serie parallel RLC circuit current

Question

I got a problem with how to calculate the current going through R and L in this circuit.

I have calculated the following.

Xc= 79.6Ω at -90°

Xl = 251.3Ω at 90°

ZRL = 128.8Ω at 30.8° (R||L)

ZT = 111.48Ω at -7.03°

Vc = 10.8V at - 83°

Vr = 17.33V at 37.83°

Total current(IT): 0.135 A at 7.03°

everything above this point ive checked in my answer key and they seem to be correct. Now to where im not correct, the current IR and IL.

I use IR = ((XL/(R+XL) * IT) = 0.084A at 97.03° IL = ((R/(XL+R) * IT) = 0.05A at 7.03°

Where am i going wrong? IL and IR are supposed to be IR = 0.115A and IL = 0.069A

Answer 1

The current thru the inductor is simply the voltage across it divided by its inductive reactance at 1000 Hz. So:

I = 17.33 / 251.3 = 0.06896 amps

The current thru the resistor is analogous, the voltage across it divided by its simple resistance: So:

I = 17.33 / 150 = 0.1155 amps

In a circuit like this where you are dealing with pure inductances, pure capacitances and pure resistances you simply apply the "AC" version of Ohms Law:

V = I x Xc, or V= I x Xl, or voltage = current times reactance

Answer 2

It is simpler if you use complex numbers:

octave:1> R=150
R =  150
octave:3> Zc=1/(j*2*pi()*1000*2e-6)
Zc =   0.00000 - 79.57747i
octave:4> Zl=j*2*pi()*1000*40E-3
Zl =    0.00000 + 251.32741i
octave:5> Zrl=1/(1/R+1/Zl)
Zrl =  110.603 +  66.011i
octave:6> Zt=Zrl+Zc
Zt =  110.603 -  13.566i
octave:7> I=15/Zt
I =  0.133611 + 0.016389i
octave:8> abs(I)
ans =  0.13461
octave:10> arg(I)
ans =  0.12205
octave:11> Ir=I*Zrl/R
Ir =  0.091306 + 0.070883i
octave:12> abs(Ir)
ans =  0.11559
octave:13> arg(Ir)
ans =  0.66014
octave:14> Il=I*Zrl/Zl
Il =  0.042305 - 0.054494i
octave:15> abs(Il)
ans =  0.068988
octave:16> arg(Il)
ans = -0.91066

I leave to you the conversion of the angle from radians to degrees