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Calculate the maximum current

  1. Jul 21, 2012 #1
    The problem
    The exercise goes like this:
    A sinusoidal source v(t) = 40 sin(100t) its applied to a circuit RLC with L=160 mH, C=99 μF and R=68Ω.

    Calculate
    a) the impedance of the circuit
    b) the maximum current

    My solution
    a)
    If v(t) = 40 sin(100t) --> ω=100Hz

    Z= [itex]\sqrt{R^{2} + \left(Xl-Xc\right)^{2}}[/itex]

    Z = 108,86 Ω

    b)
    I know that
    [itex]ic(t)= C\frac{dv}{dt} -> v(t)= \frac{1}{C}\int idt[/itex]
    [itex]v(t)=Ri(t)[/itex]
    [itex]v(t)= L \frac{di}{dt}[/itex]

    [itex]v(t)=Ri(t) + L \frac{di}{dt} + \frac{1}{C}\int idt[/itex]

    Ok, I'm stuck here. I know v(t), R, L and C. But I'm not sure how to get it.

    Thanks for you help
     
  2. jcsd
  3. Jul 21, 2012 #2

    TSny

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    Can you use your result from part (a)? [Hint: Is there an AC analog to Ohm's law?]
     
    Last edited: Jul 21, 2012
  4. Jul 21, 2012 #3
    I guess you could. I'be been thinking this exercise a couple of hours and I miss it. But you can use the sinusoidal variables with the rest?
    Z=V/I
     
  5. Jul 21, 2012 #4

    TSny

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    You may use Z = V/I to relate the maximum voltage and the maximum current.

    But remember that the maximum voltage does not occur at the same instant of time as the maximum current (phase shift). So, you should not try to use Z = V/I to related the current and voltage at the same instant.
     
  6. Jul 22, 2012 #5
    Ok, but if I want to calculate the maximum current I'm going to need the [itex]v(t)=Ri(t) + L \frac{di}{dt} + \frac{1}{C}\int idt[/itex] ecuation, and I'm stuck there.
    Otherwise, there is any other way to obtain the max current with the data I have?
     
  7. Jul 22, 2012 #6

    TSny

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    Yes, As I indicated above. You can use Z = V/I to relate the maximum voltage and current. Thus, solve this equation for I and plug in the values for Z and the maximum voltage.
     
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