Rlc problem: calculations of components and phasor drawing

[XYZ^2]

Homework Statement

Find impedance and Io.
draw phasor diagram for rlc circuit
V = 120cos(2π525t) where R = 10Ω, C = 1μF, L = 100mH, Vo = 120V

Homework Equations

Z = √(R2 + (ωL-1/(ωC))^2) = √(10^2 + (2π * 525 Hz * 100 mH - 1/(2π * 525 Hz * 1 μF ))^2) = 28.525Ω

Io = Vo/Z = 120 V / 28.525 Ω = 4.207 A

tan ϕ = (ωL - 1/(ωC))/R = (2π525 * 100 mH - 1/(2π525 *1 μF)) / 10 Ω = -15882.508
ϕ = -1.5707 rad = -89.996°

V_Ro = IoR = 4.207 A * 10Ω = 42.07 V
V_Co = Io/(ωC) = 4.207 A / (2π525 * 1 μF) = 1275.361611 V
V_Lo = IoωL = 4.207 A * 2π525 * 100 mH = 1387.751431 V

Vo = √(V_R0² + (V_Lo- V_Co)^2 ) = 120.0056523 V

The Attempt at a Solution

I get a negative phase angle which should mean that capacitance is greater than inductance but this is not the case based on values above.
when I draw out a rough phasor diagram I have voltage ahead of current.
not sure where i went wrong.
note: diagram attached is rough and not to perfect scale

 

[technician]

I calculated Xl and Xc separately and got Xl = 329Ω and Xc = 303Ω
This gave me an impedance of 28Ω and I = 4.3A
My (Xl-Xc) = 26Ω which gives Tan∅ = 2.6 (leading)
My phasor diagram would be 329Ω on the +y axis, 10Ω on the + x-axis and 303Ω on the -y axis (I don't know how to get drawings on here yet !
My values are pretty much the same as yours !
I prefer to work out individual quantities rather than lump every thing together in one equation.
I cannot see where our answers differ !
Hope this helps
Just ocurred to me... did you change mH into H and μF into F in your phase angle calculation?

 

[XYZ^2]

thanks
i did convert mH and μF in my calculations to H and F...
I finally realized that I had left out the 525 in calculating XC. now i get tanϕ = 2.671 , which is voltage leading...which soothes my brain

working out individual quantities probably would have saved me the headache. next time i won't plug in the whole thing in excel.

 

[technician]

Well done... no lack of understanding, that is the main thing.