Procedure:
We start with a real inductor connected in series with an external resistance.
2 Multimeters are taking measurements.
The resistor RL is the resistance in the inductor.
(note that a real inductor is consider an ideal inductor in series to a resistor)
The measurement for circuit elements:
RL = 8.3 Ω
Rext = 67.2 Ω
Vrms = 5V at 1000 Hz.
 |
| The original circuit. |
Iin,rms = 65.0 mA
The voltage reading differs from the FG because internal resistane drop 0.7 V
Vin/Iin = ZL = 75.8 Ω
substituding ZL in the following equation:
ZL = √((Rext + RL)^2 + (ωL)^2)
we get:
ω = 6280 rad/s
L = 1.16 mH
Adding a capacitor to cancel inductance impedance
1/ωC = ωL
C = 1/(ω^2*L) = 2.19 *10^-5 F
 |
| The remodified circuit |
The signal of the circuit displayed in a oscilloscope.
Vpp, Channe1 = 1.5V
Vpp, Channel2 = 20mV
ΔT = 0.4 ms
Phase difference = (ΔT) *6240* 360 ° = 184.32 °
Data & Data Analysis:
Frequency (kHz)
|
V_in (V)
|
I_in (A)
|
|Z_in| (Ω)
|
5
|
4.6
|
0.073
|
63.0137
|
10
|
4.2
|
0.0719
|
58.41446
|
20
|
3.51
|
0.0691
|
50.79595
|
30
|
3.51
|
0.0653
|
53.75191
|
50
|
4.5
|
0.0558
|
80.64516
|
Questions:
1) It is not the largest at 20 kHz as we used 1 kHz for calculations. The largest current should be at 1 kHz as imaginary part cancels out.
2)
VL = ZL/Z*Vin
VL = 2.8574 + 1.2737i
|VL| = 3.1284
Phasor = 24.0243 °
3) The circuit looks more inductive as imaginary part is always positive.
4) The circuit looks more inductive as imaginary part is always positive.
Conclusion:
The experiment was quite a success. Yet, some answers to the questions were off from the lab's original expectations as the original lab suggested a frequncy of 20kHz instead of 1 kHz (the actual frequency used) in the circuit, and in the calculations.
No comments:
Post a Comment