Question

S When two unknown resistors are connected in series with a battery, the battery delivers total power Ps and carries a total current of I. For the same total current, a total power Pp is delivered when the resistors are connected in parallel. Determine the value of each resistor.

Answer 1

The value of each resistor is equal to (Ps/Pp) - 2 Ohms.

How to determine the value of each resistor?

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:

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{-(\frac{2P_p -P_s)}{P_p} )\; \pm \;\sqrt{(\frac{2P_p -P_s)}{P_p} )^2 - 4 }}{2}$

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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