# Homework Help: Circuit/Back EMF question

1. Jan 4, 2009

### maccha

I realize this is probably a very simple question, but for some reason I'm getting very confused.

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

A motor is connected to a 12 V dc supply and draws 5.0 A when it first starts. What will be the back emf when the motor is operating at full speed and drawing 1.2 A?

2. Relevant equations

R=V/I

Source voltage= IR + Back EMF

3. The attempt at a solution

I realize you just have to use Ohm's law and solve for resistance, and then plug in that resistance and current and source voltage into the second equation and solve for Back EMF. What I'm confused about is which current I plug in where. I think I would plug in the initial current (5.0 A) to solve for resistance and then use the second curren (1.2 A) in the second formula to find back EMF, but I'm not completely sure as to why, so I would really appreciate an explanation.

2. Jan 4, 2009

### rl.bhat

Your reasoning is correct, because when the coil rotates at full speed, it produces back emf which is in the opposite direction to the source voltage. Hence there is a decrease in the current. The resistance of the coil does not change.

3. Jan 5, 2009

### Phrak

It's confusing because it's called 'back EMF'.

4. Jan 5, 2009

### rl.bhat

A motor consist of a coil and a magnetic field. when you connect a dc source, the coil experiences a force due to which it starts rotating. As the speed of rotation increases, change in flux in the coil will induce an emf which will try to slow down the speed. Due to this emf the effective emf across the coil will be reduced and hence the current will decrease.

5. Jan 6, 2009

### Gnosis

I’ll assume that you’re likely referring to a commonplace PMDC (Permanent Magnet Direct Current) motor.

First, divide the applied 12 volts by the initial “start-up” current (5 amps). This yields the armature’s winding resistance. “Start-up current” (also referred to as “stall current”) is the current drawn by the armature windings while the armature is completely motionless. On small motors, they can easily be held motionless for a couple seconds, just long enough to get a reading from the current meter.

Note: Don’t hold the armature motionless for any more than a few seconds or your could overheat the winding insulation and render the motor useless. Keep in mind that 12 volts x 5 amps = 60 watts, so a motionless armature begins to heat up fairly quickly.

R = E / I, yields the resistance of the windings in the armature, so:

12 volts applied / 5 amps = 2.4 ohms (armature winding resistance)

To calculate back emf:

Find the effective voltage that’s actually driving the armature at full speed. This is accomplished by multiplying the 1.2 amps that you measured (or were given) at full speed by the armature resistance of 2.4 ohms (1.2 amps x 2.4 ohms = 2.88 volts of effective voltage). You now subtract the effective voltage of 2.88 volts from the 12 volts applied (12 volts – 2.88 volts = 9.12 volts of back emf voltage is present).

6. Jan 6, 2009

### Phrak

Missing among the "know data" is the fact that the so called back emf is proportional to rotor frequency. Without this datum, R cannot be solved.