Alternate current problem

Discussion in 'Homework Help' started by ghoasted, Feb 11, 2010.

  1. ghoasted

    Thread Starter New Member

    Feb 11, 2010
    See picture to see the circuit. Both resistors are worth R.
    V(t) is an emf with angular frequency \omega and amplitude V_0.
    1)Find the current that passes through the emf.
    2)Let V_{AB} be worth V_B-V_A. Demonstrate that |V_{AB}|^2=V_0^2 for all \omega.
    3)Find the angle difference of phase between the current passing through the emf and the current reaching a capacitor.

    My attempt:
    V(t)=V_0 \cos (\omega t). Also, V(t)=Zi(t) where Z is the impedance of the circuit.
    As both branches of the circuit are in parallel, the admittance of both branches sum up. G=\frac{1}{Z}.
    So G=G_1+G_2 \Rightarrow Z=\frac{1}{G}=\frac{1}{G_1+G_2}=\frac{1}{\left ( \frac{1}{Z_1}+\frac{1}{Z_2} \right )} where Z_1 and Z_2 are the impedances of each branch, respectively.
    But Z_1=Z_2. Thus Z=\frac{Z_1}{2}.
    Z_1=R-\frac{1}{i\omega C}, thus Z=\frac{R}{2}+\frac{i}{2\omega C}.
    Finally, i(t)=\frac{2V_0 \omega C \cos (\omega t)}{R\omega C+i}, where the i in the denominator is the complex number.
    Am I right for question 1)? Did I make an error?
    For 2), I'm not really sure about how to proceed.
    I believe that V_B-V_A=\frac{i(t)R}{4}-\frac{q(t)}{C}. But I'm unsure and even if it were right, I don't know how to continue. I know that i=\frac{dq(t)}{dt}, but does this help?

    Any help is greatly appreciated! Thanks in advance.
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  2. mik3

    Senior Member

    Feb 4, 2008
    Find the impedance of all the elements in the circuit. Then find Va and Vb with the voltage divider rule or whatever you like. Then do Va-Vb to find the answer.