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AM-GM Inequality for Minimum Power Dissipation

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Problem: AM-GM Inequality for Minimum Power Dissipation

An electrical source is connected to a circuit containing two variable resistors ##R_1## and ##R_2## in series. If the product of the two resistances is held constant such that ##R_1 R_2 = C##, where ##C## is a positive constant, determine the minimum value of the total series resistance ##R_s = R_1 + R_2## using the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Provide a rigorous mathematical proof.

Worked Solution & Step-by-Step Explanation

1. Understanding the Physical and Mathematical Constraints

In passive electrical circuits, resistance is a physical quantity that represents the opposition to the flow of electric current. By physical laws, the resistance of standard passive components must be non-negative real numbers. Therefore, we establish our domain constraints as:

###R_1 > 0, \quad R_2 > 0###

We are given that the product of these two positive resistances is constrained to a constant value ##C##:

###R_1 R_2 = C \quad (\text{where } C > 0)###

The total equivalent resistance of the two resistors connected in series, denoted as ##R_s##, is given by the sum of their individual resistances:

###R_s = R_1 + R_2###

Our objective is to find the minimum possible value of ##R_s## under the given product constraint using the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

Variable/Parameter Physical Meaning Mathematical Domain
##R_1## Resistance of the first variable resistor ##R_1 \in \mathbb{R}^+##
##R_2## Resistance of the second variable resistor ##R_2 \in \mathbb{R}^+##
##C## Constant product constraint of resistances ##C \in \mathbb{R}^+##
##R_s## Total equivalent series resistance ##R_s = R_1 + R_2##

2. Rigorous Proof of the AM-GM Inequality for Two Variables

To ensure mathematical rigor, we first prove the AM-GM inequality for two positive real variables ##x_1## and ##x_2##. Let ##x_1, x_2 \in \mathbb{R}^+##.

Consider the square of the difference between their square roots. Since the square of any real number is non-negative, we can write:

###(\sqrt{x_1} - \sqrt{x_2})^2 \ge 0###

Expanding the left-hand side of this inequality using the algebraic identity ##(a - b)^2 = a^2 - 2ab + b^2##, we obtain:

###(\sqrt{x_1})^2 - 2\sqrt{x_1}\sqrt{x_2} + (\sqrt{x_2})^2 \ge 0###

Since ##x_1## and ##x_2## are positive, ##(\sqrt{x_1})^2 = x_1## and ##(\sqrt{x_2})^2 = x_2##. Substituting these back into the expression yields:

###x_1 - 2\sqrt{x_1 x_2} + x_2 \ge 0###

Adding ##2\sqrt{x_1 x_2}## to both sides of the inequality:

###x_1 + x_2 \ge 2\sqrt{x_1 x_2}###

Dividing both sides by ##2##, we arrive at the standard formulation of the AM-GM inequality for two variables:

###\dfrac{x_1 + x_2}{2} \ge \sqrt{x_1 x_2}###
Analyzing the Condition for Equality

The equality holds if and only if the initial squared term is zero:

###(\sqrt{x_1} - \sqrt{x_2})^2 = 0###

Taking the square root of both sides:

###\sqrt{x_1} - \sqrt{x_2} = 0 \implies \sqrt{x_1} = \sqrt{x_2}###

Squaring both sides yields the final equality condition:

###x_1 = x_2###
Inequality Type Mathematical Expression Condition for Equality
Arithmetic Mean (AM) ##\dfrac{R_1 + R_2}{2}## Holds for all ##R_1, R_2 > 0##
Geometric Mean (GM) ##\sqrt{R_1 R_2}## Holds for all ##R_1, R_2 > 0##
AM-GM Relation ##\dfrac{R_1 + R_2}{2} \ge \sqrt{R_1 R_2}## ##R_1 = R_2##

3. Application of AM-GM to the Circuit Problem

We map our physical variables directly to the proven inequality. Let ##x_1 = R_1## and ##x_2 = R_2##. Since ##R_1 > 0## and ##R_2 > 0##, we apply the AM-GM inequality:

###\dfrac{R_1 + R_2}{2} \ge \sqrt{R_1 R_2}###

Multiplying both sides of the inequality by ##2##:

###R_1 + R_2 \ge 2\sqrt{R_1 R_2}###

We substitute the expression for total series resistance ##R_s = R_1 + R_2## and the given constraint ##R_1 R_2 = C## into the inequality:

###R_s \ge 2\sqrt{C}###

This inequality establishes that the total series resistance ##R_s## is bounded below by ##2\sqrt{C}##. Therefore, the minimum value of ##R_s## is:

###R_{s,\text{min}} = 2\sqrt{C}###

4. Determining the Minimizing Resistance Values

The minimum value ##R_{s,\text{min}}## is achieved when the condition for equality in the AM-GM inequality is satisfied. As proven in Section 2, equality holds if and only if the terms are equal:

###R_1 = R_2###

Using the constraint equation ##R_1 R_2 = C##, we substitute ##R_2## with ##R_1##:

###R_1 \cdot R_1 = C \implies R_1^2 = C###

Since resistance must be positive (##R_1 > 0##), we take the positive square root:

###R_1 = \sqrt{C}###

Consequently, because ##R_2 = R_1##, we find:

###R_2 = \sqrt{C}###

Thus, the minimum total series resistance is achieved when both variable resistors are set to equal values of ##\sqrt{C}##.

Scenario (with ##C = 100 \ \Omega^2##) ##R_1 \ (\Omega)## ##R_2 \ (\Omega)## Product ##R_1 R_2 \ (\Omega^2)## Total Series Resistance ##R_s \ (\Omega)##
Highly Asymmetric 1 100 100 101
Moderately Asymmetric 4 25 100 29
Symmetric (Optimal) 10 10 100 20 (Minimum)

5. Physical Significance in Power Dissipation

In electrical engineering, if this series combination is connected to a constant voltage source ##V##, the total power dissipated by the circuit is given by:

###P = \dfrac{V^2}{R_s} = \dfrac{V^2}{R_1 + R_2}###

To minimize the power dissipation of the circuit for a fixed voltage source, we must maximize the total resistance ##R_s##. Conversely, to maximize the power dissipation (or to design for maximum current flow), we minimize ##R_s##. The minimum resistance occurs when ##R_1 = R_2 = \sqrt{C}##, yielding the maximum power dissipation state for a constant voltage supply:

###P_{\text{max}} = \dfrac{V^2}{2\sqrt{C}}###

This demonstrates how algebraic optimization tools like the AM-GM inequality directly govern the limits of physical systems and energy efficiency calculations in circuit design.

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