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I am really confused with the above problem. The equations which I was able to write are

Ea1 = Vt - RseIa1 = kphi1*wm1

Ea2 = Vt - 0.5*RseIa2 = kphi2*wm2

phi1*Ia1 = phi2*Ia2 [Torque is constant]

First Approach - Assuming shunt field is strong.So, phi doesn't change much



Ea2 > Ea1

So, wm2>wm1 [option B]

Second Approach - Assuming series field has appreciable contribution to net flux

phi1 = Nsh*(Vt/Rsh) + Nse*Ia1

phi2 = Nsh*(Vt/Rsh) + Nse*Ia2*0.5

Now, there are more variables than equations. So how to conclude. Please help me with the above problem.

  • \$\begingroup\$ 1. Please read how to write the equations, as they can't be understood in actual representation. 2. I guess only the prof knows the answer. If the motor already spins at rated torque, then it is already constant. \$\endgroup\$ – Marko Buršič Jan 18 '18 at 11:07
  • \$\begingroup\$ I opt for a). The excitation field weakens, then the armature field has to rise to produce the same torque. \$M\propto I_a\cdot I_f\$ \$\endgroup\$ – Marko Buršič Jan 18 '18 at 11:16

If the torque decreases, then the speed would decrease, also. If the torque increases then the speed would increase, also. If the torque remains the same then the speed remains the same.

Now if the field is shunted, then there is less excitation flux, so the only explanation is that the armature current has to increase to compensate.

$$V=k\cdot (I_{sh}+I_{ser})\cdot\Omega+ R_a\cdot I_a$$ $$M=k\cdot (I_{sh}+I_{ser})\cdot I_a$$ $$k\cdot (I_{sh}+I_{ser1})\cdot I_{a1} = k\cdot (I_{sh}+I_{ser2})\cdot I_{a2}$$ $$\dfrac{I_{a2}}{I_{a1}}=\dfrac{I_{sh}+I_{ser1}}{I_{sh}+I_{ser1}\cdot 0.5} $$

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  • \$\begingroup\$ How can you say that there is less excitation flux ? And can you please tell what is what like M,ohm symbol that you used. A diagram would be really helpful to understand. \$\endgroup\$ – Nikhil Kashyap Jan 18 '18 at 12:28
  • \$\begingroup\$ M stands for torque. Omega is the rotational speed. \$\endgroup\$ – Marko Buršič Jan 18 '18 at 12:52
  • \$\begingroup\$ How can you say that there is less excitation flux ? \$\endgroup\$ – Nikhil Kashyap Jan 18 '18 at 13:43
  • \$\begingroup\$ You have written Ise2=0.5Ise1. Doesn't Ea(Armature voltage) change ? So, you can't write like that. \$\endgroup\$ – Nikhil Kashyap Jan 18 '18 at 13:58
  • \$\begingroup\$ please help me with this problem electronics.stackexchange.com/questions/350981/…... \$\endgroup\$ – Nikhil Kashyap Jan 19 '18 at 20:10

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