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Two resistances, R1 and R2, are connected in series across a 15-V battery. The current increases by 0.050 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.025 A when R1 is removed, leaving R2 connected across the battery.

Find R1 and R2

  • Physics - ,

    V= IR
    in this case V = I(R1+R2)

    so we have 15 = IR1 + IR2

    (15 - IR1)/I = R2 we will use this later in our subsitution.

    We also know that...

    15 =(I+.05)R1 and 15 = (I+.025)R2

    since both equal 15, they must equal each other:

    (I+.05)R1 = (I+.025)R2

    Use the substitution for R2 from above and you will be able to solve for R1. by eliminating the I.

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