Electric Generators and Motors

PH 202 SI

Session 4

*If you are no comfortable with Lenz’s Law, go back and review all right-hand-rules. If you still are not comfortable, come to me for help. This worksheet is going to skip Lenz’s Law and go over electric generators and motors.

1. A 120.0-V motor draws a current of 7.00 A, when running at normal speed. The resistance of the armature wire is 0.720 . (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to 15.0 A?

Remembering Ohm’s Law, with a current of 7.00 A running through the wire and a

resistance of 0.720 , the voltage drop: V=IR, V=(7.00 A)( 0.720 ) = 5.04 VRMS

Since the back EMF opposes the current induced by the applied emf, the back emf

can be calculated by 120 V – 5.04 V = 114.96 VRMS

When the motor is not rotating, the back emf has to be 0. So, the current at this point in time is just 120 V/0.720 )=I, I=166.7 A

For the starting current to be 15.0 A, the resistance must be 120 V/15.0 A= R, R= 8. Thus, some resistance, x, must be added to the wire’s resistance to yield an overall resistance of 8 . 8 =0.720 + x, x= 7.28 .

2. A generator has a square coil consisting on 248 turns. The coil rotates at 79.1 rad/s in a 0.170-T magnetic field. The peak output of the generator is 75.0 V. What is the length of one side of the coil?

Applying Faraday’s Law, we can make the following derivation: E0sin(t)=NABsin(t)

The sin(t) terms divide out to give 1

E0=NAB

75.0 V=(248)(A)(0.170 T)(79.1 rad/s)

Since it is a square shaped coil L=W so A= L2

L=0.150 m

3. You need to design a 60.0-Hz ac generator that has a maximum emf of 5500 V. The generator is to contain a 150-turn coil that has an area per turn of 0.85 m2. What should be the magnitude of the magnetic field in which the coil rotates?

Applying Faraday’s Law again:

E0sin(t)=NABsin(t)

We need to convert the given frequency into angular frequency

=2(60.0 Hz)

=120 rad/s

E0=NAB

5500 V=(150)(0.85 m2)B(120 rad/s)

B=0.114 T

4. A motor is designed to operate on 117 V and draws a current of 12.2 A when it first starts up. At its normal speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one third of the normal speed.

Using Ohm’s Law:

R=(117 V)/(12.2 A)

R=9.59 (a)

At normal speed: Voltage drop = IR = (2.30 A)(9.59 ) = 22.06 V

Back emf = 117 V-22.06 V = 95.0 V (b)

At one-third its normal speed, the back emf is one-third its original value because

EMF=EMF0sin(t)

So back EMF at one-third normal speed is 95 V/3 = 31.7 V

So, the new “effective voltage” is 117 V-31.7 V= 85.7 V

Using Ohm’s Law again:

85.7 V/9.59  = 8.9 A (c)