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Induction Problem Related to Cars

  • Thread starter NasuSama
  • Start date
  • #1
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Induction Problem Related to Cars....

Homework Statement



A small electric car overcomes a 260N friction force when traveling 23km/h . The electric motor is powered by ten 12-V batteries connected in series and is coupled directly to the wheels whose diameters are 57cm . The 240 armature coils are rectangular, 11cm by 12cm , and rotate in a 0.55T magnetic field.

(a) How much current does the motor draw to produce the required torque?

(b) What is the back emf?

(c) How much power is dissipated in the coils?

(d) What percent of the input power is used to drive the car?

Homework Equations



##\text{Power} = Fv##
##E = \dfrac{nBA}{t}##
##V = IR##

The Attempt at a Solution



Only answer to part b is correct.

a)

force, F = 260

v = 23 km/h = 23/3.6 = 6.39 m/s

power = F*v = 260*6.39 = 1661.1 W

Also, Power = V*I

V = 12*10 = 120 V

So, I = Power/V = 415.3/120 = 13.84 A <---------answer


b)

back emf = NBAW <--- B= magnetic field = 0.55 T,

W = angular speed = speed/radius

diameter = 57cm = 0.57 m

so, radius = 0.57/2 = 0.285 m,

So, W = 6.39/0.285 = 22.42 rad/s

N = number of turns = 240,

A = area = 0.11*0.12 = 0.0132 m2

So, back emf = 240*0.55*0.0132*22.42 = 39.1 V <-------answer


c)

power dissipated in coils = (V - back emf)*current

= (120-39.1)*13.84 = 1119.7 W <----------------answer


d)

percentage of input power = 1119.7/1661.1 = 67.4 percent <-------answer
 

Answers and Replies

  • #2
326
3
Ignore part a, b and d. I got the right answers contrast to the answers I previously posted. I really need help in part c.

My answers are:

a. 43 A
b. 39 V
c. 1300 W
d. 33%

I don't know how to answer part c.
 
  • #3
326
3
Does anyone have any idea of how to approach this problem? I'm really stuck on part (c).
 

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