An electric motor draws a current of 2.60 A from a 18.0-V battery when it runs at normal speed.If the motor's armature is "jammed" so that it cannot rotate,the current suddenly rises to 10.0 A.What is the back emf of the motor when running at normal speed?

Question

Quizplus · Accepted Answer

According to the law of conservation of energy, the energy provided by the battery must be equal to the sum of the energy dissipated by the motor and the energy stored as kinetic energy.

When the motor runs at normal speed, the back emf is given by:

V = ε + IR

where V is the battery voltage, ε is the back emf, I is the current, and R is the internal resistance of the motor.

We have V = 18.0 V, I = 2.60 A, and solving for ε we get:

ε = V - IR = 18.0 V - 2.60 A * R

To find R, we can use the fact that when the motor is jammed, the current is 10.0 A:

V = ε + IR  => R = (V - ε) / I = (18.0 V - 0 V) / 10.0 A = 1.80 Ω

Substituting into the first equation, we get:

ε = 18.0 V - 2.60 A * 1.80 Ω = 13.32 V

Therefore, the back emf of the motor when running at normal speed is 13.3 V (option D).

An Electric Motor Draws a Current of 2