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Finding Terminal Voltage

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown in the figure below the current through the 12.0V battery is measured to be 70.6 mA in the direction shown. What is the terminal voltage Vab



    2. Relevant equations
    V=E-IR
    I1=I2+I3


    3. The attempt at a solution
    First I set Vab= E-IR= 24 - 10(I1)
    I1=I2 + I3
    So I applied Kirchoff's rule on the outer loop to find I3
    -12 - 10I2 + 20I3 = 0
    I3= [12 + 10I2]/20 = 0.6000353 = 0.6A
    I1= I2 + I3 = 70.6 mA + 0.6A = 0.6000706A = 0.6A
    Vab= 24 -10I1 = 24 - 6 = 18V
    Is this right?
     

    Attached Files:

  2. jcsd
  3. Jul 12, 2010 #2

    vela

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    What happened to the 30-ohm resistor?

    The voltage due to the internal resistance of the 12-volt battery is (70.6 mA)(10 ohms) = 706 mV = 0.706 V. You seem to have divided by another factor of 1000.
     
  4. Jul 12, 2010 #3
    Can I combine the 20(ohm) with the 10(ohm) resistor?
     
  5. Jul 12, 2010 #4

    vela

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    No, because they're neither in series nor in parallel. You could, however, combine the 20-ohm and 30-ohm resistors because they're in parallel.
     
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