082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

12.2 Maximum points and minimum points 413
and therefore the systemisunderdamped because ζ < 1.
Also,
ω r
= 1 √
LC
= 5000
α = R 2L = 200
= 2500
2×4×10−2 √
β = ωr 2 −α2 = √ 5000 2 −2500 2 = 4330
Finally,
t m
=
π
4330 = 7.26 ×10−4 = 726 µs
We conclude forthis case that the risetimeis726 µs.
Engineeringapplication12.2
Maximumpowertransfer
ConsiderthecircuitofFigure12.9inwhichanon-idealvoltagesourceisconnected
toavariableloadresistorwithresistanceR L
.ThesourcevoltageisV anditsinternal
resistanceisR S
.CalculatethevalueofR L
whichresultsinthemaximumpowerbeing
transferred from the voltage source to the load resistor. This is an essential piece of
information for engineers involved in the design of power systems. Often an importantdesignconsiderationistotransferthemaximumamountfromthepowersource
tothe pointwhere the power isbeingconsumed.
i
R S
Non-ideal
voltage
source
+
V
–
R L
Figure12.9
Maximum power transferoccurs whenR L =R S .
Solution
Letibethecurrentflowinginthecircuit.UsingKirchhoff’svoltagelawandOhm’s
law gives
V =i(R S
+R L
)
LetPbethe power developed inthe loadresistor.Then,
P=i 2 R L
=
V2 R L
(R S
+R L
) 2 ➔