# Induction Problem Related to Cars

1. Oct 12, 2013

### NasuSama

Induction Problem Related to Cars....

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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

2. Oct 12, 2013

### NasuSama

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.

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

I don't know how to answer part c.

3. Oct 13, 2013

### NasuSama

Does anyone have any idea of how to approach this problem? I'm really stuck on part (c).