steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
The Max Power Transform Theorem is an important one in circuit.
Connect a battery v across two resistors in series r1 and r2. Given r1 what value of r2 mxizes the power transferred to r2?
The circuit equation is
PowerR2 = ((v/(r1 + r2))^2)*r2 in watts. Take the 1st derivative, set to 0, and solve.
The 1st 2nd derivative min max test
Derivation of the proof.
A numerical solution. Numerically take the derivatives and evaluate.
Connect a battery v across two resistors in series r1 and r2. Given r1 what value of r2 mxizes the power transferred to r2?
The circuit equation is
PowerR2 = ((v/(r1 + r2))^2)*r2 in watts. Take the 1st derivative, set to 0, and solve.
The 1st 2nd derivative min max test
Finding Maxima and Minima using Derivatives
Where is a function at a high or low point? Calculus can help ... A maximum is a high point and a minimum is a low point
www.mathsisfun.com
Derivation of the proof.
Maximum Power Transfer Theorem
Maximum Power Transfer Theorem - The amount of power received by a load is an important parameter in electrical and electronic applications. In DC circuits, we can represent the load with a resistor having resistance of RL ohms. Similarly, in AC circuits, we can represent it with a complex load...
www.tutorialspoint.com
A numerical solution. Numerically take the derivatives and evaluate.
Code:
REM Max Power
option explicit
Sub Main
dim r2max, dr, r1, amps, r ,v,val as double
dim N,i,j, flag, index as integer
r1 = 1.
v = 1.
r2max = 5.
dr = .1
N = r2max/dr
dim pr2(N),dpr2_1(N),dpr2_2(N),r2(N) as double
r = 0.
for i = 1 to N
r2(i) = r
r = r + dr
amps = v/(r1 + r2(i)) ' current
pr2(i) = (amps^2)*r2(i) 'r2 power watts
next i
flag = 0
for i = 1 to N-1
dpr2_1(i) = (pr2(i+1) - pr2(i))/dr '1st derivative
' find 1st derivative zero crossing
if flag = 0 AND dpr2_1(i)<= 0. then
flag = 1
index = i
end if
next i
for i = 1 to N-2
dpr2_2(i) = (dpr2_1(i+1) - dpr2_2(i))/dr '2nd derivative
next i
print r2(index)
select case dpr2_2(index)
case > 0.: msgbox "MINIMUN"
case = 0: msgbox "NO SOLUTION"
case < 0.: msgbox "MAXIMUM"
end select
dim Doc As Object
dim cell as object
Dim Sheet As Object
Doc = ThisComponent
Sheet = Doc.Sheets (0)
for j = 1 to n-2
Cell = Sheet.getCellByPosition(0,j)
Cell.Value = 0.
Cell.Value = r2(j)
Cell = Sheet.getCellByPosition(1,j)
Cell.Value = 0.
Cell.Value = dpr2_1(j)
Cell = Sheet.getCellByPosition(2,j)
Cell.Value = 0.
Cell.Value = dpr2_2(j)
next j
msgbox("done")
End Sub