The value of each resistor is equal to (Ps/Pp) - 2 Ohms.
<h3>How to determine the value of each resistor?</h3>
Let the numerical value of two unknown resistors be R₁ and R₂ respectively.
Based on the information provided, the total equivalent resistance of these two unknown resistors connected in series with a battery is given by:
Rt = R₁ + R₂
Also, power is given by:
P = I²R
Ps = I²(R₁ + R₂) .....equation 1.
When the resistors are connected in parallel, we have:
Rt = (R₁R₂/R₁ + R₂)
Pp = I²(R₁ + R₂) .....equation 2.
Dividing eqn. 1 by eqn. 2, we have:
Ps/Pp = (R₁ + R₂)²/R₁R₂
Ps/Pp = [R₁(1 + R₂/R₁)²]/R₁R₂
Let x = R₂/R₁;
So, the power becomes;
Ps/Pp = x(1 + x)²
xPs/Pp = x² + 2x + 1
Ppx² + 2xPp + Pp = xPs
Ppx² + 2xPp + Pp - xPs = 0
x² + [(2Pp - Ps)/Pp]x + 1 = 0
Next, we would solve for x by using the quadratic formula:

x = (Ps/Pp) - 2
Therefore, x = R₂/R₁ = (Ps/Pp) - 2 Ohms.
Read more on resistance in parallel here: brainly.com/question/23282393
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