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A 59.0-Ω resistor is connected in parallel with a 116.0-Ω resistor. This parallel group is connected in series with a 15.0-Ω resistor. The total combination is connected across a 15.0-V battery.
(a) Find the current in the 116.0-Ω resistor.
______A

(b) Find the power dissipated in the 116.0-Ω resistor.
______ W

  • physics -

    R1 = 59 Ohms.
    R2 = 116 Ohms.
    R3 = 15 Ohms

    Rt=R1*R2 / (R1 + R2) + R3 = Total Res.
    Rt = 59*116 / (59 + 116) + 15,

  • physics -

    R1 = 59 Ohms.
    R2 = 116 Ohms.
    R3 = 15 Ohms

    Rt=R1*R2 / (R1 + R2) + R3 = Total Res.
    Rt = 59*116 / (59 + 116) + 15,
    Rt = 39.1 + 15 = 54.1 Ohms.

    It = I3 = Eb / Rt = 15 / 54.1 = 0.28A

    V1 = V2 = Eb - I3*R3,
    V1 = V2 = 15 - 0.28*15,
    V1 = V2 = 15 - 4.16 = 10.84 Volts.

    a. I2 = V2 / R2 = 10.84 / 116 = 0.093A

    b. P2 = V2*I2 = 10.84 * 0.093 = 1.0W.

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