PHYSICS 2

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.99 A. When the resistors are connected in parallel to the battery, the total current from the battery is 10.7 A. Determine the two resistances.

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  1. Let the smaller resistance be R1 and the larger R2.

    1.99 = 12.0/(R1 + R2)

    10.7 = 12/R1 + 12/R2

    Solve the simultaneous equations.

    R1 + R2 = 6.03
    R2 = 6.03 - R1
    10.7 = 12/R1 + 12/(6.03 - R1)
    10.7*R1*(6.03-R1) = 12(6.03-R1) +12R1
    64.51 R1 -10.7 R1^2 = 72.36

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