Question

Courtney------->

Answer 1

\(\huge\color{green}{{\rm hey}!}\)
do you have math question ???
if u do please post that :)

Answer 2

Hey

Answer 3

okay so please post ur question first then tag them
and that's how you should tag so they can get notification
@nnesha @<--this is important :)

Answer 4

ok im here

Answer 5

great

Answer 6

So long story short it seems every had a different way of calculating the currents part. The big question is still.....what voltage phasor should you use to find current.....still not sure.

Answer 7

I did get all the values form someone elses last part, but I am not sure they are correct. Not sure about you but I am ready to just wrap this lab up (;

Answer 8

ok

Answer 9

yes for sure

Answer 10

So someone gave me numbers 13-19 which look questionable at best. And I also have another number for the voltage phasor someone else used to calculate Z, but I dont think they are correct either lol.

Answer 11

I am finished with LTSpice stuff. Later i might check thee capacitor part against ltspice to see if thats right...

Answer 12

lol ok show me what you have and we can make a decision....also i didnt get to my teacher today but ill just stay till 12...does that help since thats when the class goes till???

Answer 13

yes

Answer 14

ok good

Answer 15

So should I do lTSpice too or we just turning in what you did?

Answer 16

13.) 8.7 kHZ
14.) 1.1V @ 65 deg
15.) 1.4 V @ 0 deg
16.) .0014A @ 0 deg

Answer 17

17.) 786 ohms
18.) 1.28 mS
19.) 14.5 mH

Answer 18

From LTSPICE: THis is correct (;
21.) 2.4592938 V
22.) 2.3396784 mA
23.) 105.42169 uS

Answer 19

You dont need to do ltspice. But if you would like to it is good practice. I would be curious to do the cap circuit on lt instead to check the first part of our lab.....I dont think anyone calculated the current correctly...but I could be wrong. The people I checked somehow came up with riight inductance value..

Answer 20

hmmm ok ill look into it......so for 15 they have 0 degrees.....we still think that is not right ya??

Answer 21

I dont think its right ); But I dont really know if what we did on first part was right either. The directions for both of those parts was really bad......

Answer 22

The other value I got for the Vr across resistor from someone else was 1.56 @ -38.25 for the voltage on resistor which was alot closer to what we had for first part.

Answer 23

But they might have used different frequency. So should we just go with the more complete set of answers and move on to review questions?

Answer 24

well after ready the directions again i think the phase shift if from resistor to capacitor in part one so the resistor shouldn't have a phase shift since it was the first wave....the math function gave us a reading to find the phase shift of the current to see if it was lagging or leading.....i dont think the resistor should have a phaseshift.....

Answer 25

because we only measured one delta and it was from the 2 waves....i could be off but looking over it again it kind of seems that way...

Answer 26

k

Answer 27

the only reason number 1 has a phase shift is because that was before we had the circuit built....then after we built the circuit,,,the first oscope was attached to resistor and 2nd oscope was on the capacitor....what do u think....that would make sense as to why the numbers u have has a 0 degree phase shift for the resistor

Answer 28

Yeah I think first part is wrong.....

Answer 29

The person whos answers those were says the way to find current is to take chl and ch2 both as phasors and then divide by current so I think the first part with resistor phase is ncorrect.

Answer 30

is correct...as in a 0 degree phase shift?

Answer 31

Yes.

Answer 32

ok

Answer 33

So I think all we need to do is re calculate current in first part.

Answer 34

should be 2.5 @ 0 deg - 0.8 @ 75.6 deg all div by R

Answer 35

And in the part about lt spice number 24, we can just say our numbers are a little off since this person used a pretty high frequency of 8.7 kHZ vs the 5kHZ the simulaton has you run. It wasnt specifiied to be 5kHZ in lab so it should be fine.

Answer 36

well we need to find v of resistor ....v source would be the 5 volts i think because one full period(full sine wave) is 5v.....vcap was 1.66 so Vr would be 3.34....Q #1 was just asking for amplitude....or is it cut in half??

Answer 37

cut in half to calculate???? we are putting 5v into the circuit not 2.5.....

Answer 38

Need to cut in half to work with amplitudes. 2.5 - 0.83

Answer 39

is Vr

Answer 40

ok just checking to be sure....sorry

Answer 41

This lab is super confusing.

Answer 42

i agree

Answer 43

so do we need to put them in cartisean form to subtract then divide? or just say the degrees are -75.6 deg

Answer 44

Put into complex to add/subtract

Answer 45

2.4304 @ -19.3160 deg I believe

Answer 46

Ill double check that...

Answer 47

I think that is it. Check it against an online calculator.

Answer 48

ok sorry i had to step away for a min....im working out now to see what i get

Answer 49

ok maybe im doing this wrong putting into complex numbers but i got 1.93@80.19 deg

Answer 50

How did you put into complex?

Answer 51

http://www.electronics-tutorials.ws/accircuits/complex-numbers.html

Answer 52

should be (2.5 cos0 + 2.5sin(0)j) - (.87cos(75.6) + .87sin(75.6)j)

Answer 53

Amplitude is magniitude of answer.

Answer 54

2.29 + .8039j

Answer 55

Take magnitude to find amplitude and then arctan to find angle.

Answer 56

ok i figured it out lol but mine is a little off from yours2.63@17.81 deg...positive

Answer 57

the angle should be negative because of which one was leading. There are online calulators also to check with.

Answer 58

ok nevermind lol i got 2.5 @-17.81deg...mine signs were wrong sorry

Answer 59

which ever numbers works for me. Are we good on lab numbers now and you can fill the last few in?

Answer 60

so current is 2.5 @ -17.81 deg mA...correct?

Answer 61

yes

Answer 62

Ill bring the lt spice part. And yes corrct.

Answer 63

I hate this lab setup lol. The manual is terrible. Maybe worse than tekbots.

Answer 64

Are you doind CS391 thing tonight?

Answer 65

going to try...depends on how long this takes plus study Q's and prelab.....are you?

Answer 66

Yeah. I am not sure about how to draw that graph...and trying to google study questions now....

Answer 67

i think i do...ill see here in a sec...im fixing answers at the moment

Answer 68

We can stay on here for a few and try to knock out study questions also. BTW this site is great for getting math help in calculus once you figure out how to use that equation editor

Answer 69

k

Answer 70

ok sounds good give me a sec and we can talk about the study q's

Answer 71

ok so here is what I got for part 2:
13) 5.5 Hz
14).8 @ -66.6deg
15) 1.95 @ 0deg
16)2.54 @ 16.81 deg

Answer 72

if you think that looks good ill do the last couple real quick

Answer 73

is that suppose to be negative angle for 16?

Answer 74

Otherwise yes looks good.

Answer 75

no cause the voltage leads current this time not lags so it will be positive

Answer 76

ah ok

Answer 77

so impedance....v over i ... i got .31 @ -83.41 deg.......does this seam right?

Answer 78

o i guess i need to convert to ohms

Answer 79

Yes looks good.

Answer 80

I cant figure out how to put this in ohms....

Answer 81

Dont we just use sin and cos to break into real and imaginary?

Answer 82

So .31cos(x) + .31 sin(x) j

Answer 83

well yea but the other people have it in ohms like 763 or something

Answer 84

real and imaginary gives .036 - j .3079 ohms?

Answer 85

Yes. I am not sure what they did there );

Answer 86

I am trying to do review questions but they arent much better );

Answer 87

ok cool i just looked it up in the textbook and that is right it should be small cause the inductor is mH....so im ok with this....

Answer 88

Awesome.

Answer 89

so i got -.009 - j .001 for inductance....look ok? i used L = V/ jwI...

Answer 90

or .009 @ 6.34 deg actual inductance is 10 mH or .010.... real close yea?

Answer 91

very close looks great.

Answer 92

cool whew! lol i feel a lot better now about this

Answer 93

Me too. I hope the lab tomorrow isnt as bad. I got some ideas for the first 3 review quetions.....

Answer 94

ok give me a sec im drawing the graph then ill save and email u a copy and see what you think

Answer 95

Review Questions

1.) For the inductor if we raised the frequency the phase shift would be greater since we are increasing inductance. For the capacitor circuit the phase shift will be less since with high frequency a capacitor is like a short circuit.
2.) The value of capacitor that should put the ch2 voltage back in phase with channel 1 is 13.7mF.
3.) You need to measure voltage across inductor as a phasor. After the input voltage are known and the voltage of inductor are known, these values can be used to find current in circuit. Once current through circuit is found, we can zind the impedance Z of inductor, and from there inductance = 1/jwL.

Pre Lab Questions

Answer 96

sent you a pic through text first to see what u think before I add to paper

Answer 97

Looks great.