The first derivative is used to maximize (optimize) the power delivered to a load in electronic circuits.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
In the electronic circuit shown below, the voltage \( E \) (in Volts) and resistance \( r \) (in Ohms) are constant. \( R \) is the resistance of a load. In such a circuit, the electric current \( i \) is given by:
\[ i = \dfrac{E}{r + R} \]and the power \( P \) delivered to the load \( R \) is given by:
\[ P = R i^2 \]Given that \( r \) and \( R \) are positive, determine \( R \) so that the power \( P \) delivered to \( R \) is maximum.
Show Solution to the Problem
We first express power \( P \) in terms of \( E \), \( r \), and the variable \( R \) by substituting \( i = \dfrac{E}{r + R} \) into \( P = R i^2 \):
\[ P(R) = \frac{R E^2}{(r + R)^2} \]We now differentiate \( P \) with respect to the variable \( R \) using the quotient rule:
\[ \frac{dP}{dR} = \frac{E^2 \left[(r + R)^2 - R \cdot 2(r + R)\right]}{(r + R)^4} = \frac{E^2 \left[(r + R) - 2R\right]}{(r + R)^3} = \frac{E^2 (r - R)}{(r + R)^3} \]To find out whether \( P \) has a local maximum, we find the critical points by setting \( \dfrac{dP}{dR} = 0 \) and solving for \( R \):
\[ \frac{E^2 (r - R)}{(r + R)^3} = 0 \implies R = r \]Since \( r \) and \( R \) are both positive (resistances), \( \dfrac{dP}{dR} \) has only one critical point at \( R = r \). Furthermore, for \( R < r \), \( \dfrac{dP}{dR} \) is positive and \( P \) increases; for \( R > r \), \( \dfrac{dP}{dR} \) is negative and \( P \) decreases. Hence, \( P \) has an absolute maximum value at \( R = r \).
The maximum power is found by substituting \( R = r \) into \( P(R) \):
\[ P(r) = \frac{r E^2}{(r + r)^2} = \frac{r E^2}{(2r)^2} = \frac{r E^2}{4r^2} = \frac{E^2}{4r} \]Therefore, to achieve maximum power transfer from the electronic circuit to the load \( R \), the load resistance \( R \) must equal the internal resistance \( r \).
As an example, the plot of \( P(R) \) for \( E = 5 \) volts and \( r = 100 \) Ohms is shown below. It clearly illustrates that \( P \) is maximized when \( R = 100 \) Ohms (\( R = r \)).
Let us examine \( P(R) \) further: if \( R \) approaches zero, \( P(R) \) approaches zero. If \( R \) increases indefinitely, \( P(R) \) approaches zero since the horizontal axis is a horizontal asymptote of the graph of \( P(R) \). Thus, \( P(R)\) attains a maximum at a finite value of \( R = r \).