- #1
Sirsh
- 267
- 10
Hey guys, Just need some help with this question, It asks for the answers to be in circuit parameters:
http://tinypic.com/r/k0k2ed/5
For some reason I don't think my answers are correct as I'm finding it hard to make sense of my answers to the questions.
a)Find the voltage Vab (magnitude and angle)
Vab = Va-Vb
current going to the first two resistors: I1=(V[itex]\angle0[/itex])/(2R)
current going through resistor and inductor: I2 = (V[itex]\angle0[/itex])/(R+Xl[itex]\angle90[/itex])
Itotal = I1+I2 = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex])
Va=It*2R = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex])*2R = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(R+2RXl[itex]\angle90[/itex])
Vb = It*(R+Xl[itex]\angle90[/itex]) = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+RXl[itex]\angle90[/itex])
Vab = Va-Vb
Therefore: (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(R+2RXl[itex]\angle90[/itex]) - (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+RXl[itex]\angle90[/itex]).
Not sure if that's right.. A little insight would be appreciated!
b) Calculate the active power dissipated in each reistance.
Power dissipated in resitor/s (R) = I^2*R = ((V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex]))^2*R
Power dissipated in inductor = 0.
c) Calculate the reactive power dissipated in Xl.
Q = I^2*jXl = I^2*Xl[itex]\angle90[/itex] = ((V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex]))^2*Xl[itex]\angle90[/itex]
d) calculate the total current
Now this is where I am confused it asks me to calculate the total current now even though i thought I already have.. Is this wrong?
Thanks a lot guys.
http://tinypic.com/r/k0k2ed/5
For some reason I don't think my answers are correct as I'm finding it hard to make sense of my answers to the questions.
a)Find the voltage Vab (magnitude and angle)
Vab = Va-Vb
current going to the first two resistors: I1=(V[itex]\angle0[/itex])/(2R)
current going through resistor and inductor: I2 = (V[itex]\angle0[/itex])/(R+Xl[itex]\angle90[/itex])
Itotal = I1+I2 = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex])
Va=It*2R = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex])*2R = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(R+2RXl[itex]\angle90[/itex])
Vb = It*(R+Xl[itex]\angle90[/itex]) = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+RXl[itex]\angle90[/itex])
Vab = Va-Vb
Therefore: (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(R+2RXl[itex]\angle90[/itex]) - (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+RXl[itex]\angle90[/itex]).
Not sure if that's right.. A little insight would be appreciated!
b) Calculate the active power dissipated in each reistance.
Power dissipated in resitor/s (R) = I^2*R = ((V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex]))^2*R
Power dissipated in inductor = 0.
c) Calculate the reactive power dissipated in Xl.
Q = I^2*jXl = I^2*Xl[itex]\angle90[/itex] = ((V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex]))^2*Xl[itex]\angle90[/itex]
d) calculate the total current
Now this is where I am confused it asks me to calculate the total current now even though i thought I already have.. Is this wrong?
Thanks a lot guys.