# AC Circuit using circuit parameters - HELP!

1. Apr 16, 2012

### Sirsh

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$\angle0$)/(2R)
current going through resistor and inductor: I2 = (V$\angle0$)/(R+Xl$\angle90$)
Itotal = I1+I2 = (V$\angle0$(3R+Xl$\angle90$))/(3R+2RXl$\angle90$)

Va=It*2R = (V$\angle0$(3R+Xl$\angle90$))/(3R+2RXl$\angle90$)*2R = (V$\angle0$(3R+Xl$\angle90$))/(R+2RXl$\angle90$)

Vb = It*(R+Xl$\angle90$) = (V$\angle0$(3R+Xl$\angle90$))/(3R+RXl$\angle90$)

Vab = Va-Vb
Therefore: (V$\angle0$(3R+Xl$\angle90$))/(R+2RXl$\angle90$) - (V$\angle0$(3R+Xl$\angle90$))/(3R+RXl$\angle90$).

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$\angle0$(3R+Xl$\angle90$))/(3R+2RXl$\angle90$))^2*R

Power dissipated in inductor = 0.

c) Calculate the reactive power dissipated in Xl.

Q = I^2*jXl = I^2*Xl$\angle90$ = ((V$\angle0$(3R+Xl$\angle90$))/(3R+2RXl$\angle90$))^2*Xl$\angle90$

d) calculate the total current
Now this is where Im confused it asks me to calculate the total current now even though i thought I already have.. Is this wrong?

Thanks alot guys.

2. Apr 16, 2012

### Staff: Mentor

Is there a schematic to accompany this?