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Reactive power control

  1. Mar 28, 2014 #1
    Friends,i was asked once the relationship between reactive power and voltage.ques is"if i am providing reactive power at some particular bus,voltage at that bus is gonna improve".why??i answered that if u will provide reactive power remotely to that bus by some shunt compensation device then the reactive power would not be taken from source.Hence it does not have to travel distance and IXL drop wont be created and desired voltage profile would be maintained.However examiner was not satisfied by the answer.Can anyone plz help me out here to provide me a proper explanation between reactive power and voltage????..plzzzz
  2. jcsd
  3. Mar 28, 2014 #2
    The voltage drop[ per phase] is [approximate]:
    VS-VR=I*[R*cos(a)+X*sin(a)] where :
    VS=voltage at source
    VR=voltage at receiver
    R and X-the resistance and reactance of the cable from source up to receiver.
    If a capacitor is inserted in series with cable then:
    X=XL-XC where XL=cable reactance XC= capacitor reactance.
    If X decreases then VS-VR decreases also.
    If a capacitor is inserted in parallel with the receiver then the capacity current will reduce sin(a) canceling the inductive current of the receiver and then if sin(a) decreases even cos(a) increases the total voltage drop will decrease. For instance:
    R=1 ohm X=1 ohm cos(a)=0.5 sin(a)=sqrt(1-0.5^2)=0.866 I=10 A
    VS-VR=10*(1*0.5+1*.866)=13.66 V
    Now let’s say a capacity inserted in series will delete the cable reactance. Then:
    VS-VR=10*(1*0.5)=5 V
    If the capacitor is in parallel with the receiver then cos(a)=1 and sin(a)=0 .Then:
    VS-VR=10*1*1=10 V
    Actually the current I will decrease also since I=SQRT(Iactive^2+Ireact^2).
    Iactive=I*cos(a)=10*.5=5 A
    Ireact=I*sin(a)=10*.866=8.66 A
    I=SQRT(5^2+8.66^2)=10 A
    Now, if sin(a)=0 then I=Iactive=5 A and the voltage drop will be VS-VR=5*1=5 V.
  4. Mar 28, 2014 #3
    Thankz for d response babadag. AS far as mathematical xplanations are concerned u are absolutely right.But i am not able to understand the theoretical insight.For example inductor which stores energy in magnetic field is said to be reactive power absorbers and capacitor which stores energy in electric field are said to be reactive power providers.WHY????.it is reactive power which builds magnetic field and which builds the voltage.How electric field by the capacitor helps to solve the purpose
  5. Mar 28, 2014 #4
    The alternative current flowing through the inductor generates a variable magnetic field. This magnetic field will generate a variable EFM and it will draw a power from the supply source but only in a quarter of a cycle and in the second quarter will return the power in the circuit. The total power remains not changed only the current rms will grow up with the reactive current. As expected this reactive current will reappear in the circuit retarding 1/4 cycle. So only 90 degrees later the magnetic field will generate an EFM opposing to the supply voltage.
    The electric field generated in a capacitor will increase gradually as the electric charge will be accumulated. That means from the beginning of the supply process a voltage returned by the capacitor will diminish the current up to zero when the potential between the electrodes will be equal to supply voltage. At the beginning the current is elevated and will decrease as supply voltage increase. As supply voltage decreases-from the peak toward the zero-the capacitor starts discharging-the current flows in the opposite sense.
    If the supply voltage is positive and in rising the capacitive current decreases and so the current returned by magnetic field but it flows in opposite sense. When the supply voltage decreases the capacitive current change the sense and start to grow. The magnetic produced current start to grow also but in opposite sense and so will oppose the capacitive current all the time.
  6. Mar 28, 2014 #5


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    Staff: Mentor

    The essence of power factor correction is that it results in less current being drawn from the source, while suppying a given load. The lower line current results in less voltage loss over the transmission line, so your load sees its intended voltage and this with improved regulation. The lower current from the source makes the power generator and utility companies happy for they need to provide less current while receiving the same revenue.
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