Hecter Cruz
Quoc Tran
Ivan Kuang
Thursday, June 13, 2013
Tuesday, May 28, 2013
Lab 15: Impedance and AC Analysis I
Introduction:
Impedance can affect to the AC circuits a lot by changing the frequency. By investigating the RL and RLC circuit, we can determine the impedance of a inductor and a capacitor. In this lab, we use the measurement of the RL to find the impedance of the real inductor. Then, we vary frequency of the RLC circuit to observe how the impedance affect to AC ciruit.
Procedure:
Obtain an inductor and measure its resistance:
RL =2 Ω
Set build the circuit same as the diagram below:
Obtain a resistor:
Rext = 67.6 Ω
Set up the FG at 1k Hz and RMS voltage at 5 V.
DMM reading: 4.99V
Take measurments:
The voltage reading is different from the FG displace value. This is due to the large resistance of the FG and the impedance of the inductor.
Make calculation:
Z= V/I = (3.06V)/(42.3mA) = 72.2 Ω
Rewrite the expression for the input impedance:
Z=Rext + RL + ZLj
Write an equation for the magnitude of the impedance:
Angular frequency:
w = 2*pi*f = 2*3.14*1000 = 6283 rad/s
Set the expression for the magnitude of the impedance equal to the first calculation to solve for the inductance:
Investigating the LRC circuit:
The source is 1kHz, we wish to cancel the inductive part of the real inductor impedance. The value of capacitance required is:
wL = 1/Cw
Set up the circuit:
Vary the frequency and make measurements:
1. The input current is largest at 1kHz because the capacitor and the inductor cancel out at 1kHz. The impedance is smallest at 1kHz, so the current is largest.
2. At 1kHz, the theoretical voltage across the inductor is :
ZL = RL + jωL
Zin = (Rext + RL) + jωL
V =5V
VL = (ZL/Zin)*V = 0.1479 + 0.9231i = 0.935 V at 80.9 degree.
Vpp,ch2/√2 = 1.06 V
% error = (1.06 - 0.935)/1.06 = 11.8%
3. It looks more capacitive at frequencies below 1kHz since the capacitive impedance increases when decreasing frequency.
4. It looks more inductive at frequencies above 1kHz since the inductive impedance increases when increasing frequency.
Impedance can affect to the AC circuits a lot by changing the frequency. By investigating the RL and RLC circuit, we can determine the impedance of a inductor and a capacitor. In this lab, we use the measurement of the RL to find the impedance of the real inductor. Then, we vary frequency of the RLC circuit to observe how the impedance affect to AC ciruit.
Procedure:
Obtain an inductor and measure its resistance:
RL =2 Ω
Set build the circuit same as the diagram below:
Rext = 67.6 Ω
Set up the FG at 1k Hz and RMS voltage at 5 V.
DMM reading: 4.99V
Take measurments:
Vin,RMS = 3.06 +/- 0.01 V Iin, RMS =
42.3 +/- 0.1 mA
The voltage reading is different from the FG displace value. This is due to the large resistance of the FG and the impedance of the inductor.
Make calculation:
Z= V/I = (3.06V)/(42.3mA) = 72.2 Ω
Rewrite the expression for the input impedance:
Z=Rext + RL + ZLj
Write an equation for the magnitude of the impedance:
Z = ((Rext
+ RL)2 + ZL2)1/2
Angular frequency:
w = 2*pi*f = 2*3.14*1000 = 6283 rad/s
Set the expression for the magnitude of the impedance equal to the first calculation to solve for the inductance:
72.22 = (67.6 + 2)2
+ ZL2
ZL = 21.63 Ω = wL
L = 3.44 mH
Investigating the LRC circuit:
The source is 1kHz, we wish to cancel the inductive part of the real inductor impedance. The value of capacitance required is:
wL = 1/Cw
C= 1/Lw2 = 1/(3.44 mH)(36282) = 7.36 mF
Set up the circuit:
Vpp,CH1 = 2.0 mV
Vpp,CH2 = 1.5V
Δt = 10 ms
The phase different: 90osca
Vary the frequency and make measurements:
Frequency (Hz)
|
Vin (V)
|
Iin (mA)
|
Zin (Ω)
|
100
|
4.63
|
19.9
|
232.7
|
500
|
3.21
|
39.2
|
81.9
|
1 k
|
2.96
|
40.6
|
72.9
|
5 k
|
2.90
|
38.6
|
75.2
|
10 k
|
2.96
|
32.8
|
90.2
|
1. The input current is largest at 1kHz because the capacitor and the inductor cancel out at 1kHz. The impedance is smallest at 1kHz, so the current is largest.
2. At 1kHz, the theoretical voltage across the inductor is :
ZL = RL + jωL
Zin = (Rext + RL) + jωL
V =5V
VL = (ZL/Zin)*V = 0.1479 + 0.9231i = 0.935 V at 80.9 degree.
Vpp,ch2/√2 = 1.06 V
% error = (1.06 - 0.935)/1.06 = 11.8%
3. It looks more capacitive at frequencies below 1kHz since the capacitive impedance increases when decreasing frequency.
4. It looks more inductive at frequencies above 1kHz since the inductive impedance increases when increasing frequency.
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