Problem 4 - Calculus with Concepts in Calculus

Question

The power output of an electric generator is \(V I\), where \(V\) is the constant terminal voltage and \(I\) is the variable current. The power loss is \(P+I^{2} R\), where \(P\) is constant and \(I^{2} R\) is the power lost to heat through the internal resistance \(R\). The efficiency \(E\) of the generator is given by $$E=\frac{\text { power output }}{\text { power input }}=\frac{V I}{V I+P+I^{2} R}$$ Assume that \(V, P\), and \(R\) are positive constants. Find the current for which the efficiency is maximum.

Step-by-Step Solution

Answer

The current that maximizes efficiency is \( I = \sqrt{\frac{VP}{R}} \).

Step 1: Understanding the Efficiency Formula

The efficiency formula given is \[ E = \frac{VI}{VI + P + I^2 R} \]where \(VI\) is the power output and \(VI + P + I^2 R\) is the power input. Our goal is to maximize this efficiency \(E\) by finding the current \(I\) that achieves this.

Step 2: Identify the Maximization Problem

To find the maximum efficiency, we need to find the maximum value of \(E\). This involves taking the derivative of \(E\) with respect to \(I\) and finding when this derivative equals zero.

Step 3: Differentiate the Efficiency Function

Differentiate \(E\) with respect to \(I\): Apply the quotient rule: if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\).Here, \(u = VI\) and \(v = VI + P + I^2 R\), so:\[ E'(I) = \frac{(VI)'(VI + P + I^2 R) - VI(VI + P + I^2 R)'}{(VI + P + I^2 R)^2} \]

Step 4: Apply Derivatives

Calculate derivatives:\( (VI)' = V \) (since \(V\) is constant and derivative of \(I\) is 1)\( (VI + P + I^2 R)' = V + 2IR \)Substitute these into the derivative formula to get:\[ E'(I) = \frac{V(VI + P + I^2 R) - VI(V + 2IR)}{(VI + P + I^2 R)^2} \]

Step 5: Simplify and Solve for Critical Points

Simplify the numerator:\[ V(VI + P + I^2 R) - VI(V + 2IR) = V^2I + VP + VI^2R - V^2I - 2IV^2R \]\[ = VP - I^2R \]Set this equal to zero to find critical points:\[ VP - I^2R = 0 \]\[ I^2R = VP \]\[ I^2 = \frac{VP}{R} \]\[ I = \sqrt{\frac{VP}{R}} \]

Step 6: Verify Maximum with Second Derivative or Endpoint Analysis

To confirm that \(I = \sqrt{\frac{VP}{R}}\) gives maximum efficiency, check second derivative or endpoints if applicable. Here, examine behavior as \(I\) varies from 0 to larger values, confirming that around the derived point, efficiency increases and then decreases.

Key Concepts

Efficiency in Calculus

When dealing with issues related to efficiency in calculus, especially in real-world scenarios like that of an electric generator, it is crucial to grasp the importance of maximizing output while minimizing losses. In the case of a generator, the efficiency formula is given by: \[ E = \frac{VI}{VI + P + I^2 R} \].
Here's what happens:

• The numerator \(VI\) represents the power output.
• The denominator \(VI + P + I^2 R\) sums up the total input power, including losses.

To ensure peak efficiency, one must determine the optimal current \(I\) that maximizes \(E\). This process is a crucial part of calculus optimization as it involves analyzing and improving functional performance using mathematical tools and principles.
Understanding how each term in the efficiency formula affects the overall system helps in identifying where improvements and adjustments can be made within the constraints of the constants \(V\), \(P\), and \(R\).

Differentiation Techniques

Differentiation techniques play a key role in finding the maximum or minimum values of functions, which is the heart of solving optimization problems like the one at hand. To find the current that maximizes efficiency, differentiation of the efficiency function \(E(I)\) is required.
This problem specifically uses the quotient rule, which is pivotal when dealing with ratios such as power efficiency:

• Assume the function is \( f(x) = \frac{u(x)}{v(x)} \).
• The derivative \( f'(x) \) is given by \( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

For our efficiency function, we set:

• \( u = VI \)
• \( v = VI + P + I^2 R \)

The differentiation process leads us from defining these derivatives to simplifying the expressions in a manner that reveals critical points. Differentiation not only helps find these critical points but also confirms optimality through subsequent checks.

Maximization Problem

Solving a maximization problem involves finding the critical point where a function attains its highest value. In the context of generator efficiency, the practical task is to find the current \(I\) that maximizes the efficiency \(E\).
After using differentiation techniques to derive \(E'(I)\), the next step is solving: \[ E'(I) = \frac{V(VI + P + I^2 R) - VI(V + 2IR)}{(VI + P + I^2 R)^2} \].
Simplifying the problem to find the critical point reveals: \[ I = \sqrt{\frac{VP}{R}} \]. This shows the current value that maximizes efficiency based on the constants \(V\), \(P\), and \(R\).
Verifying this result involves assessing the behavior of efficiency around this value. This can be checked using a second derivative test or by analyzing the speed at which efficiency increases or decreases relative to changes in \(I\). By evaluating these factors, one can confidently confirm that \(I = \sqrt{\frac{VP}{R}}\) indeed maximizes the generator's efficiency, striking a balance between power output and power loss.