A 250 V shunt motor has armature current 100 A runs at a speed of 300 rpm. The armature resistance is 0.10Ω. If the shunt field is reduced to 50% of its normal value and the armature current to 50A, then the new speed of the shunt motor will be :

Correct option is A

​$$\text{Given: } R_a = 0.10 \, \Omega, \quad I_{a1} = 100A$$

$$N_1 = 300 \text{ rpm}, \quad I_{a2} = 50A$$

$$\text{From EMF equation of DC motor-}$$

$$\boxed{V_t = E_b + I_a R_a}$$

$$E_{b1} = 250 - 0.10 \times 100$$

$$E_{b1} = 250 - 10$$ = $$\boxed{E_{b1} = 240 \text{ volt}}$$

$$E_{b2} = 250 - 0.10 \times 50$$

$${E_{b2} = 245 \text{ volt}}$$​​​​

​​​​​​​According to question-

​$$\frac{E_{b2}}{E_{b1}} = \frac{\phi_2 \times N_2}{\phi_1 \times N_1}\quad$$

$$\frac{245}{240} = \frac{\phi_1 / 2}{\phi_1} \times \frac{N_2}{300}\quad$$

$$N_2 = \frac{245 \times 300 \times 2}{240}\quad$$

$$N_2 = \frac{147000}{240}$$

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