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A capacitor in an AC circuit. What is the amplitude of the current?

  1. Apr 23, 2014 #1
    A capacitor in an AC circuit. What is the amplitude of the current??

    1. The problem statement, all variables and given/known data

    I really do not know how to approach this problem. What confuses me is that when calculating the % error I'm suppose to have a theoretical or accepted value and a measured value. So how do I determine the amplitude of the current without using the 2 curve fit equations from the lab quest.

    My measured capacitance is 1.7E-6F
    My working voltage measured at the start of the lab is 4.02V

    2. Relevant equations

    LabQuest2 data

    deltaV=5.6cos(376.87x+1.31)+.074
    I=.0048cos(379.57x+2.8)-6.1E-4




    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 24, 2014 #2
    Did you measure or do you know the resistance in the circuit?

    Can you define a capacitor as a function of time t (or x in this case)?
     
  4. Apr 24, 2014 #3
    Firstly, thank you paisiello for taking the time out to reply.

    This particular procedure in the lab just consisted of the capacitor and the power supply.

    I don't know if this is right but I divided the voltage by the capacitive reactance to get 3.6E-3

    (5.59V)/(Xc) Xc=1/wC =1/(2[itex]\pi[/itex]60Hz*(1.7E-3)=1560.34 Ohms
     
  5. Apr 24, 2014 #4
    I don't think that is going to do it.

    Back to my 2nd question: Can you define a capacitor as a function of time t (or x in this case)?
     
  6. Apr 24, 2014 #5
    Vc=Vcos(wx)

    Ic=[itex]\omega[/itex]CVcos([itex]\omega[/itex]x+[itex]\pi[/itex]/2)

    ?????
     
    Last edited: Apr 25, 2014
  7. Apr 25, 2014 #6
    Start from the basic definition of a capacitor and relate this to the basic definition of current. From this you should be able to get a theoretical current based on the voltage measurement you made.
     
  8. Apr 25, 2014 #7

    BiGyElLoWhAt

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    @mrknowknow
    What paisiello is getting at involves a derivative of something with respect to something, and you normally use it in time dependent analysis. ;)
     
  9. Apr 25, 2014 #8
    A capacitor in an AC circuit. What is the amplitude of the current??

    I=dQ/dt
     
  10. Apr 25, 2014 #9

    BiGyElLoWhAt

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    There you go. And whats that equal to?
     
  11. Apr 25, 2014 #10
    Ok, well, with an alternating current circuit, you know that your ##\omega## values will always be constant when running between current and voltage, so that's pretty helpful. Also, it looks like you are mentioning complex reactance, so you should be able to use Euler's formula to help you solve this problem pretty easily.

    http://en.wikipedia.org/wiki/Euler's_formula

    (If you don't know this formula, get very handy with it if you are studying circuits)

    So, using this information, I assume you already know the equation for the current over a capacitor.

    $$ I = C \frac{dV}{dt}$$

    And this is all you need in order to solve this problem. Euler's Formula and the equation for current over a capacitor.

    Theoretically, the voltage/current in a circuit is defined as the real portion of Euler's formula, which is why the cosine function is always used. So, if you are given a function in terms of just the cosine function, you can rewrite this using Euler's formula, keeping in mind that there is an "imaginary" portion of current that flows through a circuit.

    So, say your voltage takes the form $$V(t) = V_{m}\cos(\omega t + \phi)$$

    You can actually rewrite this in terms of Euler's formula as $$V_{m}e^{\jmath (\omega t +\phi)}$$

    From here, you know how to separate variables in an exponent I'm sure. You should try to use this voltage value to now plug into the formula for the current of a capacitor. You'll notice that, after taking your derivatives and separating your exponents, the part of your equation that is $$V_{m}e^{\jmath \omega t}$$ is actually going to cancel.

    In circuits, this is known as solving your problem in the "frequency domain" because your equations are no longer time dependent. This is the underlying principle of using phasor currents and voltages.

    Now, impedance is defined as having two parts: the real part and the complex part. The real part is called resistance, and the complex part is called reactance, which I imagine you've learned. In this circuit, you'll notice you only have reactance. Typically, your phase angle is the inverse tangent of your reactance divided by your resistance, but you can solve for your phase angle in this circuit by using the method I outlined above.

    Just keep in mind that the imaginary number ##\jmath## is the same thing as $$e^{\jmath \frac{\pi}{2}}$$ and by using simple algebra to simplify your exponents, you can solve for your phase angle, the magnitude of you impedance, and for both the function of your voltage and current.
     
  12. Apr 25, 2014 #11

    BiGyElLoWhAt

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    No. $$V(t) = V_{m}\cos(\omega t + \phi)$$ grants
    ##V(t)=\frac{e^{i(\omega t + \phi)} + e^{-i(\omega t + \phi)}}{2}##

    or

    ##V(t)=Real(e^{i(\omega t + \phi)})##
    but I don't particularly like this method because this is also true:
    ##V(t)=Real(e^{-i(\omega t + \phi)})##
    Also, this site has ["spoiler"]["/spoiler"] tags for a reason.
     
  13. Apr 25, 2014 #12

    BiGyElLoWhAt

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    I used i vs. j just to clarify. Same thing, but I think engineers prefer j and physicists/mathmeticians prefer i at least in my experience.
     
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