A small electric car overcomes a 273 N friction force when traveling 29.4 km/hr.

Question

A small electric car overcomes a 273 N friction force when traveling 29.4 km/hr. The electric motor is powered by ten 12.0 V batteries connected in series and is coupled directly to the wheels whose diameters are 49.6 cm. The 289 armature coils are rectangular, 10.7 cm by 14.6 cm, and rotate in a 0.500 T magnetic field.

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

What is the back emf?

How much power is dissipated in the coils?

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

Comment

Explanation / Answer

1)let current produced be I.

then voltage*current=toruqe*angular speed

here torque=force*radius of the wheel=273*0.5*0.496=67.704 N.m
angular speed=speed/radius=32.93 rad/sec
voltage=12*10=120 volts(as 10 batteries are connected in series)

hence current=67.704*32.93/120=18.58 A

2)induced emf =number of turns*magnetic field*area=2.257 volts

3) power dissipated in coils=voltage across coils*current through coils=2.257*18.58=41.93 W

d)input power=120*18.58=2229.6 W

power dissipated in coils=41.93 W

power used to drive the car=2229.6-41.93=2187.67 W

then percentage of input power=2187.67*100/2229.6=98.12%

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