Problem 3 - Calculus with Concepts in Calculus

Question

Consider the circuit shown in Figure 4.52, consisting of a battery having a constant source voltage \(E\), constant internal resistance \(r\), and a variable external resistance \(R\). When current flows through the circuit, the power \(P\) dissipated in the external resistance is given by $$P=\frac{E^{2} R}{(R+r)^{2}}$$ Assume that \(E\) and \(r\) are positive constants. Show that the largest power dissipation occurs when \(R=r\).

Step-by-Step Solution

Answer

Maximum power occurs when the external resistance equals internal resistance, \(R = r\).

Step 1: Understand the Formula for Power Dissipation

You are given the power dissipation formula: \(P = \frac{E^2 R}{(R+r)^2}\). In this formula, \(E\) and \(r\) are constants, and \(R\) is the variable external resistance. The goal is to maximize \(P\).

Step 2: Differentiate the Power Function

To find the maximum power dissipation, differentiate \(P\) with respect to \(R\). Apply the quotient rule: \[ \frac{dP}{dR} = \frac{(R+r)^2 \cdot E^2 - E^2 R \cdot 2(R+r)}{(R+r)^4} = \frac{E^2 (R+r - 2R)}{(R+r)^3} = \frac{E^2 (r - R)}{(R+r)^3} \]

Step 3: Set the Derivative to Zero and Solve for R

To find the critical points, set the derivative \(\frac{dP}{dR}\) to zero and solve for \(R\):\[ \frac{E^2 (r - R)}{(R+r)^3} = 0 \]This simplifies to \(r - R = 0\), hence \(R = r\).

Step 4: Verify that the Critical Point is a Maximum

After finding the critical point \(R = r\), verify it is a maximum by testing the second derivative of \(P\) or by analyzing the sign change of the first derivative. Setting \(R = r\) maximizes \(P\) as varying \(R\) near \(r\) shows that power decreases.

Key Concepts

Differentiation in Calculus

Differentiation in calculus is a powerful tool used to find the rate of change of a function as its input changes. Imagine plotting a curve on a graph, with differentiation allowing us to calculate the slope at any particular point in time. This concept is crucial when analyzing systems to find where extreme values occur.

In our exercise, we use differentiation to find when the power dissipation function reaches its maximum. By differentiating the function with respect to the variable external resistance \(R\), we can identify the point where it changes direction. This is known as finding the critical points. Differentiation informs us how quickly the power \(P\) is changing as \(R\) changes.

To do this, we take the derivative of \(P = \frac{E^2 R}{(R+r)^2}\) with respect to \(R\). This helps us understand how the power dissipation varies and shows us the optimal spot, specifically important when designing circuits for maximum efficiency.

Maximization Problem

A maximization problem asks us to find the largest value that a function can reach. In everyday terms, you might want to maximize the money you save or maximize the number of ice cream scoops on your sundae. In a circuit, it could be about maximizing the power "burnt off" or used efficiently.

In our maximization problem, we want to find out when the power \(P\) dissipated in the circuit is at its highest. To solve this, mathematicians typically rely on calculus, using differentiation to spot the highest points on a graph.

By taking the derivative of the power function and solving it, we identify at which resistance \(R\) the power is maximized. The solution tells us that when the external resistance \(R\) equals the internal resistance \(r\), power dissipation achieves its maximum. This is an elegant feature sometimes used in designing and optimizing electrical circuits, ensuring we are not wasting precious energy unnecessarily.

Quotient Rule

The quotient rule is a formula for taking the derivative of a quotient of two functions. When you have two functions divided by each other, like \(\frac{f(x)}{g(x)}\), the quotient rule comes to the rescue.

In simple terms, the rule helps us differentiate such expressions where one function is divided by another. In our power function \(P = \frac{E^2 R}{(R+r)^2}\), using the quotient rule was essential to find the derivative \(\frac{dP}{dR}\).

According to the quotient rule, if \(u(x)\) and \(v(x)\) are two functions, the derivative of their quotient is given by: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
Applying this to our function, we find the critical points needed to maximize the power. This mathematical tool is just one of the many ways calculus equips us to explore and solve real-world problems efficiently.