GATE EE · Electrical Machines

Transformers – GATE EE Electrical MachinesPractice Questions & PYQs

Generate GATE-level questions on Transformers. Focus on: 1. Equivalent circuit, Phasor diagram, and Efficiency/Regulation. 2. Losses: Hysteresis and Eddy current losses. 3. Three-phase transformers: Star/Delta connections, Parallel operation, and Autotransformers.

65 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2024MCQ

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.

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Q2GATE 2023NAT

When the winding $c-d$ of the singlephase, 50 $\mathrm{Hz}$ , two winding transformer is supplied from an AC current source of frequency $50 \mathrm{~Hz}$ , the rated voltage of $200 \mathrm{~V}$ (rms), $50 \mathrm{~Hz}$ . is obtained at the open-circuited terminals a-b. The cross sectional area of the core $5000 \mathrm{~mm}^{2}$ and the average core length traversed by the mutual flux is $500 \mathrm{~mm}$ . The maximum allowable flux density in the core is $B_{\max }=1 \mathrm{~Wb} / \mathrm{m}^{2}$ and the relative permeability of the core material is $5000$ . The leakage impedance of the winding $a-b$ and winding c-d at $50 \mathrm{~Hz}$ are $(5+j 100 \pi \times 0.16) \Omega$ and $(11.5+j 100 \pi \times 0.36) \Omega$ , 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).

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📖 Explanation

\begin{aligned} \mathrm{A} & =5000 \mathrm{~mm}^{2} \\ \mathrm{I} & =500 \mathrm{~mm} \\ \mathrm{~B}_{\max } & =1 \mathrm{~Wb} / \mathrm{m}^{2} \\ \mathrm{~V}_{\mathrm{ab}} & =200 \mathrm{~V} \\ \sqrt{2} \pi \mathrm{fN}_{1} \phi_{\max } & =200 \\ \sqrt{2 \pi} \times 50 \times \mathrm{N}_{1} \times 1 & \times 5000 \times 10^{-6}=200 \quad[\because \phi=\beta \mathrm{A}] \\ \Rightarrow \quad \mathrm{N}_{1} & =181 \end{aligned}

Now, mutual inductance,

\begin{aligned} M&=\frac{N^{2}}{I / \mu A} \\ & =\frac{4 \pi \times 10^{-7} \times 5000 \times 5000 \times 10^{-6} \times (181)^{2}}{0.5} \\ & =2.05843 \mathrm{H} \end{aligned}

Therefore, self inductance of winding.

\begin{aligned} L & =\text { Mutual + leakage } \\ & =2.05843+0.16 \\ & =2.21843 \mathrm{H} \text { or } 2218.43 \mathrm{mH} \end{aligned}

Q3GATE 2021MCQ

In a single-phase transformer, the total iron loss is $2500\:W$ at nominal voltage of $440\:V$ and frequency $50\:Hz$ . The total iron loss is $850\:W$ at $220\:V$ and $25\:Hz$ . Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are

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📖 Explanation

\begin{aligned} W_{i 1} &=2500 \mathrm{~W} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ W_{i 2} &=850 \mathrm{~W} \text { at } 220 \mathrm{~V}, 25 \mathrm{~Hz} \\ W_{i 3} &=R_{e_{3}}+P_{H_{3}} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ R_{\theta_{3}} &=?, \quad P_{h_{3}}=?, \quad \frac{v}{f}=\mathrm{constant} \\ \Rightarrow \qquad \qquad\quad 2500 &=A f+B f^{2} &\left\{\frac{400}{50}=\frac{220}{25}=\text { Constant }\right\} \\ \text { or, }\qquad \qquad \quad \frac{2500}{f} &=A+B f \\ \text{or},\qquad \qquad \frac{2500}{50}&=A+B(50) &\ldots(i)\\ \text{and}\qquad \qquad \frac{850}{25}&=\mathrm{A}+\mathrm{B}(25)&\ldots(ii) \end{aligned}

Solving (i) and (ii), we get

\begin{aligned} 25 B &=\frac{2500}{50}-\frac{850}{25}=\frac{2500-1700}{50} \\ &=\frac{800}{50}=16 \\ B &=\frac{16}{25} \\ \text{and from (i)},\qquad A &=50-\frac{16}{25} \times 50=50-32=18\\ \text{So, at} 50 \mathrm{~Hz}\\ P_{h}&=A f=18 \times 50=900 \mathrm{~W} \\ P_{e}&=B f=\left(\frac{16}{25}\right) \times (50)^{2}=1600 \mathrm{~W} \end{aligned}

Q4GATE 2020MCQ

The figure below shows the per-phase Open Circuit Characteristics (measured in V) and Short Circuit Characteristics (measured in A) of a 14 kVA, 400 V, 50 Hz, 4-pole, 3-phase, delta connected alternator, driven at 1500 rpm. The field current, $I_f$ is measured in A. Readings taken are marked as respective (x,y) coordinates in the figure. Ratio of the unsaturated and saturated synchronous impedances ( $Z_{s(unsat)}/Z_{s(sat)}$ ) of the alternator is closest to

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📖 Explanation

For unsaturated synchronous impedance

\begin{aligned} \text{At } I_f&=2A, \; V_{OC}=210V \\ \text{From SCC}\\ \text{at } I_f&=4A, \; I_{SC}=20A \\ \therefore \; I_f&=2A, \; I_{SC}=10A \\ Z_s, \text{unsaturated }&=\left.\begin{matrix} \frac{V_{OC}}{I_{SC}} \end{matrix}\right|_{I_f=const.} \\ &= \frac{210}{10}=21 \Omega \\ \text{For } Z_s, \text{ saurated}:\\ \text{when, } V_{OC}&=V_{rated}=400V\\ I_f&=8A \\ \text{From SCC, } I_f&=8A, \; I_{SC}=40A \\ Z_s, \text{saturated }&=\frac{400}{40}=10\Omega \\ \text{Hence,}\\ \frac{Z_s, \text{unsaturated }}{Z_s, \text{ saturated }}&=\frac{21}{10}=2.1\Omega \end{aligned}

Q5GATE 2020NAT

Windings 'A', 'B' and 'C' have 20 turns each and are wound on the same iron core as shown, along with winding 'X' which has 2 turns. The figure shows the sense (clockwise/ anti-clockwise) of each of the windings only and does not reflect the exact number of turns, If windings 'A', 'B' and 'C' are supplied with balanced 3-phase voltages at 50 Hz and there is no core saturation, the no-load RMS voltage (in V, rounded off to 2 decimal places) across winding 'X' is _________ .

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📖 Explanation

As per GATE official answer key MTA (Marks to ALL) $V_{X}=\frac{2}{20}(230\angle 0^{\circ}-230\angle 120^{\circ}-230\angle -120^{\circ})=46\angle 0^{\circ}\: V$

Q6GATE 2020NAT

A conducting square loop of side length 1 m is placed at a distance of 1 m from a long straight wire carrying a current l=2 A as shown below. The mutual inductance, in nH (rounded off to 2 decimal places), between the conducting loop and the long wire is __________.

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📖 Explanation

\begin{aligned} \phi &\propto I \\ \phi &=MI \\ \vec{B}&=\frac{\mu _{o}I}{2\pi \rho }\hat{a_{\phi }} \;\; (\vec{B} \text{ due to infinite long line}) \\ \text{Magnetic flux}& \text{ crossing square loop is}\\ \phi &=\int \int \vec{B}\cdot \vec{ds} \\ &=\int \int \frac{\mu _{o}I}{2\pi \rho }\hat{a_{\phi }}\cdot (d\rho dz)\hat{a_{\phi }}\\ &=\frac{\mu _{0}I}{2\pi }\int_{\rho =1}^{2}\frac{d\rho }{\rho }\int_{z=0}^{1}dz \\ \phi &=\frac{\mu _{o} I}{2\pi }(\ln \rho )_{\rho =1}^{2}(z)_{z=0}^{1} \\ \phi &=\frac{\mu _{o} I}{2\pi }(\ln 2)\\ M&=\frac{\phi }{I} \\ M&=\frac{\phi_{o}(\ln2)}{2\pi }=\frac{4\pi \times 10^{-7}(\ln _{2})}{2\pi } \\ M&=1.386\times 10^{-7}\, Henry \simeq 138.63\, nH \end{aligned}

Q7GATE 2020NAT

A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated no-load loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the highvoltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac source, the no-load losses will be _________W (rounded off to 2 decimal places).

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📖 Explanation

\begin{aligned}200 V,50 Hz, P_{c}&=450 Watt\\ 160 V,40 Hz, P_{c}&=320 Watt \\ 100 V,25 Hz,]P_{c}&=? Watt \\ \frac{v}{f}=\text{constant}&=\frac{200}{50}=\frac{160}{40}=\frac{100}{25}\\ \text{So, } P_{c}&=Af+Bf^{2} \\ 450&=A\times (50)+B\times (50)^{2} \;\; ...(i)\\ 320&=A\times (40)+B\times (40)^{2} \;\; ...(ii)\\ \text{From (i) and (ii),}\\ \frac{450}{50}&=A+B(50) \;\;...(iii) \\ \frac{320}{40}&=A+B(40) \;\;...(iv) \\ \text{Equation (iii)-(iv),} \\ (9-8)&=B(10)\\ B&=\frac{1}{10} \\ A&=9-\frac{1}{10}\times 50=4 \\ \text{Now at 100V,25 Hz,} \\P_{c}&=4\times 25+\frac{1}{10}\times (25)^{2}\\ &=100+62.5=162.50 Watt \end{aligned}

Q8GATE 2019NAT

A single-phase transformer of rating 25 kVA, supplies a 12 kW load at power factor of 0.6 lagging. The additional load at unity power factor in kW (round off to two decimal places) that may be added before this transformer exceeds its rated kVA is __________.

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📖 Explanation

For a 12 kW, 0.6 pf lagging load,

\begin{aligned} P_L &= 12kW\\ Q_L&= \frac{12}{0.6} \times \sin (cos^{-1} 0.6)\\ &= 16kVAR \end{aligned}

Trasformer rating , S=25kVA Let us assume load thet can be added is P kW then,

\begin{aligned} S^2 &=(P+P_L)^{2+}Q_L^2 \\ 25^2&= (P+12)^{2+}16^2\\ P &= 7.21kW \end{aligned}

Q9GATE 2019MCQ

A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is

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📖 Explanation

\begin{aligned} \text{Voltage regulation } &=\frac{V_{NL}-V_{FL}}{V_{NL}} \times 100 \\ &= \frac{100-95}{100} \times 100=5\% \end{aligned}

Q10GATE 2018NAT

A 3-phase 900kVA, $3kV/\sqrt{3}kV(\Delta /Y)$ , 50Hz transformer has primary (high voltage side) resistance per phase of 0.3 $\Omega$ and secondary (low voltage side) resistance per phase of 0.02 $\Omega$ . Iron loss of the transformer is 10 kW. The full load % efficiency of the transformer operated at unity power factor is _______ (up to 2 decimal places).

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📖 Explanation

$900kVA, \Delta /Y, 3-$ phase transformer

\begin{aligned} \text{Given,}\;\; R_1 &=300W/ph \\ R_2&= 0.02 W/ph\\ \text{Iron loss}&=10kW \end{aligned}

Primary line current, $I_{l_1}=\frac{900 \times 1000}{\sqrt{3} \times 3000}=173.2A$ Primary phase current, $I_{ph_1}=100A$ Secondary line current, $I_{l_2}=\frac{900 \times 1000}{\sqrt{3} \times 1732}=300A$ Secondary phase current $I_{ph_2}=300A$ Cu loss of H.V. side $=3 \times I_1^2 R_1=3(100)^2 (0.3) =9000 Watts$ Cu loss of L.V. side $= 3 \times _2^2 \times R_2 =3(300)^2(0.02)=5400Watts$ Full load efficiency at u.p.f.

\begin{aligned} \eta &=\frac{1 \times 900 \times 1000 \times 1}{1 \times 900 \times 1000+9000+5400+10000} \\ &= \frac{900000}{924400}\\ \therefore \;\;\eta &=\eta 0.9736 \; \text{or} \; 97.36\% \end{aligned}

Explanation Figure 1

Q11GATE 2018MCQ

A single-phase 100 kVA, 1000 V / 100 V, 50 Hz transformer has a voltage drop of 5% across its series impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with 0.8 lagging power factor is

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📖 Explanation

Percent voltage regulation $=\left ( \frac{I_2R_{02}}{E_2} \times 100 \right )\cos \phi _2\pm \left ( \frac{I_2X_{02}}{E_2} \times 100 \right ) \sin \phi _2$ $\% \; VR.=V_r \cos \phi _2\pm V_x \sin \phi _2$ (where, '+' lag p.f. and '-' lead p.f.) At full load: Given, $V_r=3\%$ Impedance drop, $V_z=5\%$ $\therefore \;$ Reluctance drop, $V_x=\sqrt{5^{2-}3^2}=4\%$ Voltage regulation at full load at 0.8 p.f. lagging $V.R.=3(0.8)+4(0.6)=4.8\%$

Q12GATE 2017NAT

If the primary line voltage rating is 3.3 kV (Y side) of a 25 kVA. Y- $\Delta$ transformer (the per phase turns ratio is 5:1), then the line current rating of the secondary side (in Ampere) is_____.

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📖 Explanation

From turn ratio,

\begin{aligned} V_2 &=381 V/phase \\ V_{2 \; line}&= 381 V\\ \sqrt{3} V_2I_2&= 25000\\ \therefore \;\; I_2&=\frac{25000}{\sqrt{3} \times 381}=37.88A \end{aligned}

Explanation Figure 1

Q13GATE 2017NAT

A three-phase, three winding $\Delta / \Delta / Y$ (1.1kV/6.6kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is _______.

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📖 Explanation

Load on 6.6 kV winding = 900 kVA @0.8 pf lag $\therefore \;\; I_2=\frac{900 \times 10^3}{\sqrt{3} \times 6.6 \times 10^3}$ $\;\;\;\;=78.73\angle -36.86^{\circ} A$ Transforamtion Ratio

\begin{aligned} k &=\frac{1.1}{6.6}=\frac{1}{6} \\ \therefore \;\; I'_2&=\frac{I_2}{k}=472.4\angle -36.86^{\circ} A \end{aligned}

Load on 400 V winding = 300 kVA at 0.6 p.f. $\therefore \;\; I_3=\frac{300 \times 10^3}{\sqrt{3} \times 400}$ $=433.025\angle -53.13^{\circ}A$ Transformation ratio $= \frac{1.1 \times 1000}{\frac{400}{\sqrt{3}}}=4.76$

\begin{aligned} I'_3 &=\frac{I_3}{k}=\frac{4333.025}{4.76} \\ &= 90.97\angle -53.13^{\circ}A \end{aligned}

So, line value of current

\begin{aligned} &=\sqrt{3} \times 90.97\angle -53.13^{\circ} \\ &= 157.56 \angle -53.13^{\circ} \end{aligned}

Current drawn by 1.1 kV winding will be

\begin{aligned} I_1&=I'_2+I'_3 \\ &=472.4\angle -36.86 +157.56\angle -53.13^{\circ} \\ &=377.96 -j283.37 +94.53 -126.04 \\ &= 472.49 -j409.41\\ &= 625.19\angle -40.9^{\circ}A \end{aligned}

Q14GATE 2016NAT

A single-phase, 2 kVA, 100/200 V transformer is reconnected as an auto-transformer such that its kVA rating is maximum. The new rating, in kVA, is ______.

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📖 Explanation

2 kVA, 100/200 V transformer,

\begin{aligned} A_{2 winding} &=\frac{200}{100}=2 \\ [(kVA)_{auto}]_{max} &= (a_{2winding}+1)(kVA)_{2winding} \\ &=(2+1) \times 2 =6 \end{aligned}

Q15GATE 2016MCQ

Three single-phase transformers are connected to form a delta-star three-phase transformer of 110 kV/ 11 kV. The transformer supplies at 11 kV a load of 8 MW at 0.8 p.f. lagging to a nearby plant. Neglect the transformer losses. The ratio of phase currents in delta side to star side is

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📖 Explanation

At, 100 kV, load is 8 MW, 0.8 p.f. lagging

\begin{aligned} \Rightarrow \;\; \frac{(V_{ph})_{\Delta }}{(V_{ph})_{Y}} &= \frac{(I_{ph})_{Y}}{(I_{ph})_{\Delta }}\\ \Rightarrow \;\; (I_{ph})_{\Delta } &= (I_{ph})_Y \times \frac{(V_{ph})_{Y}}{(V_{ph})_{\Delta }} \\ \frac{(I_{ph})_{\Delta }}{(V_{ph})_{Y}}&=\frac{11\sqrt{3}}{110}=1:10\sqrt{3} \end{aligned}

Q16GATE 2016MCQ

If an ideal transformer has an inductive load element at port 2 as shown in the figure below, the equivalent inductance at port 1 is

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📖 Explanation

At port 1 i.e. high voltage side impedance will be high and current will be low. So, $n^2L$ .

Explanation Figure 2

Q17GATE 2016NAT

A single-phase 400 V, 50 Hz transformer has an iron loss of 5000 W at the rated condition. When operated at 200 V, 25 Hz, the iron loss is 2000 W. When operated at 416 V, 52 Hz, the value of the hysteresis loss divided by the eddy current loss is ______.

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📖 Explanation

400V, 50Hz transformer, $P_i=5000Watt$ When, 200V, 25Hz, $P_i=2000 Watt$ 416V, 52Hz, $\frac{P_h}{P_e}=?$

\begin{aligned} P_i &=P_h+P_e \\ P_h & \propto f B_m^x \\ P_e& \propto f^2B_m^2 \end{aligned}

as in the problem

\begin{aligned} \frac{V}{f}&=\frac{400}{50}=\frac{416}{52}=8 \\ P_h&=Af \\ P_e&=Bf^2 \end{aligned}

From given data,

\begin{aligned} 2000 &=(P_{25hZ})){i} \\ &= A(25)+B(25)^2\;\;...(i)\\ 5000 &=(P_{50Hz})_i \\ &= A(50)+B(50)^2\;\;...(ii) \end{aligned}

Solving equation (i) and (ii),

\begin{aligned} A &=60,\;\; B=0.8 \\ (P_h)_{52Hz}&=Af=60 \times 52 \\ &= 3120 Watt\\ (P_e)_{52Hz}&=Bf^2=0.8 \times (52)^2 \\ &= 2163.2 Watt\\ \frac{(P_h)_{52Hz}}{(P_e)_{52Hz}}&=\frac{3120}{2163.2}=1.4423 \end{aligned}

Q18GATE 2016MCQ

A single-phase, 22 kVA, 2200 V/ 220 V, 50 Hz, distribution transformer is to be connected as an auto-transformer to get an output voltage of 2420 V. Its maximum kVA rating as an autotransformer is

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📖 Explanation

22 KVA, 2200 V/220 V, 50Hz Distribution transformer is to be connected as on auto transformer to get an output voltage of 2420V. $(kVA)_{maximum}$ as an auto transformer=? as voltage rating is 2420 i.e. (2200+220)V additive polarity,

\begin{aligned} (kVA)_{auto} &= (a_{2winding}+1) \times (kVA)_{2 winding}\\ \text{where, } \;\; a_{2winding}&=\frac{2200}{220}=10 \\ (kVA)_{auto} &= (10+1) \times 22 =242 \end{aligned}

Q19GATE 2015NAT

Two three-phase transformers are realized using single-phase transformers as shown in the figure. The phase difference (in degree) between voltages $V_1 \; and \; V_2$ is _______.

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Q20GATE 2015NAT

A 200/400 V, 50 Hz, two-winding transformer is rated at 20 kVA. Its windings are connected as an auto-transformer of rating 200/600 V. A resistive load of 12 $\Omega$ is connected to the high voltage (600 V) side of the auto-transformer. The value of equivalent load resistance (in Ohm) as seen from low voltage side is _____.

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📖 Explanation

\begin{aligned} K &=\frac{200}{600}=\frac{1}{3} \\ R_{eq} &=12 \times \frac{1}{9}=1.33\Omega \end{aligned}

Explanation Figure 1

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