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Circuit : Current Flow And Voltage Across Two Point

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Find Voltage across a and b and the missing current (i)
    so this is the circuit
    http://img3.imageshack.us/img3/184/circuitsf.jpg [Broken]

    2. Relevant equations
    KCL and KVL
    current towards the node is + away is -
    Et = E1 + E2 + E3 in series
    Et = E1 = E2 = E3 in parallel
    Ohms law
    E= IR

    3. The attempt at a solution
    KCL @ node c
    3A-i1-i2= 0
    i2=6V/3Ohms
    i2=2A
    3A-2A=i1
    i1=1A
    KCL @ node d
    i2+1A-1A-i3=0
    2A+1A-1A=i3
    i3=2A
    KCL @ node e
    i3+i4-i
    i4=24V/4Ohms
    i4=6A
    2A+6A=i
    i=8A

    KVL @ loop acdeba
    E(ab)+i1(8)-6V-i3(12)+12V=0
    E(ab)=-8V+6V+24V-12V
    E(ab)=10V

    so thats what i try..
    my question is.. is those current direction is right?.. XD still dont know if i did it right..
    so anyone who can help?
    and the Voltage across AB.. am i right in having the loop acdeba.. neglecting the resitor which have 24V?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 9, 2011 #2
    can you please give the proper problem statement and circuit diagram
    i can't understand the few basics here such as
    'from where the current from 3A current source completing the loop'
    'you said you have to find voltages at A 'AND' B and while solving you maid the loop through acdeba which suggest that point a and b are shorted resulting equal voltages at a and b
     
  4. Aug 9, 2011 #3
    the diagram is right.
    the problem statement.. "Find the voltage across a and b" and find the current "i"

    the 3A current source is placed like that.. there's nothing wrong with that

    the blue line is the original diagram
    and the current i1 , i2 , i3 and i4 is just my assumed direction..i dont know if its correct.
     
  5. Aug 9, 2011 #4
    i think all current calculations you have done are right
    i calculated the voltages at a and b separately 'b' is directly as 12 V and voltage at 'a' as 22
    so the difference is 10 and your answer is correct
     
  6. Aug 9, 2011 #5
    ahm can you please show ur computation in doing it separately.. i might have another idea in solving future circuit problems... thx
     
  7. Aug 9, 2011 #6
    considering voltage at point e is 0V i will call voltage at e as Ve
    Vd-Ve=i3*12
    so Vd=24
    Vc=Vd+6
    so Vc=24+6=30
    Vc-Va=i1*8
    Va=30-8
    Va=22
    so we have got the voltage at pont a as 22
    voltage at b is 12 volt directly

    i think this helps
     
  8. Aug 9, 2011 #7
    am i right that in ur analysis u also neglect the presence of the 24V at current i4
     
  9. Aug 9, 2011 #8
    first of all 24V is not the source it is just the voltage drop across the resistor and i have used it to calculate the current i4 itself
     
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