GATE 2023 EE Q.53: solution and answer

GATE 2023 · EEQ.53 · 2 marks · NAT

(NAT, 2 marks) When the winding c-d of the single-phase, 50 Hz, two winding transformer is supplied from an AC current source of frequency 50 Hz, the rated voltage of 200 V (rms), 50 Hz is obtained at the open-circuited terminals a-b. The cross sectional area of the core is 5000 mm^2 and the average core length traversed by the mutual flux is 500 mm. The maximum allowable flux density in the core is B_max = 1 Wb/m^2 and the relative permeability of the core material is 5000. The leakage impedance of the winding a-b and winding c-d at 50 Hz are (5 + j100π×0.16) Ω and (11.25 + j100π×0.36) Ω, respectively. Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding a-b in millihenry is ______ (Round off to 1 decimal place).

GATE 2023 EE question 53: A single-phase two-winding transformer schematic. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The self-inductance of winding a-b is $\approx 2199.8 mH$ .

Official answer: 2150 to 2250Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Winding a-b is open-circuited, so no current flows in it and its terminal voltage equals its induced rms EMF: $E_{ab} = 200 V$ at $f = 50 Hz$ .
The transformer is operating at the stated maximum allowable flux density, so the peak mutual flux is $Φ_{ma x} = B_{ma x} A_{cor e} = 1 \times 5000 \times 1 0^{- 6} = 5 \times 1 0^{- 3} Wb$ .
From the EMF equation $E_{ab} = 4.44 f N_{ab} Φ_{ma x}$ : $N_{ab} = \frac{E _{ab}}{4.44 f Φ _{ma x}} = \frac{200}{4.44 \times 50 \times 5 \times 1 0 ^{- 3}} = \frac{200}{1.11} \approx 180.18 turns$
Core permeance per turn $^{2}$ (linear magnetizing characteristic assumed): $\frac{μ _{0} μ _{r} A}{l} = \frac{( 4 π \times 1 0 ^{- 7} ) ( 5000 ) ( 5 \times 1 0 ^{- 3} )}{0.5} \approx 6.2832 \times 1 0^{- 5} H$
Magnetizing inductance referred to winding a-b: $L_{m, ab} = N_{ab}^{2} \times 6.2832 \times 1 0^{- 5} \approx (180.18)^{2} \times 6.2832 \times 1 0^{- 5} \approx 2.0398 H$
The leakage impedance of winding a-b itself is $5 + j 100 π \times 0.16 Ω$ at $ω = 100 π rad/s$ , so its leakage inductance is directly $L_{l, ab} = 0.16 H$ (the c-d winding's leakage data is not used for $L_{aa}$ ).
Total self-inductance of winding a-b: $L_{aa} = L_{l, ab} + L_{m, ab} = 0.16 + 2.0398 \approx 2.1998 H = 2199.8 mH$

KEY IDEA

The a-b winding is open-circuited, so its rated induced EMF (200 V) together with the given $B_{ma x}$ (the flux density at which the transformer is being operated) fixes its number of turns via $E = 4.44 f N Φ_{ma x}$ . The core's magnetic permeance ( $μ_{0} μ_{r} A / l$ ) then gives the magnetizing inductance referred to that same winding as $N^{2}$ times the permeance. The total self-inductance of a real winding is always (its own leakage inductance) + (its own magnetizing inductance referred to it) — the leakage impedance data for the c-d side is a distractor for this particular sub-question.

WATCH OUT FOR

Using the leakage reactance value of the c-d winding (or adding both leakage inductances) instead of only the a-b winding's own leakage inductance when forming $L_{aa}$ ; or forgetting that $B_{ma x}$ pins down the actual peak flux (via the a-b EMF equation) needed to get $N_{ab}$ , and instead trying to solve for turns ratio from incomplete data.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 14 Transformers questions, with notes

GATE 2023 EE Q.53: solution and answer

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