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physics-series and parellel circuits

posted by on .

I'll try to explain the schmatic drawing...
a 120V battery, total of 12A, leads into a 8ohm resistor (R1), then breaks off into two parallel lines, R2 and R3. the resistance is unknown in R2, and in R3 the resistance is 3 ohms. I need to find the Amperage in R1,R2 and R3 as well as the resistance (ohms) on R2
I hope this explains it... HELP

  • physics-series and parellel circuits - ,

    OOOPA and the voltage at R1, R2 and R3

  • typo? - ,

    12 amps through R1
    volts across R1 = 8*12 = 72

    That leaves 120 - 72 = 48 volts across R2 and R3

    amps through R3 = 48/3 = 16 amps
    You have a typo I think. Perhaps R3 is 30 and not 3 ohms?

  • physics-series and parellel circuits - ,

    R3 is 3 ohms

  • physics-series and parellel circuits - ,

    Well, you have set up a physically impossible situation where R2 would have to be negative. You can not have 12 amps coming out of the battery and 16 amps through R3. Please check your problem statement carefully. Maybe you have 1.2 amps out of the battery?

  • physics-series and parellel circuits - ,

    For R1 voltage .. E=I*R
    E=8*12=96 volts correct?

  • physics-series and parellel circuits - ,

    whoops, right. I multiplied wrong

  • physics-series and parellel circuits - ,

    120 - 96 = 24 volts across R2 and R3
    current through R3 = 24/3 = 8 amps

    then 4 amps through R2
    R2 = 24/4 = 6 ohms

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