GATE EE · Electrical Machines

Synchronous Machines – GATE EE Electrical MachinesPractice Questions & PYQs

Generate GATE-level questions on Synchronous Machines. Focus on: 1. Cylindrical and Salient pole machines: Performance, Regulation, and Parallel operation. 2. Synchronous Motors: V-curves and inverted V-curves, Starting methods. 3. Power-angle characteristics and Stability.

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

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Q1GATE 2025MSQ

Instrument(s) required to synchronize an alternator to the grid is/are

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Q2GATE 2025NAT

The induced emf in a $3.3 \mathrm{kV}, 4$ pole, 3-phase star connected synchronous motor is considered to be equal and in phase with the terminal voltage under no load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of $2^{\circ}$ mechanical with respect to the terminal voltage phasor. If the synchronous reactance is $2 \Omega$ , and stator resistance is negligible, then the motor armature current magnitude, in ampere, during loaded condition is closest to, ___________ (round off to two decimal places).

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📖 Explanation

$\quad \mathrm{V}_{\mathrm{ph}}=\frac{3.3 \times 10^{3}}{\sqrt{3}}=190.25 \mathrm{~V}$ $\mathrm{E}_{\mathrm{ph}}=\mathrm{V}_{\mathrm{ph}}$ [Given at no-load) $\theta_{\mathrm{e}}=\frac{\mathrm{P}}{2} \times \theta_{\mathrm{m}}=\frac{4}{2} \times 2^{\circ}$ $\theta_{\mathrm{e}}=4^{\circ}$ Due to load, $\mathrm{E} \angle 0^{\circ}$ will become $\mathrm{E} \angle-4^{\circ}$

\begin{aligned} & \mathrm{I}_{\mathrm{a}}=\frac{\mathrm{V}-\mathrm{E} \angle-\delta}{\mathrm{X}_{\mathrm{s}}} \\ & =\frac{1905.25-1905.25 \angle-4^{\circ}}{2 \angle 90^{\circ}}\\ & \mathrm{I}_{\mathrm{a}}=66.45-\mathrm{j} 2.32 \quad(\mathrm{or}) \quad 66.49 \angle-2^{\circ} \\ & \left|\mathrm{I}_{\mathrm{a}}\right|=\sqrt{2.32^{2}+(66.45)^{2}} \\ & =66.49 \mathrm{Amp}{ }_{\mathrm{A}} \end{aligned}

Explanation Figure 1

Q3GATE 2024MCQ

A 3-phase. $11 \mathrm{kV} .10$ MVA synchronous generator is connected to an inductive load of power factor $(\sqrt{3} / 2)$ via a lossless line with a per-phase inductive reactance of $5 \Omega$ . The per-phase synchronous reactance of the generator is $30 \Omega$ with negligible armature resistance. If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is

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📖 Explanation

$\mathrm{pF}=\frac{\sqrt{3}}{2}=30^{\circ} \text { lagging }$

\begin{aligned} E & =\sqrt{(V \cos 30+I R)^{2+}\left(V \sin 30^{\circ}+I X\right)^2} \\ E & =\sqrt{(V \cos +I R)^{2+}(V \sin +I X)} \\ \frac{11 \mathrm{~K}}{\sqrt{3}} & =\sqrt{\left(V \cdot \frac{\sqrt{3}}{2}\right)^{2+}\left(\frac{V}{2}+524.86(5)\right)^2} \\ \left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)^2 & =V^2 \cdot \frac{3}{4}+\left(\frac{V}{2}+2624.319\right)^2 \\ \left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)^2 & =\frac{3}{4} V^{2+}\frac{V^2}{4}+2624.319 \mathrm{~V}+2624.319^2 \\ V_t & =4618.105 \mathrm{~V} \\ (I Z)^2 & =E^{2+}V^{2-}2 E V \cos \delta \\ (2624.319)^2 & =\left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)^{2+}(4618.105)^{2-}2\left(\frac{11 \mathrm{~K}}{\sqrt{3}}\right)(4618.105) \cos \delta \\ \delta & =20.96 \\ \mathrm{pF} & =\cos (20.96+30)=0.629 \text { lagging } \end{aligned}

Explanation Figure 1

Q4GATE 2024MSQ

Which of the following option is/are correct for the Automatic Generation Control (AGC) and Automatic Voltage Regulator (AVR) installed with synchronous generators?

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Q5GATE 2023NAT

A three-phase synchronous motor with synchronous impedance of $0.1+j0.3$ per unit per phase has a static stability limit of 2.5 per unit. The corresponding excitation voltage in per unit is ____ (Round off to 2 decimal places).

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📖 Explanation

Developed power for motor, $P_{d e v}=\frac{V_{f}}{Z_{s}} \cos \left(\theta_{s}-\delta\right)-\frac{E_{F}^{2}}{2} \cos \theta_{s}$ At max. power developed, $\delta=\theta_{\mathrm{S}}$ and this is decide steady state stability of motor. $\therefore \mathrm{SSSL}, \mathrm{P}_{\text {dev }(max)}=\frac{V E_{F}}{\mathrm{Z}_{\mathrm{S}}}-\frac{\mathrm{E}_{\mathrm{F}}^{2}}{\mathrm{Z}_{\mathrm{S}}} \cos \theta_{\mathrm{s}}$ where, $V=1.0 \mathrm{pu}$ $\mathrm{Z}_{\mathrm{S}}=0.1+\mathrm{j} 0.3=0.316 \angle 71.6^{\circ} \mathrm{pu}$ and $\mathrm{SSSL}=2.5$ $\therefore \quad 2.5=\frac{1 \times E_{F}}{0.316}-\frac{E_{F}^{2}}{0.316} \cos 71.56^{\circ}$ $1.001 \mathrm{E}_{F}^{2}-3.16 \mathrm{E}_{F}+2.5=0$ $\Rightarrow \quad \mathrm{E}_{\mathrm{F}}=1.595 \mathrm{pu}$

Q6GATE 2022NAT

A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in volts, is __________. (round off to nearest integer).

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📖 Explanation

Power input $= 10 + 2 = 12 kW$

\begin{aligned} P&=\sqrt{3}VI \cos\phi \\ I_a&=\frac{10 \times 10^3}{\sqrt{3} \times 400 \times 0.8}\\ &=21.65\angle 36.86^{\circ}A \end{aligned}

We have, Emf equation for motor

\begin{aligned} \bar{E_{ph}}&=\bar{V_{ph}}-\bar{I_{a}}\bar{Z_{s}}\\ &=\frac{400}{\sqrt{3}}-21.65\angle 36.83^{\circ}\times 1\angle 90^{\circ}\\ &=244.542\angle -4.062^{\circ}V \end{aligned}

Q7GATE 2021NAT

An alternator with internal voltage of $1\angle \delta _{1}\text{p.u}$ and synchronous reactance of $\text{0.4 p.u}$ is connected by a transmission line of reactance $\text{0.1 p.u}$ to a synchronous motor having synchronous reactance $\text{0.35 p.u}$ and internal voltage of $0.85\angle \delta _{2}\text{p.u}$ . If the real power supplied by the alternator is $\text{0.866 p.u}$ , then $\left( \delta _{1} -\delta _{2}\right)$ is _________ degrees. (Round off to 2 decimal places.) (Machines are of non-salient type. Neglect resistances.)

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📖 Explanation

Given real power in (p.u.), $P=0.866$

\begin{aligned} P &=\frac{E V \sin \left(\delta_{1}-\delta_{2}\right)}{X_{\mathrm{eq}}}=\left|\frac{1 \times 0.85}{0.4+0.1+0.35}\right| \sin \left(\delta_{1}-\delta_{2}\right) \\ \left(\delta_{1}-\delta_{2}\right) &=60^{\circ} \end{aligned}

Explanation Figure 1

Q8GATE 2021MCQ

Consider the table given:

\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}

The correct combination that relates the constructional feature, machine type and mitigation is

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Q9GATE 2020NAT

A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now _______ A (rounded off to 2 decimal places).

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📖 Explanation

At $P_{constant},$ $I_{a1}\cos \phi _{1}=I_{a2}\cos \phi _{2}$ $\cos \phi _{1}=0.9$ $\tan \phi _{1}=0.484=\frac{Q}{P}$ $\Rightarrow \, \, \frac{2Q}{P}=0.9686=\tan \phi _{2}$ $\cos \phi _{2}=0.7182$ $\therefore \, 100\times 0.9=I_{a2}\times 0.7182$ $\Rightarrow \, \, I_{a2}=125.29\: A$

Q10GATE 2020MCQ

A three-phase cylindrical rotor synchronous generator has a synchronous reactance $X_s$ and a negligible armature resistance. The magnitude of per phase terminal voltage is $V_A$ and the magnitude of per phase induced emf is $E_A$ . Considering the following two statements, P and Q. P : For any three-phase balanced leading load connected across the terminals of this synchronous generator, $V_A$ is always more than $E_A$ . Q : For any three-phase balanced lagging load connected across the terminals of this synchronous generator, $V_A$ is always less than $E_A$ . Which of the following options is correct?

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📖 Explanation

For lagging p.f. load : For all lagging power factor loads: $E_A \gt V_A$ For unity p.f. load: Still we can see : $E_A \gt V_A$ For 'slighlty' load, phasor diagram will be quite similar to that of unity p.f. load, thus $E_A$ will be greater than $V_A$ . Thus P is false.

Explanation Figure 1

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

Q11GATE 2020NAT

A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW. as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is ________ (rounded off to 2 decimal places)

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📖 Explanation

Assumed full load frequency is 50 Hz

\begin{aligned}\tan \theta &=\frac{50-49.75}{110-100} \\ &=\frac{49.75-49.25}{(x-(110))} \\ \frac{0.25}{10}&=\frac{0.5}{(x-150)} \\ x-110&=\frac{0.5\times 10}{0.25} \\ x&=110+20=130\, MW\end{aligned}

Explanation Figure 1

Q12GATE 2020NAT

A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and subtransient reactance of 0.2 pu. It is operating at (1+j0) pu terminal voltage with an internal emf of (1+j0.7) pu. Following a three-phase solid short circuit fault at the terminal of the generator, the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is ________ pu.

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📖 Explanation

\begin{aligned} &\text{Prefault current,}\\ I_{0}&=\frac{E_{f}-V_{t}}{jX_{d}} \\ &=\frac{1+j0.7-1}{j0.7}=1 \\ &\text{Subtransient induced emf,} \\ {E_{f}}''&=V_{0}+j{X_{d}}''I_{0} \\ &=1+j0.2\times 1=1+j0.2 \\ | {E_{f}''}|&=\sqrt{1^{2}+0.2^{2}}=1.02 \end{aligned}

Q13GATE 2019NAT

A 220 V (line), three-phase, Y-connected, synchronous motor has a synchronous impedance of (0.25+j2.5) $\Omega$ /phase. The motor draws the rated current of 10 A at 0.8 pf leading. The rms value of line-to-line internal voltage in volts (round off to two decimal places) is __________.

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📖 Explanation

For synchronous motor,

\begin{aligned} E_g &=V_t-IZ \\ V_t &=\frac{220}{\sqrt{3}} \; \text{V (phase)} \\ Z &=(0.25+j2.5) \Omega \\ I &= 10\angle \cos ^{-1} 0.8A\\ E_g&= \frac{220}{\sqrt{3}}-(0.25+j2.5) \times 10\angle \cos ^{-1} 0.8\\ E_g&= 141.658 \angle -8.728^{\circ} \; \text{V (phase)}\\ E_g &= 245.36 \; \text{V(line)} \end{aligned}

Q14GATE 2019MCQ

A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at a power factor of 0.5 leading is

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📖 Explanation

$P=VI \cos \phi$ For the same voltage and load, $I \cos \phi =constant$ $I_1 \cos \phi _1=I_2 \cos \phi _2$ at unity power factor,

\begin{aligned} I_1&=200A \\ 200 \times 1 &=I_2 \times 0.5 \\ I_2&=400A \end{aligned}

Q15GATE 2018MCQ

In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is

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📖 Explanation

Salient pole synchronous motor power and torque relations per phase: $P=\frac{E_bV}{X_t}\sin \delta +\frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \;Watts$ $T=\frac{60}{2 \pi N_S}\left[ \frac{E_bV}{X_d} \sin \delta + \frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \right]\$ ; N-m $The second term is reluctance power or reluctance torque, which is directly proportional to$ \sin 2\delta $. Therefore, reluctance torque will be maximum. When,$ \delta =45^{\circ}\because ;;\sin2 (45^{\circ})\Rightarrow \sin 90^{\circ}=1$ (Maximum)

Q16GATE 2017NAT

A 25 kVA, 400 V, $\Delta$ -connected, 3-phase, cylindrical rotor synchronous generator requires a field current of 5A to maintain the rated armature current under short-circuit condition. For the same field current, the open-circuit voltage is 360 V. Neglecting the armature resistance and magnetic saturation, its voltage regulation (in % with respect to terminal voltage), when the generator delivers the rated load at 0.8 pf leading, at rated terminal voltage is _________.

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📖 Explanation

25 kVA, 400 V, $\Delta -$ synchronous generator

\begin{aligned} \therefore \;\; I_{rated} &=\frac{25000}{\sqrt{3}\times 400}=36.085A \\ I_{phase}&=\frac{36.085}{\sqrt{3}}=20.83A \\ X_3 &=\left.\begin{matrix} \frac{V_{OC}}{I_{a_{rated}}} \end{matrix}\right|_{I_{f_{same}}} \\ \therefore \;\;X_s &= \frac{360}{20.83}=17.28\Omega /phase\\ E &=\sqrt{(V \cos \phi +I_aR_a)^{2+}(V \sin \phi -I_aX_s)^2} \\ &=\sqrt{(400 \times 0.8)^2 +(400 \times 0.6 - 20.83(17.28))^2} \\ &=\sqrt{(102400)+(240-360)^2} \\ &= 341.76V\\ \therefore \;\; \% V.R. &= \frac{E-V}{V} \times 100\\ &=\frac{341.76-400}{400} \times 100=-14.56\% \end{aligned}

Q17GATE 2017NAT

A three-phase, 50Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30 $\Omega$ per phase. The load angle is $30^{\circ}$ . The power delivered to the motor in kW is _______.

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📖 Explanation

For synchronous motor, $\bar{V_t}=\bar{E_f}+j\bar{I_a}X_s$ As currrent is drawn at unity power factors. Therefore,

\begin{aligned} E_f \cos \delta &=V_t \\ E_f&= \frac{V_t}{\cos \delta }=\frac{6.6kV}{\sqrt{3}} \times \frac{1}{\cos 30^{\circ}}\\ E_f&= \frac{6.6 \times 10^3}{\sqrt{3} \times \sqrt{3}} \times 2V\\ V_t &=\frac{6.6 \times 10^3}{\sqrt{3}} \\ \therefore \;\;P_e &=\frac{3V_{ph}E_{ph}}{X_s} \sin \delta \\ &= \frac{3 \times \frac{6.6 \times 10^3}{\sqrt{3}} \times \frac{6.6 \times 10^3}{\sqrt{3} \times \sqrt{3}} \times 2 }{30} \times \sin 30^{\circ}\\ &=838.3kW \end{aligned}

Q18GATE 2017NAT

A 3-phase 50 Hz generator supplies power of 3MW at 17.32 kV to a balanced 3-phase inductive load through an overhead line. The per phase line resistance and reactance are 0.25 $\Omega$ and 3.925 $\Omega$ respectively. If the voltage at the generator terminal is 17.87 kV, the power factor of the load is ________

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📖 Explanation

\begin{aligned} V_S-V_R&=I(R \cos \phi +X_L \sin \phi) \\I &=\frac{P_R}{V_R \cos \phi } \\ V_S-V_R&= \frac{P_R}{V_R \cos \phi } (R \cos \phi +X_L \sin \phi )\\ \tan \phi &= \left ( \frac{V_SV_R-V_R^{2-}RP_R}{X_LP_R} \right )\\ \phi &=\tan ^{-1} \left[\frac{V_SV_R-V_R^{2-}RP_R}{X_LP_R} \right]\$\\ &\;\Rightarrow \;\text{substitute all values} \\ \text{power factor} &= \cos \phi =0.8 \; \text{lag} \end{aligned}

Explanation Figure 1

Q19GATE 2017MCQ

If a synchronous motor is running at a leading power factor, its excitation induced voltage ( $E_f$ ) is

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📖 Explanation

Synchonnous motor-leading p.f. over excited. $\therefore \;\;E_f \gt V_t$

Q20GATE 2017NAT

Two parallel connected, three-phase, 50Hz, 11kV, star-connected synchronous machines A and B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactances of machine A and machine B are 1 $\Omega$ and 3 $\Omega$ respectively. Assuming the magnetic circuit to be linear, the ratio of excitation current of machine A to that of machine B is ________.

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📖 Explanation

Synchronous machine A: 11 kV, 50 Hz, Y connected, $X_s=1\Omega$ Synchronous machine B: 11 kV, 50 Hz, Y connected, $X_s=3\Omega$ Given both supplying only reactive power 50MVAR tp 11 kV grid equally

\begin{aligned} \therefore \; Q_A &=\sqrt{3}VI_A \sin \phi =25mVAR \\ Q_B &= \sqrt{3}VI_B \sin \phi =25mVAR\\ \therefore \;\;I_A &=\frac{25 \times 10^6}{\sqrt{3} \times 11 \times 10^3}=1312.2A \\ \therefore \;\; I_B&= 1312.2A\\ E_A &= V-jI_AX_s\\ &=\frac{11 \times 10^3}{\sqrt{3}}-j1312.2\angle 90 \times 1 \\ E_B &=V_jI_BX_s \\ &= =\frac{11 \times 10^3}{\sqrt{3}}-j1312.2\angle 90 \times 3 \end{aligned}

Given magnetic circuit to be linear,

\begin{aligned} \therefore \;\; E &\propto I_f \\ \therefore \;\; \frac{I_{f_A}}{I_{f_B}}&=\frac{E_A}{E_B} \\ E_A&=7663.05V \\ E_B &= 10287.45V \end{aligned}

Therefore, ratio of excitation current of machine A to machine B $\frac{E_A}{E_B}=0.745$

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