GATE 2015 EE – Question 58
A separately excited DC motor runs at 1000 rpm on no load when its armature terminals are connected to a 200 V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is $1\ \Omega$. The no-load armature current is negligible. With the motor developing its full rated torque, the armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rotational losses, the full load armature current (in Ampere) is ________.
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Correct answer: 8
Explanation
On no load the back emf is 200 V at 1000 rpm, so the constant is 0.2 V per rpm. At full load, $V_a - I_a \times 1 = 0.2 \times 500 = 100$ V. At half torque the armature current is $\frac{I_a}{2}$ and $V_a - \frac{I_a}{2} = 0.2 \times 520 = 104$ V. Subtracting gives $\frac{I_a}{2} = 4$, so $I_a = 8$ A.