AC LR-C parallel circuit

Discussion in 'Homework Help' started by monkeyhead, Feb 29, 2008.

  1. monkeyhead

    Thread Starter Active Member

    Mar 5, 2007
    45
    0
    Hi there!
    Basically I'm ploughing through my assignment and trying to work out the phase angles of a capacitor, of the coil and of the supply.

    I was given the values as follows:

    Supply: 260V @ 50Hz
    Inductor: 31.831mH
    Resistor: 24 Ohm

    I had to work out the value of the cap which I worked out as: 47.1 uF


    My attempt
    Note: Results given rounded to 2dp.

    I worked out that the current through the coil and its phase angle as follows:
    Current: 10A
    Angle: 23º

    ( I did this by working out Xl, then the impedance then using ohms law and results were similar from simulation)


    I also worked out the current in the cap and its phase angle as follows:
    Current: 3.85A
    Angle= 90º (as it leads)

    ( I did this by working out Xc, then using ohms law and results were similar from simulation)

    Now for the supply angle and its current I did the following.

    I worked out the the total horizontal component which was = 9.21A

    Then did the same for the vertical components and that calculated at = -0.06

    Then calculated I by doing the following I= √((9.21)²+(-0.06)²)= 9.21

    To get the angle I used inverse tan as follows:
    tan-1(0.06/9.21)=0.37º

    To me this angle seems way to small?! Have I gone wrong somewhere?

    Many thanks,
     
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