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Problem: Optimizing Heat Dissipation via Cauchy-Schwarz Inequality
In an electrical circuit, a total constant current ##I## is split between two parallel resistors ##R_1## and ##R_2##, such that ##I_1 + I_2 = I##. The total rate of heat generation in the parallel network is given by the power dissipation formula:
Using the Cauchy-Schwarz inequality, determine the minimum value of ##H## in terms of ##I##, ##R_1##, and ##R_2##. Furthermore, identify the relationship between the currents ##I_1## and ##I_2## when this minimum dissipation is achieved.
Worked Solution & Step-by-Step Explanation
To solve this optimization problem, we utilize the Cauchy-Schwarz inequality, which provides a powerful tool for establishing lower bounds on sums of squares.
The Cauchy-Schwarz inequality states that for any two sets of real numbers ##(a_1, a_2)## and ##(b_1, b_2)##:
Equality holds if and only if the sequences are proportional, i.e., ##\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2}##.
Step 1: Mapping the variables
To apply the inequality to our heat dissipation expression ##H = I_1^2 R_1 + I_2^2 R_2##, we must map our physical terms to the mathematical variables ##a_i## and ##b_i##. We define:
- ##a_1 = I_1 \sqrt{R_1}##
- ##a_2 = I_2 \sqrt{R_2}##
Now, we choose ##b_1## and ##b_2## such that the linear combination ##a_1 b_1 + a_2 b_2## yields the constraint ##I_1 + I_2 = I##. We choose:
- ##b_1 = \dfrac{1}{\sqrt{R_1}}##
- ##b_2 = \dfrac{1}{\sqrt{R_2}}##
Step 2: Constructing the components
We compute the squares and products required by the inequality:
1. The sum of squares of the first set:
2. The sum of squares of the second set:
3. The product sum:
Step 3: Applying the inequality
Substituting these components into the inequality ##(a_1^2 + a_2^2)(b_1^2 + b_2^2) \ge (a_1 b_1 + a_2 b_2)^2##:
Solving for ##H##, we obtain the lower bound:
Since the equivalent resistance of two resistors in parallel is ##R_p = \dfrac{R_1 R_2}{R_1 + R_2}##, we can write:
Step 4: Finding the condition for minimum dissipation
Equality occurs when ##\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2}##. Substituting our definitions:
Simplifying this expression:
This result is precisely the condition for parallel circuits where the voltage drop across both resistors must be equal (##V = I_1 R_1 = I_2 R_2##). Thus, physical systems naturally distribute current to satisfy the condition that minimizes energy dissipation.
| Quantity | Expression | ||
|---|---|---|---|
| :--- | :--- | ||
| Minimum Power | ##H_{\text{min}} = I^2 R_p## | ||
| Parallel Resistance | ##R_p = \dfrac{R_1 R_2}{R_1 + R_2}## | ||
| Current Condition | ##I_1 R_1 = I_2 R_2## |
RESOURCES
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