Impedance and AC Analysis 1

Introduction 
             Inductor is one of circuit element that we will explore in this lab today. A real inductor model must include a series resistance to account for the many resistance of the many turns of wire. In this lab, we will try to characterize a real inductor. We will also observe what happen to the AC signal if we have an RLC circuit

Experimental
The first step that is done is to measure the resistor value of the inductor which appears to be 9.5 ohms

Next, the external resistance was measured and it appear to be 69.1 ohms


The FG was then energized with a sinusoid output of 5000Hz frequency.

The voltage of this setting turns out to be 2.01V

Once the components have been measured, the following circuit was built

Circuit diagram for the RC circuit

The o-scope graph of the RC circuit

Data for the RC circuit:

Vin,rms = 0.884V
Iin,rms = 5.4 mA

The voltage reading differ from the FG display value probably due to the FG's internal resistance.

The impedance can also then be calculated as follow:

Z = V/I = 164ohms

Once impedance is solved, the value of the unknown inductor can be solved:
Z = Rr +Rl +jwL
L = (Z -Rr -Rl) / jw
w = angular frequency = 2*pi*f = 31416rad/s
L = 0.00272H

Part 2- RLC Circuit

First we have to calculate the value of the capacitor required to cancel out the inductance from before. The calculation is as follow:

jwL = -1/jwC
jwL = j/wC

canceling j gives the expression:
wL= 1/wC
C = 1/w^2*L
C= 23.3 nF

Once the capacitance is calculated, the following circuit is built:

RLC circuit

o-scope projection of the RLC circuit at 20kHz

 
Data:
Vpk-pk,ch1 = 2.66V
Vpk-pk,ch2 = 6.04V

Time difference = 17.57µs

From the data phase angle can be calculated

The data for different frequency is also tabulated below:

Frequency	Vin	Iin	|Zin|
5kHz	1.95V	0.1mA	19.5kΩ
10kHz	1.84V	0.1mA	18.4kΩ
20kHz	0.92V	0.4mA	2.3kΩ
30kHz	1.31V	0	∞
50kHz	1.84V	0	∞

Questions

1. The input current is largest in 20kHz because we set the capacitive and inductive impedance to be equal in this frequency, which makes the imaginary part of the circuit impedance equal to zero
2. ZL = RL + jωL
   
   Zin = (Rext + RL) + jωL
   
    V = 6.04V
   
   VL = (ZL/Zin)*V = 4.22 at 86.2°
   
   Vpp,ch2/√2 = 4.27V
3. when we decrease the frequency, the circuit will look more capacitive since decreasing frequency means increasing capacitive impedance.
4. When we increase frequency, the circuit will look more inductive since increasing frequency means increasing the inductive impedance