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1 December, 23:35

Three resistors of 4.0, 6.0, and 10.0 Ω are connected in parallel. If the combination is connected in series with a 12.0-V battery and a 2.0-Ω resistor, what is the current through the 10.0-Ω resistor? Group of answer choices

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  1. 1 December, 23:38
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    0.536 A

    Explanation:

    From the question,

    The combined resistance of the parallel connection is given as

    1/Rt = 1/4+1/6+1/10

    Rt = 120/15

    Rt = 8 ohms.

    Since the the combination is connected in series with a 10 ohms resistor,

    Rt' = 8+10 = 18 ohms.

    Current flowing through the series circuit

    I = V/Rt'

    V = 12 V, Rt' = 18 ohms

    I = 12/18

    I = 0.67 A.

    Therefore voltage across the combined resistor

    V' = 8 (0.67)

    v' = 5.36 V.

    Since the resistors a connected in parallel, the same voltage drop across them.

    Therefore, current through the 10 ohms

    Ix = 5.36/10

    Ix = 0.536 A
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