I still seem to be off!Solving RCL Circuit w/Iron Core at Resonance

[ParoXsitiC]

Homework Statement

The AC power supply (maxium voltage 20.0V) in RCL circuit is run at resonance for the following parameters:

R = 20.0 Ω
L = 88.0 mH
C = 79.9 μF

When an iron core is added to the inductor, the self-inductance is doubled

A) Find the frequency of the generator
B) After the core had been added, find the phase angle and impedance in the circuit.
C) Draw phasor diagram at time 4.1267 ms
D) Find the voltages across each circuit element at this time and show that they follow Kirchoff's voltage rule.

Homework Equations

σ = tan^-1 ([itex]\frac{X_{L}-X_{C}}{R}[/itex])

X_L = ωL
X_C = [itex]\frac{1}{ωC}[/itex]

@ Resonance X_L = X_C
frequency (f) = [itex]\frac{ω}{2\pi}[/itex]
impedance (z) = [itex]\sqrt{(R)^2+(X_{L}-X_{C})^2}[/itex]

The Attempt at a Solution

A)

ω= [itex]\frac{1}{\sqrt{LC}}[/itex] = 377 rad/s
frequency (f) = [itex]\frac{ω}{2\pi}[/itex] = [itex]\frac{377 rad/s}{2\pi}[/itex] = 592 Hz

B)

Assuming ω stays the same, L is now 2*88.0 mH = 176 mH



X_L = ωL = (377 rad/s)(176 mH) = 66.3

X_C = [itex]\frac{1}{ωC}[/itex] = [itex]\frac{1}{(377 rad/s)(79.9 μF)}[/itex] = 33.2

σ = tan^-1 ([itex]\frac{X_{L}-X_{C}}{R}[/itex]) = tan^-1 ([itex]\frac{66.3-33.2}{20 Ω}[/itex]) = 58.9 degrees

impedance (z) = [itex]\sqrt{(R)^2+(X_{L}-X_{C})^2}[/itex] = [itex]\sqrt{(20 Ω)^2+(66.3-33.2)^2}[/itex] = 38.7 Ω

C)

E = E_max sin (ωt) = 20.0 v sin(377 rad/s 4.167 ms) ≈ 20 v

Thus E_max should be vertical, and the phase angle of 58.9 degrees is between Vr (also Imax) and Emax

D)

This is where I am stuck...

I know the vertical components should all equal 0.

Emax = 20 v
V_L = X_L Sin(180-59.8) = 66.3 Sin(120.2) = 57.3 v


V_C = X_C Sin(0-59.8) = 33.2 Sin(-59.8) = -28.7 v


V_R = R sin(90-59.8) = 20 Ω sin(30.2) = 10.1 v

-20 + 57.3 - 28.7 + 10.1 ≠ 0








OK, so this isn't the right way... maybe using Imax..

I_max = [itex]\frac{E_{max}}{z}[/itex] = [itex]\frac{20.0 v}{38.7 Ω}[/itex] = 0.517 A

Emax = 20 v
V_L = X_L I_max = 66.3 * 0.517 = 34.3 v


V_C = X_C I_max = 33.2 * 0.517 = 17.2 v


V_R = R I_max = 20 Ω * 0.517 = 10.3 v

-20 + 34.3 - 17.2 + 10.3 ≠ 0

 

[ParoXsitiC]

I realized my mistake. Combination of both attempts really...

I was assuming since the graph of reactance was similar to voltages that it would make sense to use them in the calculation but now I see I should first find the max voltage using the impedance multipled by max current, and then getting max voltage use it with trig.

For future googler's, here is how it's done.

I_max = [itex]\frac{E_{max}}{z}[/itex] = [itex]\frac{20.0 v}{38.7 Ω}[/itex] = 0.517 A

E_max= 20 v
V_Lmax = X_L I_max = 66.3 * 0.517 = 34.3 v


V_Cmax = X_C I_max = 33.2 * 0.517 = 17.2 v


V_Rmax = R I_max = 20 Ω * 0.517 = 10.3 v


vertical components should all equal 0.

E = E_Max Sin(0) = 20.0 Sin(0) =20.0 v
V_L = V_Lmax Sin(180-59.8) = 34.3 v Sin(120.2) = 29.6 v


V_C = V_Cmax Sin(0-59.8) = 17.2 v Sin(-59.8) = -14.9 v


V_R = V_Rmax sin(90-59.8) = 10.3 v sin(30.2) = 5.18 v

-20 + 29.6 - 14.9 + 5.18 = -0.12 (Due to significant figures and rounding) ≈ 0