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Oh joy dc circuits

  1. Mar 13, 2005 #1
    Been stumped on these 2 questions for at least a week.
    Both diagrams scanned + uploaded.
    23.Find the current in the 8 ohm resistor.
    1/Req1=1/8+1/4
    2.67+6=8.67
    1/Req2=1/8.67+1/10
    R=4.27+.5+5=10.14
    I first found the voltage drop in each resistor.
    I=V/R
    I=9/10.14=.89A
    volt.drop .5 resistor=.5*.89=.445V
    volt.drop 5 resistor=5*.89=4.45V
    volt.drop 4.27 resistor=4.27*.89=3.8V
    Then I solved for current in the 8 ohm resistor.
    I=[9V-(.89A)*(.5*5)]/x ohm
    And that's where I'm stuck. Don't know what ohm to divide by. I think 25 but I don't know why. :confused:
    35.What would current I1 be if the 12 ohm resistor were shorted out(r=1 ohm)?
    OK...this is where I'm just screwed by Kirchhoff's Laws.
    I'm thinking:
    I3=I2+I1
    Loop 1: 12=(1)I1+(8)I2
    Loop 2: 12=(1)I2+(10)I2+(12)I2
    Loop 3: 6=(1)I3+(18)I3+(15)I3+(10)I2+(1)I2
    :frown: Completely stuck. Help please.
     

    Attached Files:

  2. jcsd
  3. Mar 13, 2005 #2
    Pro 23.

    "I=[9V-(.89A)*(.5*5)]/x ohm"
    No quite.
    Use Req2 to find the voltage drop over that system of resistors. Or
    VReq2=[9V-(.89A)*(.5*5)]

    Then find the current going through Req1 + 6. Now you can find the voltage drop over Req1. Once you have the voltage drop you can find the current.

    Does this make sense?
     
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