Need Direction for a couple of AC circuit problems

Discussion in 'Homework Help' started by wildnixon, May 28, 2008.

  1. wildnixon

    Thread Starter Member

    May 1, 2008
    I need direction to start 3 problems on my midterm:
    question #4: what is the series-parallel relationship with the reactants? My first inclination is to group the capacitor and the resistor as parallel, and then add them in series with the inductor, and then go from there. What do you think?

    question #5: I am at a loss here; I think I would first use the superposition theorem and remove the left Vs and then do the same for the other and add the two; I'm just not sure what is going over V0 between points a and b?? Do I treat it as 3 different voltage sources?

    question #6: My first inclination here is to treat this as two different circuits and do branch currents.

    Any advice is appreciated!
  2. prussell226

    New Member

    May 15, 2008
    4.Reactance are not added. The impedance of the capacitor(-jxc=-j*(1/wc)) or the impedance of the inductor(jXL or Jwl) can be added if the elements are in series or simplified if they are in parallel.[w stands for the greek letter omega. w=2*pi*f where f is the frequency].The 4 ohms resistor is in parallel with the impedance of the capacitor -j8.
    4|| -j8= 32ohms at -90 degrees divided by 8.944 ohms at -63.43 degrees which is 3.578 ohms at -26.57 degrees. So the impedance of 4|| -j8= 3.578 ohms at -26.57 degrees. Now add the impedance of 3.578 ohms at -26.57 degrees with j10. Convert 3.578 ohms at -26.57 degrees to rectangular form first. After adding this place it back in polar form and use Ohm's law to find the current Ix. To find the current Iy use the current divider rule.

    5. you can use superposition like you said. If you remove the 10v sourcefirst your replacing it by a short circuit. Doing this Vo doesnt exist anymore because the circuit has been modified. You can call it Vo1 if you would like. Vo1 is across the terminals ab which further is across the impedance of the inductor j10 ohms. Finding the voltage across the impedance of the inductor(j8ohms) will result in Vo1. After this go back to the original circuit and short the voltage source with value 8v. The voltage across ab can now be called Vo2 .Vo2 is now across the impedance of the capacitor -j8ohms. Finding the voltage across the impedance of the capacitor (-j8ohms) will result in Vo2. Vo=Vo1 + Vo2. The current Ix can be found using KVL and then ohms law.

    6. Superposition can be used again. If the 10v source is shorted out the impedance of the inductor j10 will be in parallel with the short circuit placed from a to b. The resistor of a short circuit is consisdered to be 0ohms. This means that the impedance j10ohms will be shorted out. Now just solve the simple series circuit for the current Iy. Do the same to find Ix. Short the 8v source. The impedance of the capacitor will then be in parallel with the short circuit. This will result in a short circuit
    Now solve the simple series network to find Ix.Now that two current are known use KCL to solve for Iz.
  3. wildnixon

    Thread Starter Member

    May 1, 2008
    Thanks!! I am working with a group of my classmates on this right now, and we are all matching up!