GATE 2022 EE Q.62: solution and answer

GATE 2022 · EEQ.62 · 2 marks · NAT

A 280 V, separately excited DC motor with armature resistance of 1 Ω 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 Ω is connected in series with the armature, is ________. (round off to nearest integer)

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Answer and solution

ANSWER

The motor speed with the additional 10 Ω resistance in the armature circuit is 483 rpm.

Official answer: 480 to 485Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

Back EMF at the first operating point: $E_{b 1} = V - I_{a 1} R_{a} = 280 - 30 (1) = 250 V$
Since the field flux is constant, back EMF is proportional to speed: $E_{b} \propto N \Rightarrow \frac{E _{b 1}}{N _{1}} = \frac{250}{1000} = 0.25 V/rpm$ , so $E_{b 2} = 0.25 N_{2}$
Since flux is constant, developed torque is proportional to armature current: $T \propto I_{a}$ . The load torque is given as proportional to speed: $T_{L} \propto N$ . At steady state $T = T_{L}$ , so $I_{a} \propto N$
Using the first operating point: $\frac{I _{a 1}}{N _{1}} = \frac{30}{1000} = 0.03 A/rpm$ , so $I_{a 2} = 0.03 N_{2}$
After adding the series resistance, the total armature-circuit resistance is $R_{a} + R_{e x t} = 1 + 10 = 11 Ω$
Applying the voltage equation at the new operating point: $V = E_{b 2} + I_{a 2} (11)$ $280 = 0.25 N_{2} + 0.03 (11) N_{2} = 0.25 N_{2} + 0.33 N_{2} = 0.58 N_{2}$
Solving: $N_{2} = \frac{280}{0.58} \approx 482.76 rpm \approx 483 rpm$

KEY IDEA

With constant field flux, back-EMF is proportional to speed (Eb = k·N) and developed torque is proportional to armature current (T = k·Ia). At steady state the developed torque equals the load torque, and since load torque is proportional to speed here, armature current also becomes proportional to speed (Ia = k'·N). Writing the voltage equation V = Eb + Ia·R for both operating points (before and after adding series resistance) and using the two proportionality constants found from the first operating point gives a single linear equation in the new speed.

WATCH OUT FOR

Assuming armature current stays the same (30 A) after adding the series resistance, or assuming speed stays constant and only current changes — both ignore that the load torque itself changes with speed, which is what links Ia to N in this problem (T ∝ Ia and T_L ∝ N together force Ia ∝ N at every steady-state operating point, not just current or speed alone).

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GATE 2022 EE Q.62: solution and answer

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