Circuit Analysis finding Voc for circuit

Discussion in 'Homework Help' started by damm1371, Apr 12, 2012.

  1. damm1371

    Thread Starter New Member

    Feb 22, 2012
    The objective of this problem is to find Vo in terms of voltage magnitude and phase angle in degrees. Given: The current generator supplies 1 amp at 60 Hz and 0 degrees.
    If 60 Hz is given the voltage will be sinusoidal.

    V = I * z
    z = R + jx ( j is -sqrt(1)) (x is reactance)
    using this converting x into ohms
    x= 1/ (j* pi* f * C1) ( now in ohms)

    KVL at far left loop:
    0 = Vc1/(R+ 1/(j*2pi* f * C1) + I R4
    with this I think you can find V@C1
    Then KCL at far left node:
    0= Vc1/1 + V4/4 - V1/R1

    I am not sure how to attack the inductor. Please help.
    Last edited: Apr 12, 2012
  2. WBahn


    Mar 31, 2012
    I really looks like you are just throwing equations against the wall and hoping something sticks. Several of your equations are not even dimensionally consistent nor are the components involves clearly indicated.

    In your first post, you say z = R + jx but then you say x is 1/(j*pi*f*C1). If this is the case, the the j's cancel out. Also, you are missing a factor of 2 (as in, 2pi). But which are are you using here?

    You next give:
    Aside from the missing close paren (which I assume goes after the C1), the first term is current (voltage divided by impedance) while the second term is voltage (current times resistance). And what's Vc1? Is it the voltage across the capacitor? If so, what polarity?

    Then you have:

    Were do the 1 and 4 come from? What are they? For that matter, what is V4 and V1? Are they the voltages across R4 and R1? If so, what's the polarity. Or are they the voltages at certain nodes? If so, which nodes?

    In your second post, you have:

    Y is admittance and, like conductance, has units of 1/ohm (or mho or seimans) while reactance has units of ohms. You can't add them. and I = V*Y.

    You are also being very sloppy with the j's.

    Reactance is the imaginary part of the impedance. Hence:

    <br />
Z = R + jX<br />
<br />
X_L = 2 \pi f L<br />
X_C = \frac{-1}{2 \pi f C}<br />
<br />

    In light of all of this, my first question to you is: if this were a DC circuit and the capacitor and inductor were replaced with resistors, are you confident that you could solve for Voc? If so, then simply replace each component with its complex impedance (either 1/(jwC) for a capacitor or jwL for an inductor), replace the current source with its phasor equivalent, and solve it exactly the same way you would have in the DC case.

    If not, then spend some quality time working DC analysis problems until you are very comfortable with them.

    And make a point of not being sloppy with your notation. Make it clear what voltage and/or current you are refering to and what the polarities are. This is best done my annotating the schematic.
    damm1371 likes this.
  3. damm1371

    Thread Starter New Member

    Feb 22, 2012
    my apologies. I will clean it up. Thank you.