Question:
A DC shunt generator running at 300 RPM delivers 100 kW at 200 V. When belt breaks, it becomes a motor taking 10 kW from 200 V supply. Armature resistance = 0.025 \(\Omega\), field resistance = 50 \(\Omega\), brush drop = 2 V. Neglect armature reaction. Speed of motor is \(\underline{\hspace{2cm}}\) RPM. (Round off to 2 decimals.)
A DC shunt generator running at 300 RPM delivers 100 kW at 200 V. When belt breaks, it becomes a motor taking 10 kW from 200 V supply. Armature resistance = 0.025 \(\Omega\), field resistance = 50 \(\Omega\), brush drop = 2 V. Neglect armature reaction. Speed of motor is \(\underline{\hspace{2cm}}\) RPM. (Round off to 2 decimals.)
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In DC shunt machines, speed is directly proportional to back EMF when flux is constant.
Updated On: Dec 29, 2025
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Correct Answer: 273
Solution and Explanation
Generator mode EMF:
\[ E_g = V + I_a R_a + V_b \] Load current:
\[ I_L = \frac{100000}{200} = 500\,A \] Thus:
\[ E_g = 200 + (500)(0.025) + 2 = 214.5\text{ V} \] In motor mode, input power = 10 kW:
\[ I_a = \frac{10000}{200} = 50\,A \] Motor back-EMF:
\[ E_m = V - I_aR_a - V_b \] \[ E_m = 200 - (50)(0.025) - 2 = 196.75\text{ V} \] Speed proportional to back EMF:
\[ \frac{N_m}{N_g} = \frac{E_m}{E_g} \] \[ N_m = 300 \times \frac{196.75}{214.5} = 275.2\text{ RPM} \] \[ \boxed{275.20\ \text{RPM}} \]
\[ E_g = V + I_a R_a + V_b \] Load current:
\[ I_L = \frac{100000}{200} = 500\,A \] Thus:
\[ E_g = 200 + (500)(0.025) + 2 = 214.5\text{ V} \] In motor mode, input power = 10 kW:
\[ I_a = \frac{10000}{200} = 50\,A \] Motor back-EMF:
\[ E_m = V - I_aR_a - V_b \] \[ E_m = 200 - (50)(0.025) - 2 = 196.75\text{ V} \] Speed proportional to back EMF:
\[ \frac{N_m}{N_g} = \frac{E_m}{E_g} \] \[ N_m = 300 \times \frac{196.75}{214.5} = 275.2\text{ RPM} \] \[ \boxed{275.20\ \text{RPM}} \]
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