When two unknown resistors are connected in series with a ba

When two unknown resistors are connected in series with a battery, the battery delivers 230 W and carries a total current of 5.00 A. For the same total current, 35.5 W is delivered when the resistors are connected in parallel. Determine the values of the two resistors.

Solution

When in series:

Req = R1 +R2

P = I^2*Rs = 230 W

Rs = P/I^2 = 230/5^2 = 9.2 ohm

R1 + R2 = 9.2

R2 = 9.2 - R1

When in parallel

Req = R1*R2/(R1+R2)

P = 35.5 W = I^2*Rp

Rp = 35.5/5^2 = 1.42 ohm

R1*R2/(R1 + R2) = 1.42

R1*(9.2 - R1)/9.2 = 1.42

9.2*R1 - R1^2 = 9.2*1.42

R1^2 - 9.2*R1 + 9.2*1.42 = 0

R1 = 7.45 ohm

R2 = 1.75 ohm

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