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EEET 2249 - Tutorial 5 - Problems

Circuits with voltage and current sources *Unsolved*

Q6

Considering the circuit in Fig.Q6, calculate the voltage across the 50 Ω resistor and the power supplied by the sources.

No matter how I try it, I can’t get the right values out of this mesh analysis… What am I doing wrong?

Mesh #1

Mesh #2

Solve V1A and i2

(Ans: 75 V, P1A = 175 W, P100V = 50 W)

Circuits with controlled sources

Q7

Calculate the current through the controlled voltage source in Fig.Q7.

I’ve tried this with mesh and got the wrong answer, so I tried node voltage analysis which still doesn’t work… What am I doing incorrectly?

Node #1 KVL (Leaving)

Node #2 KVL (Leaving)

Node #3KVL (Leaving)

Substitute V3 into Equations

Solve ix

Solve

[Ans: 3 A downward]

Q8

Calculate the equivalent resistance as seen from the terminals a and b for the network shown in

Why does a 1A connection between a and b allow me to solve the equivalent resistance? Where is this in our lecture notes?

{Hint: Connect a 1A current source between terminals a and b. Then solve for the mesh currents using the mesh current analysis. Having calculated the mesh currents, find the voltage across a-b.

This voltage is equal to the value of the equivalent resistance between terminals a and b.}

[Ans: -15Ω]

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This entry was posted on Saturday, September 13th, 2008 at 10:15 am and is filed under Circuit Theory. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

12 Responses to “EEET 2249 - Tutorial 5 - Problems”

  1. Paul Says:
    September 15th, 2008 at 8:00 am

    In Q7, I1 != 1A.

    I think you need to form an equation that says i1 + i2 == 1A.

  2. Paul Says:
    September 15th, 2008 at 8:01 am

    Actually it’s i1 - i2 == 1A because of the assumed direction of i2.

  3. Darryl Says:
    September 19th, 2008 at 12:54 pm

    Hi Tim,

    Sorry for the late reply, I didn’t bother facebook until today I saw your msg.

    Q6:

    Logically, if you say i1 = 1A, ask yourself why not i2 = -1A?

    Hence, this problem should be solved using Super Mesh current.

    Take the biggest square circuit path (the outer wire passing through both 50ohm R and Voltage source) as a mesh current

    take any node and you’ll get Paul’s equation: i1 = 1 + i2

    Write one (1) equation and substitute i1 or i2 to solve the question, getting

    i1 = 1.5 A
    i2 = 0.5 A

  4. Darryl Says:
    September 19th, 2008 at 1:41 pm

    Q7:

    I dont know, i tried it a few times, couldn’t get it, i’ll tell you when i get it

  5. Darryl Says:
    September 19th, 2008 at 3:24 pm

    Q8:

    this question is a bit tricky. After a long hard thinking, I finally accepted my not-so-good answer to explain why that hint was given.

    reasons:
    1) The circuit simply do not have enough information to proceed, or may be there are, but I never bother to try because I’m doing something else.

    2) here comes the explanation:

    - It’s not in lecture note
    - It is if you look carefully
    - Req = Voc/Isc
    - You cannot determine Voc since you simply cant.. (or i cant)
    - this is doing it FROM the answer
    - setting Isc as 1A enables us to determine Voc
    - at the same time determine Req because Isc = 1A (easy)
    - setting Isc as other values will get you the same Req

    Even though the hint is given, I still cant solve it, got to finish up my C++ before i tough circuit theory.. T.T, i’m so dead

  6. Darryl Says:
    September 19th, 2008 at 4:04 pm

    oh, also, please do inform me and teach me if you get them. I wonder the boy above is you.. :)

  7. Tim Cinel Says:
    September 20th, 2008 at 4:57 am

    Thanks for the replies Paul, I only just realised my fallacy after your help… Wtf was I thinking? i1 and i2 = 1!?!?!

    Cheers.

  8. Tim Cinel Says:
    September 20th, 2008 at 4:58 am

    Hey Darryl thanks for the replies. It’s comforting to know I’m not alone in being challenged by this subject.

    Regarding Q7, Q8:
    I solved it and understood it a the time but I’ve forgotten now! When I post the full solution I’ll notify you.

    Regarding Q6:
    Thanks to you an Paul I’m aware of my that ridiculous assumption I was making, solved now. Cheers :)

    Nice blog, too - You’re going on my blogroll.

  9. Jason Says:
    September 21st, 2008 at 3:51 pm

    I think RMIT should put your site up as one of the discussion links in the learning hub!

  10. Darryl Says:
    October 19th, 2008 at 7:44 am

    I dont know where to get you, but i think here is the best way..

    Now that I’ve started Circuit Theory studies, I found myself stuck at Q8, (Q9 in tute 5).. can you please post an answer if you have one? LOL.. sorry for the trouble but i’m really stucked.. LOL

  11. Darryl Says:
    October 19th, 2008 at 8:26 am

    oh.. i’ve got it now.. HAHA.. did a careless mistake. no need to post.. i understand now and solved it too.. thanks anyway.

    all the best in your exam

  12. Tim Cinel Says:
    October 28th, 2008 at 5:21 am

    Darryl - Even though this is obviously now irrelevant:
    “Yes” I did get them right in the end, I just haven’t had time to post them here. Wish I did though :(

    I probably still will after eggs hams are over.

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