GATE 2024 EE Q.18: solution and answer

GATE 2024 · EEQ.18 · 1 mark · MCQ

Which one of the following options represents possible voltage polarities in a single phase two winding transformer? Here, $V_{p}$ is the applied primary voltage, $E_{p}$ is the induced primary voltage, $V_{s}$ is the open circuit secondary voltage, and $E_{s}$ is the induced secondary voltage.

GATE 2024 EE question 18: Each option (A)-(D) shows the same core transformer schematic: two windings on a common rectangular core, primary winding… — Figure from the GATE 2024 EE paper. It is part of the question.

(A)
Figure — polarities as listed above for (A)
(B)
Figure — polarities as listed above for (B)
(C)
Figure — polarities as listed above for (C)
(D)
Figure — polarities as listed above for (D)

Answer and solution

ANSWER

(B). With the induced voltages read as potential differences across each winding, $V_{p}$ and $E_{p}$ are both $+$ at the primary dot, and $V_{s} = E_{s}$ is $+$ at the secondary dot. (The official key awarded marks to all candidates, because the counter-emf convention makes (D) consistent instead.)

Official answer: marks to all candidatesSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Convention used. The question calls $E_{p}$ and $E_{s}$ induced voltages, so we read every $+ / -$ pair as a potential difference across the same two terminals. The official key awarded marks to all candidates. Step 7 explains why.
Primary. By KVL around the primary, $V_{p} = E_{p} + I_{p} Z_{p}$ , where $I_{p} Z_{p}$ is the small leakage-impedance drop of the no-load current. So $V_{p} \approx E_{p}$ : at any instant the applied voltage and the induced voltage have the same polarity at the same terminal.
Secondary. The secondary is on open circuit, so $I_{s} = 0$ and there is no drop in the winding: $V_{s} = E_{s}$ exactly. They too must have the same polarity at the same terminal.
Dots. Both windings link the same mutual flux $ϕ$ , so $e_{p} = N_{p} \frac{d ϕ}{d t}$ and $e_{s} = N_{s} \frac{d ϕ}{d t}$ rise and fall together. The dots mark the terminals whose induced voltages are positive at the same instant. In every option the dots are at the primary's bottom terminal and the secondary's top terminal. So if $E_{p}$ is $+$ at the bottom of the primary, $E_{s}$ must be $+$ at the top of the secondary.
(C) and (D) fail the primary condition. At the top primary terminal they show $V_{p}$ as $+$ but $E_{p}$ as $-$ . Opposite signs at the same terminal contradict $V_{p} \approx E_{p}$ . (C) also shows $V_{s}$ as $+$ and $E_{s}$ as $-$ at the top, which contradicts $V_{s} = E_{s}$ .
(A) fails the dot condition; (B) meets every condition. In (A), $V_{p}$ and $E_{p}$ agree and $V_{s}$ and $E_{s}$ agree. But $E_{p}$ is $-$ at the primary dot while $E_{s}$ is $+$ at the secondary dot, which violates the dot rule. In (B), $V_{p}$ and $E_{p}$ are both $+$ at the primary dot (bottom), and $V_{s}$ and $E_{s}$ are both $+$ at the secondary dot (top). Every condition holds, so (B) is the answer.
Why marks were awarded to all. Many textbooks draw the primary induced emf as a counter-emf opposing the applied voltage. In phasors that is $\overset{ˉ}{V}_{p} = - \overset{ˉ}{E}_{p} + \overset{ˉ}{I}_{p} Z_{p}$ , with $E_{p}$ marked opposite to $V_{p}$ . Under that convention (D) is self-consistent. $E_{p}$ is $+$ at the primary dot, so the primary dot is actually at $-$ potential. The secondary dot must then also be at $-$ , so $V_{s} = E_{s}$ is $+$ at the undotted bottom, as (D) shows. The answer therefore depends on how the question means the $E_{p}$ marking. With $E_{p}$ read as the voltage across the winding, as the question words it, the answer is (B).

KEY IDEA

Check three things for each option. On the primary, $V_{p} \approx E_{p}$ because the leakage drop is small. On the open-circuit secondary, $V_{s} = E_{s}$ . And because one mutual flux induces both windings, the induced voltages at the two dotted terminals have the same instantaneous polarity. Only (B) passes all three when $E$ is read as the voltage across the winding.

WATCH OUT FOR

One mistake is to apply the dot rule to $V_{p}$ alone and ignore how the $E$ markings relate to the $V$ markings. Another is to mix conventions: drawing $E_{p}$ as a counter-emf opposing $V_{p}$ while treating $E_{s}$ as the secondary's terminal voltage. That can make (D) look right. Fix one sign convention first, then apply it to both windings.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 14 Transformers questions, with notes

GATE 2024 EE Q.18: solution and answer

Answer and solution

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