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Circuits Problem (Physics)

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Given the circuit above:
a) Solve for unknown currents.
b) What is the potential difference between points a and b?
c) If both batteries start with 20 A*h of charge, how long does it take the first of the batteries to run out? Which battery is it (left or right)?
d) If both 12V batteries are replaced by batteries of identical voltage, what is the smallest voltage which will result in integer currents throughout the circuit?

  • Circuits Problem (Physics) - ,

    Sorry, can not open your image :(

  • Circuits Problem (Physics) - ,

    hi,
    i put spaces between the "h t t p :" because it wouldnt let me post it otherwise. if you delete the spaces it should load (i hope!).

  • Circuits Problem (Physics) - ,

    I call the current coming out the top of the left battery i1
    I call the current coming out the top of the right battery i2
    Then the current headed down from a to b is (i1+i2) (Kirkhoff)
    In the left loop
    12 = 4 i1 + 6(i1+i2)
    In the right loop
    12 = 3 i2 + 6(i1+i2)
    so
    12 = 10 i1 + 6 i2
    12 = 6 i1 + 9 i2
    solve those two. I will multiply the first by three and the second by 2
    36 = 30 i1 + 18 i2
    24 = 12 i1 + 18 i2
    subtract
    12 = 18 i1
    i1 = 2/3
    then i2
    12 = 10(2/3) + 6 i2
    12 = 20/3 + 6 i2
    36 = 20 + 18 i2
    16 = 18 i2
    i2 = 8/9

  • Circuits Problem (Physics) - ,

    b)
    i1+i2 = 2/3 + 8/9 = 14/9
    times the resistance which is 6 = 28/3 volts a above b

  • Circuits Problem (Physics) - ,

    2/3 = 6/9 = i1
    8/9 = i2
    so i2>i1 so the one on the right will run out first
    (8/9) t = 20 amp hr
    t = 9*20/8 = 9*5/2 = 22.5 hr

  • Circuits Problem (Physics) - ,

    at present i1 = 6/9 and i2 = 8/9
    clearly if I multiply the voltages by 9 I will get 6 amps and 8 amps
    then if I divide both voltages by two I will get 3 amps and 4 amps
    so multiply the voltage by 9/2
    (9/2)12 = 54 volts

  • Circuits Problem (Physics) - ,

    omg! thankss!!

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