Question

When two unknown resistors are connected in series with a battery, the battery delivers 255 W and carries a total current of 5.00 A. For the same total current, 40.5 W is delivered when the resistors are connected in parallel. Determine the values of the two resistors.

_____Ω (higher resistance)
_____Ω (lower resistance)

Answer 1

Suppose the battery voltage is V and R1, R2 be the resistors' resistances.

when the resistance are connected in series, total resistance = R1 + R2

V = I(R1+R2) = 5(R1+R2)

now, Power, P = VI = 25(R1+R2) = 255

=> R1 + R2 = 255 / 25 = 10.2------------------------------------------------(i)

Again -
When the resistors are connected in parallel, the total resistance is R!R2/(R1+R2),
and the power P = VI = I^2R = 25R1R2/(R!+R2) = 40.5

=> R1R2/(R1+R2) = 1.62

=> R1R2 = 1.62*(R1+R2) = 1.62*10.2 = 16.52-----------------------------------(ii)

Now solve these two equations to find the answers.

R1 + 16.52/R1 = 10.2

=> R1^2 - 10.2R1 + 16.52 = 0

R1 = (10.2 - sqrt(10.2^2 - 4*16.52))/2

=> R1 = ((10.2 - sqrt(37.96)) / 2 = 2.02 ohm
R2 = 10.2 - 2.02 = 8.18 ohms

So, higher resistance = 8.18 ohms

lower resistance = 2.02 ohms.