Calculating Impedance in an RLC circuit

[lpau001]

Homework Statement

Calculate the impedance Z_ab across ab.
w=0.75 Hz
w_o=1 Hz.
L=1 H
R=1 Ohm
C=1 F

Inductor and Capacitor are in series, and together are in parallel with the resistor.

circuit looks like a capital A, but square on top, and rotated 90 degrees clockwise.. if that makes sense.

Homework Equations

Z_eq=sqrt(R² +(X_L-X_C)²)
X_C=1/(wC)
X_L=wL

The Attempt at a Solution

Using the equations above, I found X_C to be 4/3 and X_L to be .075.
inputting those in the impedance equations, I get
Z=sqrt(1² + (4/3-.75)²)
which is 1.1577 which is wrong..

 

[gneill]

If I understand your circuit layout, you've got a resistance in parallel with a reactance (the net result of the sum of the capacitive and inductive reactances). You don't just sum them to get the result, they combine like parallel resistances.

It might be easier to use the complex form for the impedance and do the math.

Zr = 1Ω, Zx = -0.583jΩ, then the net impedance is

Z = Zr*Zx/(Zr + Zx)

 

[lpau001]

If I understand your circuit layout, you've got a resistance in parallel with a reactance (the net result of the sum of the capacitive and inductive reactances). You don't just sum them to get the result, they combine like parallel resistances.

It might be easier to use the complex form for the impedance and do the math.

Zr = 1Ω, Zx = -0.583jΩ, then the net impedance is

Z = Zr*Zx/(Zr + Zx)

I actually did try this in an earlier attempt, because I thought this was right, but apparently not.
1/Z=1/Zr + 1/Zx

I'm stuck, and I actually ran out of attempts on the HW, but this is more out of curiosity now than anything.

Thanks, Gneill

 

[gneill]

Z = Zr*Zx/(Zr + Zx)

= 1*(-0.583j)/(1 + -0.583j) Ω

= -0.583j/(1 - 0.583j) Ω

Normalizing,

= 0.254 - 0.435j Ω

This has magnitude |Z| = 0.504 Ω, and phase angle -59.7°