# Calculate values of externally excited DC generator/motor

We use a DC externally excited generator/motor as a generator.We excite the input of the generator and get at the output 10V at 1Wb flux and at 5rps generator speed without connecting the generator to a load.

Now we use the same DC externally excited generated/motor as a motor with 10V input voltage.The 10V voltage source provides 60W of power.The armature resistance is 1Ω.Find the rpm of the motor.

Solution:

Since we havent connected the generator to a load the output voltage Vo will be equal to the voltage produced by the generator Vg = 10V.

The input power of the voltage source is 60W so P= VI->I = P/V = 6A.

The voltage of the motor Vm will be equal to Vin-IarRar = 10-6 = 4V

rpm = rps*60 = 180rpm.

Is that correct?

## 1 answer

Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.

## Added

I didn't have much time when I wrote the above, but can delve into this more now.

From your first measurement, we can see that the pure generator inside this unit produces 2 V/(Wb Hz). Since you are using the fixed exitation of 1 Wb throughout this question, we can simplify that to 2 V/Hz for this case.

Now the question is how fast the unit will spin without load when 10 V are applied. You know that the armature resistance is 1 Ω and that the unit draws 60 W when the 10 V is applied.

We'll use the simple model of an ideal motor/generator with the armature resistance in series. To get an idea of speed, we need to know the voltage applied to the internal ideal motor. To get that we find the voltage drop across the resistance, then subtract that from the applied 10 V.

The current from a 10 V source delivering 60 W is (60 W)/(10 V) = 6 A. The voltage drop from that current across the internal resistance is then (6 A)(1 Ω) = 6 V. That means the voltage across the internal ideal motor/generator is (10 V) - (6 V) = 4 V.

Since the ideal generator inside this motor is producing 4 V, we know that the motor speed is (4 V)/(2 V/Hz) = 2 Hz = 120 RPM.

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