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- Homework Statement
- A 500 W, 200 V compound motor has two field windings, one has a resistance of

0.03Ω and the other a resistance of 50 Ω.

i) If the armature resistance is 0.15 Ω, calculate the emf when it draws an

armature current of 5A whilst running at 3000 rpm.

ii) When the load increases so that the motor draws a current of 10A, what is

the new speed of the machine? (assume the field shunt current does not

change – just the series field)

- Relevant Equations
- ...

I am assuming the motor is cumulative compound wound and that it is long shunt. I also assume that the shunt field resistance ##R_{shunt}## = 50Ω, while the series field resistance is ##R_{series}## = 0.03Ω.

i) calculate EMF with armature current ##I_{a}## = 5A and armature resistance ##R_{a}## = 0.15Ω.

For a motor with long shunt compound winding, the terminal voltage is

##V_{T} = E + I_{a}(R_{a} + R_{series})##, where E is emf. Therefore

$$E = V_{T} - I_{a}(R_{a} + R_{series})$$

E = 200V - 5A * (0.15Ω + 0.03Ω) = 199.1V

ii) calculate motor speed ##n## when ##I_{a}## = 10A

This is where I get really stuck. We're told the shunt field remains constant and only the series field changes with the increased ##I_{a}##.

With indexes 1 and 2 referring to intial current and increased current respectively, I know that in general:

$$\frac{n_{2}}{n_{1}} = \frac{E_{2}}{E_{1}}\frac{\Phi_{1}}{\Phi_{2}}$$

If we had either a shunt or series wound machine, the unknown ##\Phi## would cancel out, but for our compound wound machine

$$\Phi_{2} = \Phi_{1} + \Delta\Phi = \Phi_{shunt} + \Phi_{series} + \Delta\Phi_{series}$$

with ##\Phi \propto I## this becomes

$$\Phi_{2} = \Phi_{shunt} + 3\Phi_{series}$$

This means we can't cancel the magnetic flux.

i) calculate EMF with armature current ##I_{a}## = 5A and armature resistance ##R_{a}## = 0.15Ω.

For a motor with long shunt compound winding, the terminal voltage is

##V_{T} = E + I_{a}(R_{a} + R_{series})##, where E is emf. Therefore

$$E = V_{T} - I_{a}(R_{a} + R_{series})$$

E = 200V - 5A * (0.15Ω + 0.03Ω) = 199.1V

**E = 199.1V**ii) calculate motor speed ##n## when ##I_{a}## = 10A

This is where I get really stuck. We're told the shunt field remains constant and only the series field changes with the increased ##I_{a}##.

With indexes 1 and 2 referring to intial current and increased current respectively, I know that in general:

$$\frac{n_{2}}{n_{1}} = \frac{E_{2}}{E_{1}}\frac{\Phi_{1}}{\Phi_{2}}$$

If we had either a shunt or series wound machine, the unknown ##\Phi## would cancel out, but for our compound wound machine

$$\Phi_{2} = \Phi_{1} + \Delta\Phi = \Phi_{shunt} + \Phi_{series} + \Delta\Phi_{series}$$

with ##\Phi \propto I## this becomes

$$\Phi_{2} = \Phi_{shunt} + 3\Phi_{series}$$

This means we can't cancel the magnetic flux.

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