Maximize Power Delivered to Circuits
Optimization Problem

The first derivative is used to maximize (optimize) the power delivered to a load in electronic circuits.

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 = \(R i^2\)

\(r\) and \(R\) being positive, determine
R so that the power \(P\) delivered to \(R\) is maximum.
maximize power problem 1

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 = \(R i^2\).
\[ \text{P(R)} = \frac{R E^2}{(r + R)^2} \]
We now differentiate \(P\) with respect to the variable \(R\) \[ \frac{dP}{dR} = \frac{E^2 [(r + R)^2 - R 2 (r + R)]}{[(r + R)^4]} = \frac{E^2 [(r + R) - 2 R]}{(r + R)^3} = \frac{E^2 (r - R)}{(r + R)^3} \] To find out whether \(P\) has a local maximum, we need to find the critical points by setting \(\dfrac{dP}{dR} = 0\) and solve for \(R\).
Since \(r\) and \(R\) are both positive (resistances), \(\dfrac{dP}{dR}\) has only one critical point at \(R = r\). Also for \(R \lt r\), \(\dfrac{dP}{dR}\) is positive and \(P\) increases, and for \(R > r\), \(\dfrac{dP}{dR}\) is negative and \(P\) decreases. Hence \(P\) has a maximum value at \(R = r\). The maximum power is found by setting \(R = r\) in \(P(R)\)
\[ \text{P(r)} = \frac{r E^2}{(r + r)^2} = \frac{E^2}{4r} \]
So in order to have maximum power transfer from the electronic circuit to the load \(R\), the resistance of \(R\) has to be equal to \(r\).
As an example, the plot of \(P(R)\) for \(E = 5\) volts and \(r = 100\) Ohms is shown below and it clearly shows that \(P\) is maximum when \(R = 100\) Ohms = \(r\).

plot of power P(R) in problem 1

Let us examine \(P(R)\) again. If \(R\) approaches zero, \(P(R)\) also approaches zero. If \(R\) increases indefinitely, \(P(R)\) approaches zero since the horizontal asymptote of the graph of \(P(R)\) is the horizontal axis. So that somewhere for a finite value (found to be \(r\)) \(P(R)\) has a maximum value.

References and Links

calculus problems

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