Optimizing Heat Dissipation via Cauchy-Schwarz Inequality

Problem: Optimizing Heat Dissipation Using Cauchy-Schwarz Inequality

In an electrical circuit, a total constant current ##I## is split between two parallel resistors ##R_1## and ##R_2##. Let the currents through the two branches be ##I_1## and ##I_2## respectively. Therefore,

###I_1 + I_2 = I###

The total rate of heat generation, or power dissipation, in the two resistors is given by Joule’s law:

###H = I_1^2R_1 + I_2^2R_2###

Using the Cauchy-Schwarz inequality, determine the minimum possible value of ##H## in terms of ##I##, ##R_1##, and ##R_2##. Also identify the relationship between ##I_1## and ##I_2## when this minimum heat dissipation is achieved.

Worked Solution and Step-by-Step Explanation

We use the Cauchy-Schwarz inequality to obtain a lower bound for the expression ##I_1^2R_1 + I_2^2R_2## under the condition ##I_1+I_2=I##.

The Cauchy-Schwarz inequality states that for any two real sequences ##(a_1,a_2)## and ##(b_1,b_2)##,

###(a_1^2+a_2^2)(b_1^2+b_2^2) \geq (a_1b_1+a_2b_2)^2###

Equality holds when the two sequences are proportional.

Step 1: Choose Suitable Variables

We want the expression ##a_1^2+a_2^2## to become the heat dissipation expression ##H##. Therefore, we choose:

###a_1 = I_1\sqrt{R_1}, \qquad a_2 = I_2\sqrt{R_2}###

Then,

###a_1^2+a_2^2 = I_1^2R_1 + I_2^2R_2 = H###

Now we choose ##b_1## and ##b_2## so that the product sum ##a_1b_1+a_2b_2## gives the current constraint ##I_1+I_2=I##. For this, take:

###b_1 = \frac{1}{\sqrt{R_1}}, \qquad b_2 = \frac{1}{\sqrt{R_2}}###

Step 2: Compute the Components of the Inequality

First, compute the sum of squares of the first sequence:

###a_1^2+a_2^2 = (I_1\sqrt{R_1})^2 + (I_2\sqrt{R_2})^2###

Therefore,

###a_1^2+a_2^2 = I_1^2R_1 + I_2^2R_2 = H###

Next, compute the sum of squares of the second sequence:

###b_1^2+b_2^2 = \left(\frac{1}{\sqrt{R_1}}\right)^2 + \left(\frac{1}{\sqrt{R_2}}\right)^2###

Thus,

###b_1^2+b_2^2 = \frac{1}{R_1} + \frac{1}{R_2}###

Combining the fractions gives:

###b_1^2+b_2^2 = \frac{R_1+R_2}{R_1R_2}###

Finally, compute the product sum:

###a_1b_1+a_2b_2 = \left(I_1\sqrt{R_1}\right)\left(\frac{1}{\sqrt{R_1}}\right) + \left(I_2\sqrt{R_2}\right)\left(\frac{1}{\sqrt{R_2}}\right)###

Hence,

###a_1b_1+a_2b_2 = I_1+I_2 = I###

Step 3: Apply the Cauchy-Schwarz Inequality

Now substitute these results into:

###(a_1^2+a_2^2)(b_1^2+b_2^2) \geq (a_1b_1+a_2b_2)^2###

This gives:

###H\left(\frac{R_1+R_2}{R_1R_2}\right) \geq I^2###

Solving for ##H##, we obtain:

###H \geq \frac{I^2R_1R_2}{R_1+R_2}###

Therefore, the minimum possible heat dissipation is:

###H_{\min} = \frac{I^2R_1R_2}{R_1+R_2}###

Since the equivalent resistance of two resistors in parallel is:

###R_p = \frac{R_1R_2}{R_1+R_2}###

we can also write:

###H_{\min} = I^2R_p###

Step 4: Find the Condition for Minimum Dissipation

Equality in the Cauchy-Schwarz inequality occurs when the two chosen sequences are proportional. That is,

###\frac{a_1}{b_1} = \frac{a_2}{b_2}###

Substituting the chosen values:

###\frac{I_1\sqrt{R_1}}{\frac{1}{\sqrt{R_1}}} = \frac{I_2\sqrt{R_2}}{\frac{1}{\sqrt{R_2}}}###

Simplifying both sides gives:

###I_1R_1 = I_2R_2###

This is exactly the condition that the voltage across both parallel resistors must be the same:

###V_1 = V_2###

Since ##V_1=I_1R_1## and ##V_2=I_2R_2##, the condition becomes:

###I_1R_1 = I_2R_2###

Thus, the heat dissipation is minimized when the branch currents are distributed inversely proportional to the resistances.

From ##I_1R_1=I_2R_2##, we get:

###\frac{I_1}{I_2} = \frac{R_2}{R_1}###

Therefore, the smaller resistance carries the larger current, and the larger resistance carries the smaller current.

Physical Interpretation

This result has a direct physical meaning. In a parallel circuit, the current does not divide equally unless the resistances are equal. Instead, more current flows through the smaller resistance and less current flows through the larger resistance.

The condition

###I_1R_1 = I_2R_2###

is simply the statement that both branches of a parallel circuit have the same potential difference across them. The Cauchy-Schwarz inequality shows that this natural current distribution is also the one that minimizes total heat dissipation for a fixed total current.

Final Conclusion

For two parallel resistors ##R_1## and ##R_2## carrying a total current ##I##, the total heat dissipation is:

###H = I_1^2R_1 + I_2^2R_2###

Using the Cauchy-Schwarz inequality, we obtain:

###H \geq \frac{I^2R_1R_2}{R_1+R_2}###

Therefore, the minimum possible heat dissipation is:

###\boxed{H_{\min} = \frac{I^2R_1R_2}{R_1+R_2}}###

Equivalently, since ##R_p=\dfrac{R_1R_2}{R_1+R_2}##,

###\boxed{H_{\min}=I^2R_p}###

The equality condition is:

###\boxed{I_1R_1 = I_2R_2}###

This means that minimum heat dissipation occurs when the voltage drops across the two parallel resistors are equal.