Impedance and AC analysis I

Purpose: analyze impedance the inductor and capacitor in a AC circuit.

Background: Introduce  that a real inductor model must include a series resistance to account for the resistance of the many turns of wire and thus appears as series.

Zl.real = RL+ jωL

 Vin/Iin= Zreal=sqrt(RL^2+ jωL^2)

steps:
step 1. RL=3.4Ω
step 2.
Rext=68.5Ω
DMM reading: 4.54V

Vin= 4.5V
I in,rms=87.3mA
f=20kHz

-make computations to get inductor value.
Z=V/I=4.5/87.3m= 51.546Ω
Z=sqrt(Rext^2+RL^2+(jωL)^2)
ω=20k*2π=1.257*10^5
Z=sqrt((68.5+3.4)^2+(jωL)^2), 51.55^2=71.9^2-(1.257*10^5)^2*L^2
L^2=1.59*10^-7, L=4.0*10^-4H=0.4mH
L=0.4mH

step 3. Investigating a series RLC circuit

make the inductor and capacitor's impedance cancel up.
X= ωL-1/ωC
ωL=1/ωC.
C=1/(ω^2*L)= 1/((1.257*10^5)^2*0.4m)=158nF
The capacitot we use in the circuit is 151nF

The frequence we get max frequence is 12.7kHz
Vpp.ch1=21.2
Vpp.ch2=19.5
Δt=19.46μs
ΔΦ=19.46μ*2*12.7k*360=88.971

	Vin(V)	Iin(mA)	Zin
5kHz	5.78	29.3	197.2696
10Khz	5.22	65.1	80.18433
12.7kHz	4.95	73.5	67.34694
30kHz	5.56	23.8	233.6134
50kHz	7.62	1.7	4482.353

Follow-up question.
1.ans: when in maximum current, the ωL=1/ωC and make the Zeq be the minimum. Therefore, I=V/R and make the current be the maximum.
2.
3.ans:The circuit will me more capacitive(ωL<1/ωC)
4.ans:The circuit will me more inductive(ωL>1/ωC)