GATE 2025 EE Q.26: solution and answer

GATE 2025 · EEQ.26 · 1 mark · MCQ

The operating region of the developed torque ( $T_{e m}$ ) and speed ( $ω$ ) of an induction motor drive is given by the shaded region OQRE in the figure. The load torque ( $T_{L}$ ) characteristic is also shown. The motor drive moves from the initial operating point O to the final operating point S. Which one of the following trajectories will take the shortest time?

GATE 2025 EE question 26: A T_em (vertical axis) vs ω (horizontal axis) plot, origin at O. — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
$O - Q - R - S$
(B)
$O - P - S$
(C)
$O - E - S$
(D)
$O - F - S$

Answer and solution

ANSWER

Option (A), the trajectory $O$ – $Q$ – $R$ – $S$ , takes the shortest time.

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

The drive obeys $J \frac{d ω}{d t} = T_{e m} - T_{L}$ , so $d t = \frac{J d ω}{T _{e m} ( ω ) - T _{L} ( ω )}$ , and the total transit time from O to S is $t = \int_{0}^{ω_{E}} \frac{J d ω}{T _{e m} ( ω ) - T _{L} ( ω )}$ .
The speed excursion is fixed (0 to $ω_{E}$ , since both E and S sit at $ω = ω_{E}$ ), so $t$ is minimized by making $T_{e m} (ω)$ as large as possible — i.e. by maximizing $T_{e m} - T_{L}$ — at every speed inside the feasible region OQRE.
The upper boundary of the feasible region is the horizontal segment $QR$ , at the constant torque level $T_{Q}$ . This is the largest torque the drive can deliver at any speed between $0$ and $ω_{E}$ .
Along the path $O$ – $Q$ – $R$ – $S$ : torque rises from $0$ to $T_{Q}$ at $ω = 0$ (negligible time, since speed is unchanged), then the drive accelerates at the constant maximum torque $T_{Q}$ from $ω = 0$ to $ω = ω_{E}$ (segment $Q$ – $R$ ), and finally torque adjusts from $T_{Q}$ down to the value needed at $S$ , again at $ω = ω_{E}$ (negligible time).
Over the whole speed range, this path keeps $T_{e m}$ pinned at its maximum feasible value $T_{Q}$ , giving the largest possible $(T_{e m} - T_{L})$ and hence the largest possible $d ω / d t$ at every speed in $(0, ω_{E})$ .
Any of the other paths ( $O$ – $P$ – $S$ , $O$ – $E$ – $S$ , $O$ – $F$ – $S$ ) has $T_{e m} (ω) < T_{Q}$ over at least part of the range $0$ to $ω_{E}$ (P, the O–E–S diagonal, and F–S along/near $T_{L}$ all lie below the QR level at intermediate speeds), so acceleration is lower there and the transit time is longer.
Hence $O$ – $Q$ – $R$ – $S$ , which rides the upper torque boundary of the feasible region for the entire speed excursion, takes the shortest time.
Therefore option (A), $O$ – $Q$ – $R$ – $S$ , is correct.

KEY IDEA

Since $J d ω / d t = T_{e m} - T_{L}$ , the transit time over a fixed speed range is minimized by keeping the developed torque $T_{e m}$ at its largest feasible value at every speed — i.e., by riding the upper boundary of the allowed operating region.

WATCH OUT FOR

Assuming the shortest-time path is the one with the shortest-looking geometric route (e.g. the diagonal $O$ – $P$ – $S$ or $O$ – $E$ – $S$ ) rather than reasoning about which path maximizes accelerating torque $(T_{e m} - T_{L})$ at every intermediate speed.

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 17 Induction machines questions, with notes

GATE 2025 EE Q.26: solution and answer

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