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Circut analysis -> ?

  • 12-10-2013 1:22pm
    #1
    Registered Users, Registered Users 2 Posts: 36


    "FInd the current through each of the resistors in the following circuit: "

    (see attached)

    from my notes: we can immediately see that :

    e1 = 3v and
    e4= 2/3 e3

    well its just a first step but i don't really get how e4 = 2/3 (e3)
    So e1 e2 e3 e4 are node voltages, also what does the part under 10k ohm's do exactly ?

    .....help
    appreciated !


Comments

  • Registered Users, Registered Users 2 Posts: 36 theLEVI


    Also another question here:


    $1----|
    $2
    o A
    |................|
    |................|
    (5V)........(cunnet source ^ 2A)
    |................|
    |__________|_______________________o B
    (ignore the dots)

    where $1 = 10ohm and $2 = 15 ohms,

    Find the thevenin voltage between A and B ?


    My anwser was 5V , but apparently 25 is correct... :P, i have set current source to zero and then added up the resistances,
    So it looked like this,

    |
    $=25ohm
    o A
    |
    |
    (5V)
    |
    |______________________o B so is Vth not 5V ?!


  • Registered Users, Registered Users 2 Posts: 1,654 ✭✭✭greedygoblin


    See the attached pdf for a run through of how to approach the problem.
    Note that the part under the 10k resistor is the ground symbol representing a potential of 0 V.


  • Registered Users, Registered Users 2 Posts: 36 theLEVI


    i can't thank you enough ! thanks so much :D !


  • Registered Users, Registered Users 2 Posts: 1,654 ✭✭✭greedygoblin


    theLEVI wrote: »
    i can't thank you enough ! thanks so much :D !

    No bother. Regarding your first question, it may also be useful to apply Kirchoff's current law to each node in the circuit. Look at the current entering and leaving each of the nodes E1, E2 and E3. It's much the same procedure as that required to find [latex] E_3 = \frac {2}{3} E_4 [/latex] in the pdf file above. This should provide you with enough equations to figure out the current flowing through each component (I hope!).


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