.

    For part c the answer is:

    [itex]V_{ab} = I_{0}Z_{C}[/itex]

    [itex]V_{ab} = I_{0}\frac{1}{ωc}[/itex]

    [itex]V_{ab} = 0.0053\frac{1}{(2000π)(10^{-7})}[/itex]

    [itex]V_{ab} = 8.44 V[/itex]

    For part d then:

    i: This stays the same.

    [itex]ω = \frac{1}{Rc}[/itex]

    [itex]ω = \frac{1}{(1000)(10^{-7})}[/itex]

    [itex]ω = 10000 rad/s[/itex]

    [itex]f ≈ 1591.55 Hz[/itex]

    ii: I need to find the current again without making the same mistake.

    [itex]I_{0} = \frac{V_{0}}{\sqrt{R^{2}+(\frac{1}{ωc})^{2}}}[/itex]

    [itex]I_{0} = \frac{10}{\sqrt{(1000)^{2}+(\frac{1}{(10000)(10^{-7})})^{2}}}[/itex]

    [itex]I_{0} = 5(10^{-6}) A[/itex]

    Then peak voltage across the resistor is:

    [itex]V_{R} = I_{0}R[/itex]

    [itex]V_{R} = 5(10^{-6})(1000)[/itex]

    [itex]V_{R} = 0.005 V[/itex]

    Also the peak voltage across the capacitor is:

    [itex]V_{C} = I_{0}\frac{1}{iωc}[/itex]

    [itex]V_{C} = 5(10^{-6})\frac{1}{(10000)(10^{-7})}[/itex]

    [itex]V_{C} = 0.005 V[/itex]


    They are not 5 V because the frequency of switching is too rapid for the capacitor to have much of a potential difference across it (not enough charging time).

    I believe I now have correctly answered every part of this question. Thank you for your comment gneill. Please let me know if I did not correctly apply your suggestions or have made other mistakes.
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