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When two unknown resistors are connected in series with a battery, the battery delivers 210 W...

When two unknown resistors are connected in series with a battery, the battery delivers 210 W and carries a total current of 5.00 A. For the same total current, 39.5 W is delivered when the resistors are connected in parallel. Determine the values of the two resistors.

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Answer #1

Let the unknown resistors be R1 & R2.

In Series,
P = I^2 * (R1+R2)
210 = 5.0^2 *  (R1+R2)
R1 + R2 = 8.4 --------------1

In Parallel,
Req = (R1*R2)/(R1+R2)

I1*R1 = I2*R2
I1 + I2 = 5.0

(5.0 - I2) * (8.4 - R2) = I2 * R2 ------1

P = I1^2 * R1 + I2^2 * R2
39.5 = (5.0 - I2)^2 * (8.4 - R2) +  I2^2 * R2 -------2

Solving two eq,
Values of the two resistors,  ​
R1 = 6.3 ohm
R2 = 2.1 ohm


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