Simple Circuits question that i cant do

Discussion in 'Homework Help' started by medontknowye, Nov 9, 2011.

  1. medontknowye

    Thread Starter New Member

    Nov 9, 2011

    Uploaded with

    Question 1:The current i(a)
    Question 2:The current flowing through 12(ohm) resistor
    Question 3: The power Delivered to the Circuit
  2. Georacer


    Nov 25, 2009
    Your thread has been moved to the Homework Help section, where it is more appropriate.

    The Homework Help section asks that you post up the work you have done so far, so we can see it and find where you went wrong, or suggest better approaches. We will not do all your work for you.

    If you are in doubt about how to structure your post and what to include please read this thread:

    Thank you.
  3. medontknowye

    Thread Starter New Member

    Nov 9, 2011
    please someone help me im new to this.. and i got a test tommorow .. and i need to figure this out asap..please someone help me.
  4. medontknowye

    Thread Starter New Member

    Nov 9, 2011
    hi. can u help me jst once..i can figure out the current for the first question which is 8 i think but im not sure how to calculate the urrent on the 12 ohms...i also know how to calculate the last question which is P=IV or P=I^2R or P=V^2/R please jst help me with second question
  5. Adjuster

    Late Member

    Dec 26, 2010
    You could use the current division rule (Google it), otherwise you know the total current, the current in the 10ohms is obvious, the remainder goes into the 6ohms AB.

    You then have X amps flowing into a parallel combination of two resistors...
  6. medontknowye

    Thread Starter New Member

    Nov 9, 2011
    so the current on the 10 ohms would be 3 and on the 6 it will be 5 right? on the first segment

    then i used I=(12/6+12)*5 to get 3.33 for the other 6 ohms so that means that 12 ohms is 1.67 ohms is this wrong or right?.. if not can u correct please
  7. thatoneguy


    Feb 19, 2009
    It might help to re-draw it.

    start with the40V Source as a battery, put 10Ω across it. Then a 6Ω resistor which leads to both a 6Ω and 12Ω in parallel.

    Remove battery, find total resistance (combining series and parallel resistances to a single resistor across power supply. (Thevenin Equivalent Circuit)

    Divide power supply voltage by that resistance and you have your current drawn from source.

    Voltage across the first resistor will be battery voltage, it is independent of the rest, divide voltage by resistance for current through that resistor.

    Subtract the current through the resistor in parallel with voltage source from the total current in the Thev Equivalent circuit and use in the followng equations:

    Use the current through the series resistor to find the voltage it drops, which will give you the voltage across the last two which are in parallel.

    Subtract voltage dropped across that series resistor from source voltage to find voltage across parallel pair.

    Divide that voltage by the resistance and you have solved for all voltages and currents.
  8. bountyhunter

    Well-Known Member

    Sep 7, 2009

    That answer is wrong.

    You need to combine to one equivalent resistor and get the source current.

    To combine any two parallel resistor to a single resistor use this:

    REQ = (R1 x R2) / (R1 + R2)

    Series resistors simply add.

    To find the current divided between two parallel resistors, it is inversely proportional to the current in the branch. In other words: if one parallel side has three times as much resistance, it will have 1/3 of the total current.

    You really need to learn the basics or switch to literature. Everything you will need to learn has to stand on the basic laws like Ohm's Law.
  9. testing12


    Jan 30, 2011
    I suppose this is late since you have the test today buy it may be of benefit in future to learn nodal and mesh analysis.
  10. EL7819

    New Member

    Apr 15, 2011
    RT=12ohm ll 6ohm =(12*6)/(12+6)=72ohm/18ohm=4ohm

    b.)4ohm + 6ohm=10ohm

    c.)10ohm ll 10ohm= 10ohm/2=5ohm



    It=8A            P=(8)^2(5)=320W


    8A-4A=4A between I2 and I3.

    4A through the resistor 6ohm

      6ohm ll 12ohm






    Last edited: Dec 7, 2011