GATE 2020 EE Q.28: solution and answer

GATE 2020 · EEQ.28 · 2 marks · MCQ

A 250 V dc shunt motor has an armature resistance of $0.2 Ω$ and a field resistance of $100 Ω$ . When the motor is operated on no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to:

(A)
1200
(B)
1000
(C)
1220
(D)
900

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Answer and solution

ANSWER

The loaded speed is closest to 1220 rpm — option C.

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

Field current is fixed by the rated voltage and field resistance: $I_{f} = \frac{V}{R _{f}} = \frac{250}{100} = 2.5 A$ (unchanged at no-load and under load, since the shunt field sees rated voltage throughout).
No-load condition: armature current $I_{a 1} = 5 A$ (given directly). Back EMF: $E_{b 1} = V - I_{a 1} R_{a} - V_{b r u s h} = 250 - 5 (0.2) - 2 (1) = 250 - 1 - 2 = 247 V$ .
Loaded condition: total line current $I_{L} = 50 A$ , so armature current $I_{a 2} = I_{L} - I_{f} = 50 - 2.5 = 47.5 A$ .
Back EMF under load: $E_{b 2} = V - I_{a 2} R_{a} - V_{b r u s h} = 250 - 47.5 (0.2) - 2 (1) = 250 - 9.5 - 2 = 238.5 V$ .
Flux under load is reduced by armature reaction: $ϕ_{2} = 0.95 ϕ_{1}$ .
Since $E_{b} \propto N ϕ$ : $\frac{N _{2}}{N _{1}} = \frac{E _{b 2}}{E _{b 1}} \times \frac{ϕ _{1}}{ϕ _{2}} = \frac{238.5}{247} \times \frac{1}{0.95}$ .
Computing: $\frac{238.5}{247} = 0.9656$ , and $\frac{0.9656}{0.95} = 1.0164$ .
$N_{2} = 1200 \times 1.0164 \approx 1220 rpm$ , matching option (C).

KEY IDEA

For a dc shunt motor, back EMF $E_{b} = V - I_{a} R_{a} - V_{b r u s h}$ , and speed is proportional to $E_{b} / ϕ$ (since $E_{b} \propto N ϕ$ ). Field current stays fixed (rated voltage across a fixed field resistance), so armature current at each condition is found by subtracting field current from line current (no-load) or total line current (loaded).

WATCH OUT FOR

Using the given no-load armature current directly (correct) but then forgetting to subtract the field current from the loaded total line current to get loaded armature current — using $I_{a} = 50 A$ instead of 47.5 A. Also common: forgetting the 2 V total brush drop (1 V/brush × 2 brushes), or applying the flux reduction the wrong way (multiplying instead of dividing by 0.95 in the speed ratio).

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GATE 2020 EE Q.28: solution and answer

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