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ktehmok

Diamond Member
Aug 4, 2001
4,326
0
76
Whats worst for you, Tripple Cheeseburger with apple smoked bacon with a huge side of chilli cheese fries or Crack?
Topic Summary: Colorado Grill combo vs Crack, whats worst?
Created On: 09/29/2005 05:27 PM
 

Fraggable

Platinum Member
Jul 20, 2005
2,799
0
0
The purpose of this lab was to gain experience with the lab equipment and to help understand how capacitors work and how they react to DC and AC current and voltage sources.

The first circuit I constructed is pictured in Figure 1. It has one 10V DC voltage source, a 470? resistor, a 1K? resistor, and a 100µF ceramic capacitor.

Figure 1


I first computed how much current was to be expected through each element of the circuit, then turned on the circuit and measured the amount of voltage across each element. The computed current and measured voltage values are listed in Table 1.

Table 1

V Measured Value I Computed Value
V1 3.24V I1 6.8mA
V2 6.73V I2 6.8mA
V3 6.73V I3 0mA
V4 6.76V

As illustrated in the table, the circuit behaves almost identically to an equivalent circuit without the capacitor. This is because the capacitor appears as an open circuit in a DC circuit. The only effect that the capacitor has on the circuit is that when the V1 is turned on, there is a very short charging time for the capacitor and when V1 is turned off, there is a short discharge time, during which time there is a small current flow through the 1K? resistor.




The second circuit I constructed is pictured in Figure 2, and has two capacitors, two resistors, and one DC voltage source. I computed the expected current flows and then measured the voltage across each component of the circuit, and recorded my findings in Table 2.

Figure 2

Table 2

V Measured Value I Computed Value
VR1 0V IR1 0A
VC2 9.95V IR2 0A
VC1 9.96V IC2 0A
VR2 0V

The table illustrates that although there is a voltage potential across the two capacitors, there is no current flow. Again, this is because capacitors do not conduct current, and there is no complete path along which current can flow. There was a small amount of current flow when the capacitors were in their charging state and electrons were moving from one plate to the other, but when the potential across the capacitor reached 10V, the current flow stopped.







The third circuit that I constructed was designed to charge a capacitor very slowly so that I could see the increase in voltage across the capacitor as positive and negative electrons were moved to opposite plates. I used a 15V DC voltage source, a 100K? resistor, and a 100µF electrolytic capacitor. The circuit is pictured in Figure 3.

Figure 3


Table 3

Time 10s 20s 30s 40s 50s 60s 70s 80s
Vc1 8.9V 12.2V 13.6V 14.2V 14.4V 14.6V 14.68V 14.7V

Table 3 illustrates the charging phase of a capacitor, and proves the RC time constant theory, t=RC. In this case, R=100K? and C=100µF, and t=10sec. Theoretically after 1t the voltage across the capacitor will be 9.45V, or 63% of V1. As can be seen in Table 3, the RC time constant theory is correct. 2t after the switch is closed the capacitor gains 63% of the voltage difference between V1 and 1t and so on. By 5t the capacitor is considered to be fully charged.

The fifth circuit I constructed had an AC voltage source and a .01µF ceramic capacitor and is pictured in Figure 4. I measured the voltage across the capacitor at three different frequencies as shown in Table 4.

Figure 4





Table 4

F (kHz) I (rms) I (peak) V (peak)
2 kHz 450mA 6.38mA 10V
10 kHz 2.0A 2.8A 10V
20kHz 3.2A 4.53A 10V

As I expected, the voltage across the capacitor did not change but the current flow did. This is because when the frequency increases; the number of times that the capacitor charges and discharges is also raised, creating current flow.

The last circuit I constructed had an AC voltage source, a DC voltage source, a 1µF electrolytic capacitor, a 4.7K? resistor and a 10K? resistor and is pictured in Figure 5. I measured the voltage at point A using a digital oscilloscope and show the results in Figure 6.

Figure 5











Figure 6



As you can see, the DC voltage source offset the AC signal to approximately 10V above the 0V line on the scope, as I expected.

Conclusion

After completing this lab I understand much better how AC circuits work, how to expect capacitors to affect a DC circuit, how to calculate capacitor charging phases and how to calculate points along a charging phase.


You asked.
 

ChiBOY83

Senior member
Dec 28, 2004
517
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i don't even understand what homecoming is for college... i mean, its the homecoming weekend right now, but i don't notice any difference
 

Krazy4Real

Lifer
Oct 3, 2003
12,221
55
91
People will only listen if they want to. A truly caring person is someone who tries to understand others, and this compassion comes from the heart. Simply understanding another's viewpoint already does a lot. It is NOT related to bickering on who is right and who is wrong over trivial things. A need to correct others and convert them to your way of thinking only reflects on your need and dependence on proving yourself right. Learn to be reflective... ego building does nothing but put you at odds with others. Tolerance of differences in opinion is the first step towards maturity.

Some people have strange away messages.
 

effee

Golden Member
Sep 4, 2004
1,797
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04:32] <elu> zzz
[04:36] * Joins: |EsMuNd| (esmund@cm53.sigma6.maxonline.com.sg)
[04:47] * Joins: TX (~UPP@bb220-255-175-99.singnet.com.sg)
[04:47] * QE sets mode: +l 35
[04:47] * Quits: |EsMuNd| (esmund@cm53.sigma6.maxonline.com.sg) (Ping timeout for |EsMuNd|[cm53.sigma6.maxonline.com.sg])
[04:47] * Quits: Xharky (~UPP@bb220-255-175-99.singnet.com.sg) (Ping timeout for Xharky[bb220-255-175-99.singnet.com.sg])
[04:47] * QE sets mode: +l 33
 

Umberger

Golden Member
Apr 13, 2005
1,710
0
76
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