My AC Circuit by Ghafour https://padlet.com/ghafouram/3e3zgo6tkb31 Transmission System en-us 2016-10-22 23:19:46 UTC 2023-03-05 06:41:35 UTC hello@padlet.com Ghafour ghafouram https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132514990 This is a transmission system to deliver electrical energy to the load.
When the load increased in the system, the total current of the system will be increased. How the power dissipated on the transmission line change?
In other words, you need to calculate the active power of the resistance R1 in the two cases and compare them. Explain your comparison. ]]> 2016-10-22 23:22:09 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132514990 Keifer https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132520422 Before : 125 W 
After : 201.69 W

Since the latter draws a higher current due to the inductor, the total power dissipated will be increased as indicated by the formula P=I^2.R]]> 2016-10-23 03:58:39 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132520422 Hew Wee Seong https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132528173 Part 1
V = 100V
R = 20 ohm
I = 5A #
P = 500W #

Part 2
V = 100V
Z = 5.067 - 0.067i
   = 5.067<-0.76
I = 19.73 + 0.26i
  = 19.74<0.76 
S= 1973.34 - 25.97i
    = 1973.52<-0.76
P = 1973.34
Q = -25.97


]]> 2016-10-23 09:10:58 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132528173 Sam https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132528889 Case 1:
Power dissipated on R1=5^2x5=125W

Case 2:
total Z=12.5+j7.5
mag of total current=mag(V/Z)=6.86 A
Power dissipated on R1=6.86^2x5 = 235.3W

By comparison, more power dissipated on the transmission line of case 2 because the circuit is inductive .]]> 2016-10-23 09:27:42 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132528889 Ea Jia Tong (0318332) https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132533876
  • 1st case:
    • V=100V ; R1=5ohm ; RL=15ohm
    Current =V/total R
                  = 100/(5+15)
                  = 5A#
    Power dissipated = I^2*R1
                                    = 5^2*(5)
                                     =125W#
    • 2nd case:
    V=100V ; R1=5ohm ; RL=15ohm ; XL=j15ohm
    Total Impedance, Z = (XL*RL)/(XL+RL)+R1
                                       = (j15*15)/(j15+15)+5
                                       =12.5+j7.5 ohm#
    Magnitude of Power = 14.58 watts#
    Current = V/total Z
                   = 100/14.58
                   = 6.86 A#
    Power dissipated = I^2*R1
                                    = 6.86^2*(5)
                                     = 235.298 W#

    • Explanation:
    In comparison, with a higher amount of current due to the inductor, the power dissipated is increased. Therefore, in a inductive circuit, there is more power dissipated. #
    ]]>     2016-10-23 11:31:40 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132533876     Diong Han Wen (0324178) https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132539268 2016-10-23 13:24:17 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132539268 Yi khai https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132544222 Part 1
    Vin = 100Vrms ; R1 = 5ohm ; RL = 15 ohm

    I = Vin/(R1 + RL)
    = 100/(5+15)
    = 5A

    Power at R1 = (I^2)*R1
                           =25*5
                           = 125W

    Part 2
    Vin=100Vrms ; R1=5ohm ; RL=15ohm ; XL=j15ohm

    Z= (XL*RL)/(XL+RL)+R1
       =(j15*15)/(j15+15)+5
        =12.5+j7.5 ohm

    I=Vin/Z
    = 100/(12.5+j7.5)
    = 6.89 A (with angle -30.1 degree)

    Power at R1 = (I^2)*R1
                           =47.4721*5
                           = 237.36W

    The power dissipated at R1 is higher for the second circuit due to the inductor in the circuit that will draw higher current.]]>     2016-10-23 14:53:36 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132544222     Loon https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132746884 2016-10-24 15:47:20 UTC https://padlet.com/ghafouram/3e3zgo6tkb31/wish/132746884