Shunt Inductor for Power Factor Correction

Discussion in 'Homework Help' started by ecm2719, Aug 22, 2015.

  1. ecm2719

    Thread Starter New Member

    Aug 22, 2015
    4
    0
    Hello, I am reviewing circuits for my upcoming semester and was curious about capacitive loads and power factor correction. I worked through the proofs for an inductive load and the power triangle where I add a capacitor in parallel to reduce the phase angle and approach unity. I am doing the same for a capacitive load and seem hung up on adding the shunt inductor. It seems like adding the inductor in parallel to the capacitive load is correct due to the reduction in current. Can anyone verify this please? It seems obvious, but I learned this at the tail end of the semester when classes were hectic and have not really returned to it before this week. Thank you in advance for any clarity and help.
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    in DC circuits, adding loads in parallel means there will be alternate paths for current and total resistance of the circuit is reduced.

    in AC circuits this generally also holds. exception are reactive parallel loads of different type. in this case it is possible to partially (or completely) cancel out current due reactive components. when circuit is purely reactive (no resistive component) and L and C have equal magnitude of reactance, circuit is said to be in resonance. parallel LC circuit in resonance has infinite impedance. since this happens at specific frequency, it is used as key factor in radio for example. in case of power transfer, we are primarily interested to reduce current to what it really needs to be so we can use thinner conductors. being off by few percent is not an issue in this case.
     
  3. ecm2719

    Thread Starter New Member

    Aug 22, 2015
    4
    0
    Thank you for the reply. I am not sure if I understand or if I was not specific enough. Given a voltage source in series with a resistor and inductor as the load, I know I can place a capacitor in parallel with the resistor and inductor with the result being a smaller angle between voltage source and current. This in effect increases the power factor by bringing it nearer to unity. This makes sense.

    If I have a voltage source in series with a resistor and a capacitor as the load, does the same concept apply? If I add an inductor in parallel, a reduction in the current occurs across the load occurs and the power factor again increases. It makes sense to me the inductor would need to be in parallel to have an effect on the circuit.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    When you talk about adding the shunt capacitor to a resistor/inductor load you talk about reducing the angle between the source voltage and current. Why do you then shift and talk about reducing the current when talking about adding a shunt inductor to a resistor/capacitor load? Why aren't you still looking at reducing the angle between the source voltage and current, since that is what matters in power-factor correction?

    The idea in both cases is to recognize that the load has both real and reactive power and to satisfy the reactive power needs via the shunt component so that the source only has to supply the real power needs.
     
  5. ecm2719

    Thread Starter New Member

    Aug 22, 2015
    4
    0
    I am looking at all of it as a whole I think. My textbook talks about both the change in angle and the change in current through the load after adding a reactive component. I think I finally figured out where I was hanging up. My real curiosity was finding out what causes the reduction in the angle and how does it relate to the reduced current. I went back to my basic formulas for apparent, complex, real and reactive powers and power factor and was able to find the relationship I needed to make it clear.

    Sometimes the textbook makes leaps and I have a hard time moving on without knowing why. For example, the text does not offer a proof for power factor correction for a resistive-capacitive load, citing the more prevalent use of inductive loads. The book just gives the formula. I searched a bit online and posted before I realized I was making a simple error in my thought processes and correcting it. This is why I review before the semester. Very rusty. Thank you for the response.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    Glad you got it figured out. Good luck on the upcoming semester.
     
  7. ecm2719

    Thread Starter New Member

    Aug 22, 2015
    4
    0
    Thank you. It's a little overwhelming at times. I'm finally at a point where I am in all EE related classes, so I am hoping it starts to come together now. I'm glad I found this website. I've been looking for a place where I can learn from experienced professionals.
     
Loading...