Maximize Power Delivered to Circuits
The first derivative is used to maximize the power delivered to a load in electronic circuits.
Problem 1: 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 = 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.
Solution to Problem 1:
We first express power P in terms of E, r and the variable R by substituting i = E / (r + R) into P = R i 2.
P(R) = R i 2 = R E 2 / (r + R) 2
We now differentiate P with respect to the variable R
dP / dR = E 2 [ (r + R) 2 - R 2 (r + R) ] / [(r + R) 4 ]
= E 2 [ (r + R) - 2 R ] / [(r + R) 3 ]
= 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
dP / dR = 0 and solve for R.
Since r and R are both positive (resistances) dP / dR has only one critical point at R = r. Also for R < r, dP / dR is positive and P increases and for R > r, 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)
P(r) = r E 2 / (r + r) 2 = 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.
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.
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