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

22 Chapter 1 Review of algebraic techniques
Asimplified circuitdiagram of a vehicle isshown inFigure 1.3.
Internal
resistance
R
I
Terminals
V
Motor and gearbox
Drive
wheels
Battery pack
Figure1.3
Electric vehicle wiringdiagram.
The total power delivered by the battery pack is
power = voltage ×current =VI
This is shared between loss due to the internal resistance and the power, P, to the
motor. The power loss due to the internal resistance isI 2 R (see Engineering application
1.1).So the equation for the power inthe circuitis
VI=I 2 R+P
This can be rewritten into the formof a quadratic equation
RI 2 −VI+P=0
whichcan besolved tocalculate the currentinthe wireforaparticular power delivered
to the motor. It is important to know the current in order to specify the size of
the fuses,the motor controller and the wirediameters used inthe vehicle.
Consider the case where the power output is 2 kW. If the circuit parameters are
V = 150 volts, R = 1.6 ,we have
1.6I 2 −150I +2000 = 0
The solutions tothe quadratic equation are
I = −b ± √ b 2 −4ac
2a
= 77.7 A, 16.1 A
= −(−150) ± √ (−150) 2 − (4 ×1.6 ×2000)
2×1.6
Therelevantsolutiondependsontheelectricalcharacteristicsofthemotorusedinthe
circuit. In practice, the larger of the two currents would correspond to a substantial
loss inthe internal resistance and would beavoided by the correctchoice of motor.
The technical computing language MATLAB ® has the function roots which
finds the solutions of a polynomial equation. In this example we would type
roots ([1.6 -150 2000]) at the command line to obtain the results calculated
above.