Industrial Electronics Assignment 1

[frankwas]

Hi Everybody

It's been a while since I was on this great site. I wanted to ask everyone for help.

I am busy with two assignments for college and I've completed the 1st one and would like to get some feedback on my answers before I submit it.

I will be posting Assignment 2 tomorrow for your perusal.

Thanks very much for any assistance. There are two documents. The containing the questions and one with my answers.

Take care.
Francois

 

[MikeMl]

Post deleted.

 

[frankwas]

Thanks for bringing that to my attention. I selected the wrong file when i attached the docs.

 

[MikeMl]

Thanks for bringing that to my attention. I selected the wrong file when i attached the docs.

Now to your work. Your equations for I1 and I2 would only be true if the voltage at the common node is zero, which it is not!

You must solve for the voltage at the common node.

If you call the voltage at the common node (the one that you apply Kirchoff to) Va, then the current through the 2Ω branch is I1 = E/R = (6-Va)/2...

I'll let you write the other two current equations...

btw: is the neg end of the 9V battery supposed to be floating?

 

[frankwas]

Hi Mike

Thanks for your feedback. I will see if I can figure it out with your recommendations. No it's supposed to be grounded but the quality of their printouts are horrible. Just to clarify you are referring to question 1 right? I redrew my sketch so that the 9V is grounded correctly.

I just saw my error that I made. I think I did it correctly in the other questions, but not with sim equations. I will update this question and resend.

 

[MikeMl]

... Just to clarify you are referring to question 1 right?...

Yes, my comments apply to question 1.

 

[WTP Pepper]

Regarding Q1 as I will come back to Q2 when I have time....

See attached for naming convention. Using nothing more than Ohm's law.

Vo=I3 x RL
I2=(V2-Vo)/R3 = (9-Vo)/4 = 9/4 - Vo/4
I1=(V1-Vo)/R2 = (6-Vo)/2 = 3 - Vo/2

I3=I1+I2...splitting the bottom of the fractions gives..

I3=9/4 - Vo/4 + 3 -Vo/2

replacing Vo with I3 x Rl

I3=9/4 - (I3 x Rl)/4 + 3 - (I3 x Rl)/2

Rl=3

So I3=9/4 - (3 x I3)/4 + 3 - (I3 x 3)/2

I3=1.615 A

 

[WTP Pepper]

I think this is how it is done. Forget R3 R5 R6.
Short V2 and using thev you get 2.25V with 0.75 ohm
Short V1 and as above you get 5.625V with 0.75 ohm
Add the voltages and parallel the resistors and you get 7.785V with 0.375ohm
Total current through the loop is 7.785/(0.375 + 3 +4 +10) = 0.4532A
Across the 10ohm using V=IR gives 4.532V

Not sure if method is correct as there is 30 years between learning and doing this, but LT Spice comes pretty close. I would be interested in an alternate method.

 

[WTP Pepper]

Q4. 0.41A. Sorry no clues yet.
Now it's time you did your homework and worked it out for yourself. You have enough with my examples. Please come back if you're stuck and I will explain further.

 

[rigel masompe]

electronic weighbridge circuit diagram

would you assist me with the designning of an electronic weighbridge circuit diagram and the required components. for a school project. thanks!

 

[WTP Pepper]

If I understand this correctly, that's ambitious for a school project. I would start with a strain guage and a google of Wheatstone bridge,