# The back EMF of a DC motor spinning at full speed

1. Apr 4, 2012

### OmegaFury

1. The problem statement, all variables and given/known data
The resistance of the armature windings of a DC motor is 5Ω. The motor is connected to a 120V line. When the motor is spinning at full speed, it draws a current of 4A. The back EMF at full speed is:

2. Relevant equations
Pav=$\frac{1}{2}$I2peakR
Pav=$\frac{1}{2}$εpeakIpeak

3. The attempt at a solution
Pav=$\frac{1}{2}$I2peakR
=$\frac{1}{2}$(4A)25Ω=40W
Pav=$\frac{1}{2}$εpeakIpeak
2Pav/Ipeakpeak
2(40W)/4A=20V.
This isn't correct. I also didn't know what to do with the 120V line.

2. Apr 4, 2012

### Staff: Mentor

Presumably in the motor the back EMF opposes the line voltage. That reduces the net voltage that appears across the winding resistance, and it's the net voltage that drives the current through the winding resistance.

3. Apr 4, 2012

### OmegaFury

Does that mean that the back emf is equal to the line voltage minus the peak emf?

4. Apr 4, 2012

### Staff: Mentor

Peak emf? At constant rotation rate ("spinning at full speed") the back emf will be a constant value.

5. Apr 4, 2012

### OmegaFury

I don't follow. I honestly don't fully understand what peak values are. Or rms values for that matter. They just come up in formulas for average power... All I know is that it has something to do with an alternating current, and they arise from moving electrical charges that periodically switch direction. It supposedly has some advantage over a direct, one directional current, but what that advantage is and why, I have no clue.

6. Apr 4, 2012

### Staff: Mentor

Here you're dealing with a DC motor. The supply voltage is DC. You can expect the back emf to be a constant DC value as well so long as RPM is constant.

If the supply voltage is V and the back emf is E, and the current is I and the coil resistance is R, (whew!), then if E opposes V, what is the current? What formula would you write for the net potential if "E opposes V"?

7. Apr 4, 2012

### OmegaFury

The current would be 120V/5Ω= 24A

.........E= -V?

8. Apr 4, 2012

### Staff: Mentor

Nope. Because 120V is the line voltage, not the net voltage across the coil. The back EMF opposes the line voltage. Also, you're GIVEN the current...
Nope. That would make the net potential across the coil zero, so that no current would flow.

9. Apr 4, 2012

### Kavik

Perhaps a simple diagram will help:

If you treat the EMF voltage E as a simple DC Voltage supply, what would the equation for Ia equal? You can then move that around to solve for E.

10. Apr 4, 2012

### OmegaFury

Okay. So. You meant find the voltage in the motor. I suppose that would be V=IR=4A x 5Ω=20V.

Umm... E= Vline-Vmotor

11. Apr 4, 2012

### Staff: Mentor

Yes, that looks good. So E = ?

12. Apr 4, 2012

### OmegaFury

120V-20V= 100V

That took me forever X__X

13. Apr 4, 2012

### Staff: Mentor

Perhaps. But then, you're probably going to remember it

14. Apr 4, 2012

### OmegaFury

Lol. Very true. Frustration is well-remembered while ease is a fleeting memory.