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GATE EE · Electrical Machines

Synchronous Machines Practice Questions

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

Q21GATE 2016MCQ

Consider a system consisting of a synchronous generator working at a lagging power factor, a synchronous motor working at an overexcited condition and a directly grid-connected induction generator. Consider capacitive VAr to be a source and inductive VAr to be a sink of reactive power. Which one of the following statements is TRUE?

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📖 Explanation

Synchronous generator working at a lagging power factor, will supply active power and lagging reactive power, so this is source. Synchronous motor working at an overexcited condition i.e. leading p.f. operation so it will take active power and leading reactve power in other words taking active power and supplying lagging reactive power. Induction generates active power but as there is no dc excitation in rotor so it takes lagging reactive power.

Q22GATE 2016NAT

A three-phase, 50 Hz salient-pole synchronous motor has a per-phase direct-axis reactance ( $X_{d}$ ) of 0.8 pu and a per-phase quadrature-axis reactance ( $X_{q}$ ) of 0.6 pu. Resistance of the machine is negligible. It is drawing full-load current at 0.8 pf (leading). When the terminal voltage is 1 pu, per-phase induced voltage, in pu, is _________.

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📖 Explanation

Synchronous motor at leading p.f.

\begin{aligned} X_d &=0.8 \\ \phi &= 36.86\\ X_q&= 0.6\\ R_q&= 0\\ \tan \psi &=\frac{V \sin \phi +I_aX_a}{V \cos \phi }\\ \psi &=56.30 \end{aligned}

For synchronous motor at leading p.f.

\begin{aligned} \psi &=\phi +\delta \\ \delta &=19.70^{\circ} \\ E&=V \cos \delta +I_dX_d \\ I_d&=I_a \sin f=0.831 \\ E&=(1)\cos (19.7)+(0.831)(0.8) \\ &=1.606 \end{aligned}

Q23GATE 2016NAT

The power consumption of an industry is 500 kVA, at 0.8 p.f. lagging. A synchronous motor is added to raise the power factor of the industry to unity. If the power intake of the motor is 100 kW, the p.f. of the motor is _____________

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📖 Explanation

\begin{aligned} P_1 &= 500 \times 0.8=400kW\\ Q_1&= 500 \times 0.6=300kVAR \end{aligned}

The power factor is to be raised to unity The motor has to supply 300kVAR The motor rating is 100 kW, 300kVAR

\begin{aligned} \phi _m &=\tan ^{-1}\left ( \frac{Q}{P} \right ) \\ \phi _m&= \tan^{-1}\left ( \frac{300}{100} \right ) =71.56 \end{aligned}

Power factor of motor $= \cos \phi _m=\cos 71.56=0.316$

Q24GATE 2015NAT

A three-phase, 11 kV, 50 Hz, 2 pole, star connected, cylindrical rotor synchronous motor is connected to an 11 kV, 50 Hz source. Its synchronous reactance is 50 $\Omega$ per phase, and its stator resistance is negligible. The motor has a constant field excitation. At a particular load torque, its stator current is 100 A at unity power factor. If the load torque is increased so that the stator current is 120 A, then the load angle (in degrees) at this load is ________.

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📖 Explanation

Currently no explanation available

Q25GATE 2014NAT

A star connected 400 V, 50 Hz, 4 pole synchronous machine gave the following open circuit and short circuit test results : Open circuit test : $V_{oc}$ = 400 V (rms, line-to-line) at field current, $I_{f}$ = 2.3 A Short circuit test : $I_{sc}$ = 10 A (rms, phase) at field current, $I_{f}$ = 1.5 A The value of per phase synchronous impedance in $\Omega$ at rated voltage is_____.

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📖 Explanation

\begin{aligned} \text{Given, for}\;\;V_{OC} &= 400V(rms)\\ &= \frac{400}{\sqrt{3}}\\ &= 230.94 \;\text{(Phase value),}\\ I_f&= 2.3A\\ \text{for}\;\; I_{sc}&=10A \;\text{(rms, phase)} \\ I_f&= 1.5A\\ & \therefore \;\;\text{Base impedance}\\ Z_{base}&=\frac{V_{OC}}{I_{sc}}\\ &=\frac{230.94}{10}=23.094\Omega \end{aligned}

For finding the synchronous impedance at rated voltage, saturation effect of the machine has to be considered (i.e. the adjusted value of $Z_s$ ) short circuit ratio,

\begin{aligned} SCR &=\frac{I_f \;for \; V_{OC}}{I_f \;for \; I_{sc}} \\ &=\frac{1}{Z_{s_{(sat)pu}}} =\frac{2.3}{1.5}\\ Z_{s_{(sat)pu}} &= 0.65 pu \end{aligned}

Therefore, actual value of $Z_s$ at rated voltage

\begin{aligned} &=Z_{s_{(sat)pu}} \times Z_{base} \\ &=0.65 \times 23.094=15.06\Omega \\ \therefore \;\;Z_s&= 15.06\Omega \end{aligned}

Q26GATE 2014MCQ

A three phase synchronous generator is to be connected to the infinite bus. The lamps are connected as shown in the figure for the synchronization. The phase sequence of bus voltage is R-Y-B and that of incoming generator voltage is R' - Y' - B'. It was found that the lamps are becoming dark in the sequence $L_a-L_b-L_c$ . It means that the phase sequence of incoming generator is

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📖 Explanation

As,

\begin{aligned} L_a \; \text{is connected between R-R'} & \\ L_b \; \text{is connected between Y-Y'} & \\ L_c \; \text{is connected between B-B'} & \\ \end{aligned}

therefore, the given method of synchronization is called "synchronization by dark lmap method". In this method of synchronization, correct phase sequence of the incoming generator is obtain when all the three lamps become dark together and bright together. Since lamps are becoming dark in the sequence $L_a-L_b-L_c$ therefore, the phase sequence of incoming generator is opposite to the infinite bus.

Q27GATE 2014MCQ

A 20-pole alternators is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generated voltage is $V_1$ . If the coils are reconnected to realize three phase star connected winding, the generated phase voltage is $V_2$ . Assuming full pitch, single-layer winding, the ratio $V_1/V_2$ is

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📖 Explanation

Currently no explanation available

Q28GATE 2014MCQ

In a synchronous machine, hunting is predominantly damped by

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📖 Explanation

Hunting is a phenomenon occuring in case of synchronous generator and motor where the machine oscillates about the equilibrium point. To reduce this phenomenon, damper bars are used which are embedded in the rotor pole faces of the synchronous machines. Thus, these damper bars causes additional copper loss in the rotor for damping the undesirable phenomenon of hunting.

Q29GATE 2014NAT

A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is supplying 1 pu power to a unity power factor load at a terminal voltage of 1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the synchronous reactance with respect to the angle of the terminal voltage in degrees is _______.

🚀 Solve in Practice Mode

📖 Explanation

Given,

\begin{aligned} X_s &= 0.8 \;\text{pu} \\ \cos \phi &=1 \\ V&= 1.1\angle 0^{\circ}\; \text{pu(reference voltage)}\\ P&=1 \;\text{pu} \end{aligned}

Let, $E_f\angle \delta$ be the excitation voltage Now, power supplied =1 pu

\begin{aligned} VI_a\cos \phi &= 1\\ I_a &= \frac{1}{V \cos \phi }\\ &= \frac{1}{1.1 \times 1}=0.91 \;\text{pu}\\ \text{Now,}\;\;\vec{E_f} \angle \delta &=\vec{V}+j\vec{I_a}X_s \\ &=1.1\angle 0^{\circ}+j(0.91) \times 0.8\\ &=1.1+j0.728\\ &=1.319\angle 33.49^{\circ}\;\text{pu}\\ \therefore \;\; E_f\angle \delta &=1.319\angle 33.49^{\circ}\;\text{pu} \end{aligned}

Hence, the angle of the voltage behind the synchronous reactance =Power angle $(\delta )$ $\;\;\;\; \delta =33.49$

Explanation Figure 1

Q30GATE 2013MCQ

The angle $\delta$ in the swing equation of a synchronous generator is the

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📖 Explanation

Currently no explanation available

Q31GATE 2011MCQ

The direct axis and quadrature axis reactance's of a salient pole alternator are 1.2 p.u and 1.0 p.u respectively. The armature resistance is negligible. If this alternator is delivering rated kVA at upf and at rated voltage then its power angle is

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📖 Explanation

Power angle is the angle between $E_f$ and $V_t$ but as the $E'$ and $E_f$ are in phase, angle between $E'$ and $V_t$ =power-angle= $\delta$ Rated power is delivered at upl, so $I_a$ and $V_t$ are in phase

\begin{aligned} I_a &=1\angle 0^{\circ} \; pu\\ &\text{Quadratur-axis function,}\\ X_q &=1 \; pu \\ E' &= V_t+jI_aX_q\\ &=1+j1 \times 1 \\ &= \frac{1}{\sqrt{2}}\angle 45^{\circ}\\ \delta &=45^{\circ} \end{aligned}

Explanation Figure 1

Q32GATE 2011MCQ

A three phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current

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📖 Explanation

As field current reduces, the flux will start reducing, to keep this flux constant, the armature draws high current from bus. Ar zero field current the motor acts as synchronous reluctance motor in this case the magnetizing current fully taken from bus. Ie we increase field current in reverse direction to keep flux constant, motor draws more current. During this process the load angle increases at one point the reversed field force dominates reluctance torque and rotor slip one pole pitch anf align to opposite pole. The instant align to opposite pole the flux will be very high, to reduce this flux current drops steeply t synchronous motor value. Here torque is reluctance + synchronous motor torque.

Q33GATE 2009MCQ

A field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is

🚀 Solve in Practice Mode

📖 Explanation

If $R_a$ is neglected, $\vec{E_f}=\vec{V_t}+j\vec{I_a}X_s$ For a given field current under short-circuit condition,

\begin{aligned} I_a &=I_{sc} \\ V_t&= 0\\ X_s &=\frac{E_f}{I_{sc}} \\ \text{But,}\;\; E_f &=V_{oc} \end{aligned}

(open circuit voltage i.e. $I_a=0$ with the same field current). Then with the linearity assumption

\begin{aligned} X_s&=\left.\begin{matrix} \frac{V_{oc}}{I_{sc}} \end{matrix}\right|_{I_f=\text{constant}} \\ &= \frac{2000}{400}=5\Omega \end{aligned}

Internal voltage drop $I_aX_s=200 \times 5=1000V$

Q34GATE 2008MCQ

Distributed winding and short chording employed in AC machines will result in

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📖 Explanation

Distribution factor in distributed winding $k_d(n^{th} \; \text{harmonic})=\frac{\sin \left ( \frac{mn\gamma }{2} \right )}{m \sin \left ( \frac{n\gamma }{2} \right )}$ $k_d$ is always less than 1, so emf reduces. If, $mn\gamma /2$ is chosen in such a way that $\sin (mn\gamma /2)=0, n^{th}$ harmonic gets eliminated. Similarly, pitch factor $=k_p(n^{th} \;\text{harmonic}) = \cos \frac{n\alpha }{2}$ $k_p$ is always less than 1, so emf reduces. If $n\alpha /2$ is chosen in such a way that $\cos \frac{n\alpha }{2}=0, n^{th}$ harmonic gets eliminated.

Q35GATE 2008MCQ

A synchronous motor is connected to an infinite bus at 1.0 pu voltage and draws 0.6 pu current at unity power factor. Its synchronous reactance is 1.0 pu resistance is negligible. Keeping the excitation voltage same, the load on the motor is increased such that the motor current increases by 20%. The operating power factor will become

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📖 Explanation

\begin{aligned} \text{Let power factor} &= \cos \phi \; \text{lagging} \\ I_a &=0.6 \times 12=0.72pu \\ E\angle \delta &=V_t-jI_a\angle -\phi \times X_s \\ E(\cos \delta +j\sin \delta ) &= \\ &=1\angle 0^{\circ}-jI_aX_s(\cos \phi -j\sin\phi ) \\ &=1-j0.72 \times 1(\cos \phi -j \sin \phi ) \\ &= (1-0.72 \sin \phi )-j0.72 \cos \phi \\ E^2&=(1-0.72 \sin \phi )^2 +(0.72 \cos \phi )^2 \\ &= 1+0.72^2 \sin ^2 \phi -1.44 \sin \phi \\ &+0.72^2 \cos ^2 \phi \\ 1.17^2 &=1+0.72^{2-}1.44 \sin \phi \\ \sin \phi &= 0.1038\\ \phi &= 5.96^{\circ}\\ \cos \phi &=0.995 \; \text{lagging} \end{aligned}

Q36GATE 2007MCQ

A three-phase synchronous motor connected to ac mains is running at full load and unity power factor. If its shaft load is reduced by half, with field current held constant, its new power factor will be

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📖 Explanation

Neglecting armature resistance $R_a$ $V_t=E_f+jI_aX_s$ It can be seen from the phasor diagram that, when load is reduced at shaft keeping field current constant the new power factor becomes leading.

Explanation Figure 1

Explanation Figure 2

Q37GATE 2007MCQ

A 100 kVA, 415 V(line), star-connected synchronous machine generates rated open circuit voltage of 415 V at a field current of 15 A. The short circuit armature current at a field current of 10 A is equal to the rated armature current. The per unit saturated synchronous reactance is

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📖 Explanation

Rated open circuit voltage is obatined at $of''=15A$ Rated armature current $=\frac{100}{\sqrt{3} \times 415}=139.12A$ Short-circuit characteristics is linear. Short-circuit armature current at field current 15A $I_{a1}=I_a \times \frac{of''}{of'}$ $=139.12 \times \frac{15}{10}=208.68A$ Phase voltage, $V_p=\frac{V_{oc(line)}}{\sqrt{3}}=\frac{415}{\sqrt{3}}$ Saturated synchronous reactance

\begin{aligned} &=X_s =\frac{V_p}{I_{a1}}\\ &= \frac{415/\sqrt{3}}{208.68}=1.148\Omega \\ &\text{Base impedance,} Z_B&= \frac{415^2}{100 \times 10^3}=1.722\Omega \end{aligned}

Per unit saturated synchronous reactance $=\frac{X_s}{Z_B}=\frac{1.148}{1.722}=0.666pu$

Explanation Figure 1

Q38GATE 2006MCQ

A 3-phase, 400 V, 5 kW, star connected synchronous motor having an internal reactance of 10 $\Omega$ is operating at 50% load, unity p.f. Now, the excitation is increased by 1%. What will be the new load in percent, if the power factor is to be kept same? Neglect all losses and consider linear magnetic circuit.

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📖 Explanation

$I_{fl}=\frac{5 \times 10^{3}}{\sqrt{3} \times 400 \times 1}=7.21A$ For synchronous motor,

\begin{aligned} E^2 &=(V \cos \theta -I_aR_a)^2 +(V\sin \theta -I_aX_s)^2 \\ E&= \sqrt{(400/\sqrt{3})^{2+}(10 \times 3.6)^2}\\ &= 233.729V \end{aligned}

If there is 1% increase in $E_2$ , then $E_2$ becomes,

\begin{aligned} E_2 &=233.729 \times 1.01=236V \\ \therefore \;\; I_aX &=\sqrt{E_2^{2-}V^2} \\ &= \sqrt{(236)^{2-}(400/\sqrt{3})^2}\\ &= 48.93\\ \therefore \;\; I_a&=4.893A \\ \therefore \;\;\% \;\text{load} &= \frac{4.893}{7.21}=67.86\% \end{aligned}

Q39GATE 2006MCQ

A 4 pole, 50 Hz, synchronous generator has 48 slots in which a double layer winding is housed. Each coil has 10 turns and is short pitched by an angle to 36 $^\circ$ electrical. The fundamental flux per pole is 0.025 Wb. The fifth harmonic component of phase emf(in volts), for a three phase star connection is

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📖 Explanation

Pitch factor due to $5^{th}$ harmonic component, $(k_p)_5=\cos \left ( \frac{5 \times 36}{2} \right )=\cos 90=0$ $[as pitch factor is zero, induced emf due to $5^{th}$ harmonic component is zero]$

Q40GATE 2006MCQ

A synchronous generator is feeding a zero power factor (lagging) load at rated current. The armature reaction is

🚀 Solve in Practice Mode

📖 Explanation

$E_f \gt V_t$ , the generator is overexcited. Therefore, armature-reaction is demagnetizing in nature i.e. $\phi _{ar} \; \text{opposes} \; \phi _f$ .

Explanation Figure 1

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