AC-circuit: Finding the current

Discussion in 'Homework Help' started by Niles, Nov 23, 2008.

  1. Niles

    Thread Starter Active Member

    Nov 23, 2008
    56
    0
    Hi all.

    Please take a look at the attachted circuit.

    I have found the following expression for the complex current in the resistor (the hat indicates that it is complex):

    \widehat_{I (t)} = \frac{{\omega ^2 LC {\widehat{U_0} }}}{{\omega ^2 LRC + i\omega L - R}}\exp ( - i\omega t),

    where U_0 is the amplitude of U(t), and the hat indicates that it is complex.

    Now I wish to find the real current, and I want to write it as:
    I(t) = I_0\cos(\omega t + \phi).

    But how do I do this? I have spent like 2 hours trying, but I don't
    know how to rewrite the complex amplitude of the current to have a
    phase.


    Thanks in advance.



    Regards

    Niles.
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    A(cos(a)+isin(a))=Aexp(ia),

    where
    a is the angle in rads and
    A is the amplitude

    use this to convert your formula to a cos+isin version.

    Or you can re-analyze the circuit but use jωL for the impedance of the inductor and 1/jωC for the impedance of the capacitor and analyze the circuit using the voltage and current divider equations, like you do with resistances.
     
  3. Niles

    Thread Starter Active Member

    Nov 23, 2008
    56
    0
    Thanks for replying.

    I have found the complex current, and it is now on the form:

    I_complex (t) = 0.0015 * exp[-i(ωt-φ)],

    where φ is the phase, and φ = 5.65.

    Does this mean that the real current is given as:

    I_real(t) = 0.0015 * cos[ωt+(-φ)],

    where φ = 5.65?

    Thanks in advance.
     
  4. blazedaces

    Active Member

    Jul 24, 2008
    130
    0
    How are you arriving at these constant values? Your circuit has only variables in it... so shouldn't your final answer contain only variables as well?

    -blazed
     
  5. Niles

    Thread Starter Active Member

    Nov 23, 2008
    56
    0
    I am given some numerical values that I've plugged in.
     
  6. blazedaces

    Active Member

    Jul 24, 2008
    130
    0
    Alright. Then I just have one comment, just in case you forgot: you have an imaginary number in the denominator of your fraction. Just make sure to remove it by multiplying the top and bottom by the conjugate...

    Cheers,

    -blazed
     
Loading...