GATE 2023 EE Q.30: solution and answer

GATE 2023 · EEQ.30 · 1 mark · NAT

A separately excited DC motor rated 400 V, 15 A, 1500 RPM drives a constant torque load at rated speed operating from 400 V DC supply drawing rated current. The armature resistance is 1.2 Ω. If the supply voltage drops by 10% with field current unaltered then the resultant speed of the motor in RPM is ____________ (Round off to the nearest integer).

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Answer and solution

ANSWER

The resultant speed of the motor is $\approx 1343 RPM$ .

Official answer: 1340 to 1345Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Rated condition: $E_{b 1} = V_{1} - I_{a} R_{a} = 400 - 15 \times 1.2 = 382 V$
Constant torque load with field current unaltered $\Rightarrow ϕ$ is constant; since $T = k ϕ I_{a}$ must also stay constant with $ϕ$ fixed, the armature current stays at its rated value $I_{a} = 15 A$ even after the supply voltage drops
New supply voltage: $V_{2} = 0.9 \times 400 = 360 V$
New back EMF: $E_{b 2} = V_{2} - I_{a} R_{a} = 360 - 15 \times 1.2 = 342 V$
Since $ϕ$ is unchanged, $E_{b} \propto N$ , so $\frac{N _{2}}{N _{1}} = \frac{E _{b 2}}{E _{b 1}}$
$N_{2} = 1500 \times \frac{342}{382} = 1342.93 RPM$
Rounding to the nearest integer: $N_{2} \approx 1343 RPM$

KEY IDEA

For a separately excited DC motor, $E_{b} = V - I_{a} R_{a}$ and $E_{b} \propto N ϕ$ . With field current unaltered, $ϕ$ is constant. A constant-torque load means $T = k ϕ I_{a}$ stays constant, so with $ϕ$ fixed, $I_{a}$ stays fixed too — it does NOT change just because the supply voltage changes. So only $V$ changes, and $N_{2} = N_{1} \times \frac{E _{b 2}}{E _{b 1}}$ .

WATCH OUT FOR

Assuming the armature current changes when the supply voltage drops, or forgetting that a "constant torque load with unaltered field current" pins $I_{a}$ at its rated value throughout, and instead recomputing $I_{a}$ from the new terminal voltage.

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GATE 2023 EE Q.30: solution and answer

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