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.

 

[asteriatic]

, but it should give a general idea of the phasor relationships in this circuit.
Based on your calculations, it seems that you have correctly determined the impedance (Z) and the current (Io) in the RLC circuit. The negative phase angle can be attributed to the fact that the inductive reactance (ωL) is greater than the capacitive reactance (1/(ωC)), leading to a net inductive effect in the circuit. This can happen even if the values of L and C are not exactly equal, as long as the inductance is significantly larger than the capacitance.

In terms of the phasor diagram, the voltage should indeed be ahead of the current, as the voltage across the inductor (VLo) and the voltage across the capacitor (VCo) are out of phase with each other. This can be seen in the equation for Vo, where the two voltages are added together using the Pythagorean theorem. The phasor diagram you have drawn is consistent with this relationship.

Overall, it seems that your calculations and phasor diagram are correct and you have a good understanding of how to solve for impedance and draw phasor diagrams in RLC circuits. Good job!