Introduction: In this laboratory session, we will be dealing with simple circuits formed of resistors, capacitors and inductors. We will be passing AC signals from a signal generator through the circuit to monitor how the respective components affect the signals' behaviour. In particular we will be monitoring phase relationships and component reactances.
Equipment: • • • • •
Circuit Board Digital Multimeter x2 Shorting Links and Connecting Leads Signal Generator (EI810322) Oscilloscope (EI211384)
Method: As the method of all three experiments are generally the same, I will describe the process of only one of the experiments. The circuit was connected up as shown in figure 1, with the ammeter connected in series and the voltmeter in parallel. The highlighted connecting link was removed as the ammeter completes the circuit. The following settings were set for the oscilloscope: • • • • •
Timebase – 0.2ms/div AC trigger Dual trace operation Ch1 vertical gain – 2V/div, AC input Ch2 vertical gain – 50mV/div, AC input
The signal generator was then connected and set to a sinusoidal wave of frequency 500Hz and amplitude 12V pk-pk. The frequency was doubled up to 4 kHz, taking down the voltage and current for each frequency value, as shown in table 1. The capacitive reactance was then worked out using the relation Xc = V/I.
1.RMS value for a square wave:
2. From graph 2.2 we can see that reactance has a relationship of inverse proportion. Table 2.1 also shows this. Xc is roughly halved when frequency is doubled.
Graph 'b' best represents the phase relationship between the voltage across a capacitor and the current through it. If we examine the graph we can see that the peak of channel 1 is leading by about 90o over channel 2.
4.
It can be concluded from the values obtained that the phase relationship between the voltage and current are not dependant the signal frequency.
5.
Frequency/kHz
Phase Difference
2
-90°
4
-90°
8
-90°
on
We can work out from the graph that at Xc=1kΩ , Capacitance is about 320nF.
Reactance against Capacitance 8 7
Reactance/kΩ
6 5 4 3 2 1 0 0
50
100
150
200
250
300
350
400
450
500
Capacitance/nF
Again, the graph shows inverse proportion, so it can be said reactance Xc is proportional to 1/C.
that
Table 2.3 – Variation of Capacitive Reactance with Capacitance Capacitance/nf
Voltage/V
Current/mA
Capacitive Reactance/kΩ
47
3.96
0.56
7.07
100
3.96
1.22
3.25
220
3.95
2.7
1.46
470
3.94
5.7
0.69
6. Graphs 3.1 and 3.2 show similarities in their plots. Both resemble straight lines, therefore showing signs of direct proportionality. This is supported by the values obtained in table 3.1.
Inductive Reactance against Frequency Inductive Reactance against Inductance 3 2.5 2 1.5 1 0.5 0 0
Current Frequency Relationship Using a 10 kΩ Resistor 0.27
13. The shapes of graphs 4.1 and 4.2 differ quite drastically. The current peak seems to spread out over a wider frequency range when the 330Ω was used. The current rises at a
0.27
Current/mA
0.26 0.26 0.25 0.25 0.24 0.24 0.23 12
14
16
18
20
Frequency/kHz
22
24
26
steadier rate too. When the 1kΩ resistor was used, the peak of the curve is a lot sharper, and the gradient much higher.
0.27 0.26 0.26 0.26 0.26 0.26 0.25 0.25 17
18
19
20
21
22
23
Conclusion: The effects that an AC signal has on inductors, capacitors and resistors has been investigated, and we found that the relationship between frequency and capacitive reactance has an opposite relationship to frequency against inductance. It can also be noted that in an RC circuit, varying the capacitance has a similar effect to varying the frequency of the signal. This can be concluded from the fact that graphs 2.2 and 2.3 have similar properties. In the final experiment using the resonant circuit, the resonant frequency could have been found more accurately if we had a more sensitive signal generator. The model used was “TG120 20MHz Function Generator” which could only increment the frequency by a minimum of about 1kHZ. Especially around the current peak, it would have been useful to plot more points. The calculated value of fr in question 11 is more than double the value we fr measured from the multimeter. Using the formula, Q= f we can see that we would have obtained a more accurate result had we used a lower 'quality factor', Q.
