Anonymous
Anonymous asked in Science & MathematicsWeather · 2 months ago

Physics Problem?

Update:

Thank you

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  • 2 months ago
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    replace the two batteries with one of 66 volts with Ri = 4Ω, oriented same as the 80 volt one

    the 3 r's in parallel equal 

    1/R = 1/6 + 1/12 + 1/4

    1/R = 2/12 + 1/12 + 3/12

    R = 12/6 = 2 Ω

    that is in series with 5+4, so total R is 11 Ω

    total I = 66/11 = 6 amps

    power in 5Ω = I²R = 36x5 = 180 watts

    voltage across the parallel combo is

    V = IR = 6x2 = 12 volts

    current thru the 6 Ω is 

    I = E/R = 12/6 = 2 amps

    drop across the two Ri's is 6x3 and 6x1 or 18 and 6 volts

    battery voltages are therefore 80–18 = 62 and 14+6 = 20 volts

    note that the 14 volt battery is actually being charged by the larger one

    as a check, voltages around the loop are

    62 – 5•6 – 20 – 12

    62 –30 – 32

    0

    ok

    note that I made a change to your drawing to avoid ambiguities.

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