An electric motor draws a current of 2.30 A from a 13.0-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 11.7 A.What is the back emf of the motor when running at normal speed?

Question

Quizplus · Accepted Answer

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

V - IaR = emf

where V is the battery voltage, Ia is the armature current, R is the resistance of the armature, and emf is the back emf.

Solving for emf, we get:

emf = V - IaR 
     = 13.0 V - 2.30 A × R

Now, when the armature is jammed and cannot rotate, the motor acts like a short circuit and the current is limited only by the resistance of the motor windings. In this case, the current is given by:

I = V / R

where I is the current, V is the battery voltage, and R is the total resistance of the motor windings.

Solving for R, we get:

R = V / I 
   = 13.0 V / 11.7 A 
   = 1.11 Ω

Now, we can use this value of R to find the back emf as follows:

emf = V - IaR 
     = 13.0 V - 2.30 A × 1.11 Ω 
     = 10.4 V

Therefore, the back emf of the motor when running at normal speed is 10.4 V, which is answer choice D.

An Electric Motor Draws a Current of 2