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Electronics assignment

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frankwas

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Hi All

Thanks to the admin for creating this section in the forum. I need some help urgently but I don't want to be a nucience. I have two electronics assignments that's I am working on for the second time, because I tried the first one from the supplied materials and I failed miserably. I wanted to know if I could ask help for basically most questions regarding the assignment so that I can get a much better mark this time.

Please let me know. I can scan all pages.

Thanks in advance!

Frank
 
Hi,

Surely you can ask questions here, that's why this forum was created :)

I do have to advise though that you try to understand all the material you have been given first, and any replies that you get you should study over carefully too so you get better marks from now on :)
 
Absolutely!

I will use the information to study for the exams. I will scan and post the documents after this post.

Once again, thank you.
 
Attached are the pages to the assignments. I have tried a few but for basically every single one, I get stuck somewhere.
 

Attachments

  • scan1.pdf
    1.7 MB · Views: 842
  • scan2.PDF
    245.4 KB · Views: 340
Hello there,

There are a lot of questions in those files, so we'll have to take them one at a time. We'll start with scan1.pdf Question number 1.

You basically have two laws to follow:

1. The sum of currents entering and leaving a node is zero.
2. The sum of voltage drops around a closed path is zero.



To start with question 1, you would first assign nodes to each junction,
then write simultaneous equations relating the currents flowing into and out of the
nodes, then solve those equations for the unknowns.

If we call the junction of 6v and 2 ohm resistor node v1, the junction of the three resistors
node v2, and the junction of the 9v and 4 ohm resistor node v3, we can start writing equations.
We will assume current flowing left to right in the 2 ohm resistor, and top to bottom
in the 4 ohm and 3 ohm resistors. That means that the current flowing through the 2
ohm resistor and into node v2 has to equal the current flowing through the other two
resistors out of node v2 using Law #1 above.

Starting with node v1:
(v1-v2)/2=(v2-0)/3+(v2-v3)/4
Note that all we did here was take the difference between the voltages at three pairs of nodes and divide by the resistance between them. Because each pair divided by a resistance is equal to a current, we could rewrite that equation as I1=I2+I3 if we wanted to show the currents entering and leaving node v2.

Now since we already know the voltage for v1 and v3 we can fill those in:
(6-v2)/2=(v2-0)/3+(v2-9)/4

and get rid of the zero:
(6-v2)/2=(v2)/3+(v2-9)/4

and now we have only one unknown so we only have to solve for v2:
v2=63/13

To calculate the current through the 3 ohm resistor now we only have to divide v2/3 to get the current.

We can double check this result by using the result for v2 and calculate the current in the other two resistors
and make sure that Law #1 is obeyed. If not, we made a mistake.

This type of analysis is also known as Nodal Analysis.
 
Last edited:
Thanks MrAl!

My original calc (before I posted here) i got was 1.35A, 0.577A and 1.615A. I will use your methodology because it makes it so much simpler. The way I tried to do it was splitting the circuit into two loops and then trying to calculate the values in each loop and then trying to get the current from that. My final answer was 1.35A flowing through Rl.

I also now see what you mean by the nodes. I was doing it by loops but by node is so much simpler!

I appreciate your help so much!
 
Last edited:
I used your calculations and I got 1.61 A. It seems from my original calculation that I almost had it, but I got it in the wrong place.
 
I will now out of desperation go out on a limb here, but I am willing to pay $50 to anyone who can help me finish this assignment. I am sorry for doing this, but I have no other choice.

If anyone is interested, pm me please. I can only make paypal payments.
 
For help with some of the Thevenin stuff try this site. It explains quite clearly with a similar example.

**broken link removed**

Hope this can help you.
 
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