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16 October, 03:00

a battery of 9V is connected in series with resistors of 0.2Ω,0.3Ω,0.4Ω,0.5Ωand 12Ω resistors. How much current would flow through th 12Ω resistors?

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  1. 16 October, 05:09
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    current flowing through 12Ω = 0.67 A

    The resistances are given by 0.2Ω,0.3Ω,0.4Ω,0.5Ωand 12Ω

    in series, we add the resistances, so the total resistance is given by

    R=0.2+0.3+0.4+0.5+12=13.4 Ω

    using ohm's law V = i R

    9 = i (13.4)

    i = 9/13.4

    i=0.67 A
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