Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(la

Question

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 1.68 A. When the resistors are connected in parallel to the battery, the total current from the battery is 11.5 A. Determine the two resistances.

Explanation / Answer

let resistors be R1 and R2 (R1 Is = V/(Rseries) = 12/(R1+R2) = 1.68 ==>R1+R2 = 12/1.68 and Ip = V/(Rparallel) = 12/{R1*R2/(R1+R2)} = 12*(R1+R2)/(R1*R2) = 144/(1.68*R1*R2) = 11.5 ==> R1R2 = 144/(1.68*11.5) on solving we get R1 = 1.269 ohm and R2 = 5.874 ohm

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