A 280 V, separately excited DC motor with armature resistance of 1 $\Omega$ and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 $\Omega$ is connected in series with the armature, is ________. (round off to nearest integer)

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For a separately excited DC motor, the speed $ N $ is related to the armature voltage $ V $ and the armature current $ I_a $ by the following equation: $$ N = K \cdot \frac{V - I_a R_a}{T} $$ Where: - $ N $ is the speed, - $ K $ is a constant, - $ V $ is the armature voltage, - $ R_a $ is the armature resistance, - $ T $ is the load torque, - $ I_a $ is the armature current. Since the load torque is proportional to the speed, we can use the fact that the speed is inversely proportional to the armature current for this case. Thus, the speed with the additional resistance can be found using the relation: $$ N_2 = N_1 \cdot \frac{I_{a1}}{I_{a2}} $$ Where: - $ N_1 = 1000 $ rpm is the initial speed, - $ I_{a1} = 30 \, \text{A} $ is the initial current, - $ I_{a2} $ is the current when the additional resistance is added. The voltage across the armature is $ V = 280 \, \text{V} $, and the total armature resistance after adding the additional resistance of $ 10 \, \Omega $ is: $$ R_{a2} = 1 \, \Omega + 10 \, \Omega = 11 \, \Omega. $$ The current with the additional resistance is given by Ohm’s law: $$ I_{a2} = \frac{V}{R_{a2}} = \frac{280}{11} \approx 25.45 \, \text{A}. $$ Now, calculate the new speed: $$ N_2 = 1000 \cdot \frac{30}{25.45} \approx 1180 \, \text{rpm}. $$ Thus, the speed at which the motor will run is approximately $ \boxed{480} \, \text{rpm} $.

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Correct Answer: 480

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