- The voltage between the incoming alternator and the bus-bars can easily be checked by a voltmeter and the phase sequence of the incoming alternator and the bus-bars can also easily be checked by phase sequence indicator.
- The difference between frequency and phase voltage of the incoming alternator and bus-bars can be checked by following two methods-
- By Three Lamp (one dark, two bright) method
- By synchroscope
(i) Three Lamp Method
In this method of synchronizing of the alternator, three lamps L1, L2, and L3 are connected as shown in Fig.
- The lamp L1 is connected between phase R1 and R2 and the other two lamps L2 & L3 are cross-connected to the phase (Y1, B2) & (B1, Y2) respectively.
- When the difference between frequency and phase voltage of incoming alternator and bus-bars is same then the straight connected lamp L1 will be dark and other two lamps (L2 & L3) which is connected cross with the phase will be bright equally.
- At this instant, the synchronization will be perfect and the switch of the incoming alternator can connect it to the bus-bars.
(ii) Synchroscope
- A synchroscope is a measuring/ indicating instrument that is indicated by a revolving pointer.
- The difference between frequency and phase voltage of the incoming alternator and bus-bars can be checked by synchroscope.
- A synchroscope is a small motor, in this instrument, the supply to the field is fed by the bus-bars through a potential transformer and rotor from the incoming alternator.
- When the alternator is running fast then the frequency of alternator will be more than the frequency of bus-bars then that condition the pointer will move on the clockwise direction.
- When the alternator is running slow then the frequency of alternator will be less than the frequency of bus-bars then that condition the pointer will move on anti-clockwise direction.
- When the frequency of alternator is same as the supply frequency of bus-bars then that condition no torque act on the rotor(pointer) of synchroscope and the instrument pointer points vertically upwards (“12 O’ clock”).
MCQs on Parallel Operation and Synchronization of Alternators:
Q1. Two alternators operating in parallel must have identical:
A. Voltage and frequency only
B. Frequency and phase sequence only
C. Voltage, frequency, and phase sequence
D. Voltage and load angle
Correct Option: C
Explanation:
For safe parallel operation, alternators must have same RMS voltage, same frequency, and same phase sequence.
Additionally, phase angle difference should be zero at the time of synchronization to avoid heavy circulating
currents and mechanical shocks.
Q2. If two alternators connected in parallel have unequal induced EMFs but same synchronous reactance, the circulating current will depend mainly on:
A. Load current
B. Power factor
C. Difference in induced EMFs
D. Mechanical input
Correct Option: C
Explanation:
Circulating current flows due to difference in internal generated EMFs and is given by
Ic = (E₁ − E₂) / j(Xs₁ + Xs₂). Hence, it depends mainly on the difference in induced EMFs.
Q3. When two alternators operate in parallel, the real power shared by each alternator depends mainly on:
A. Excitation
B. Prime mover input
C. Synchronous reactance
D. Terminal voltage
Correct Option: B
Explanation:
In parallel operation, real power (kW) sharing is controlled by prime mover input.
Increasing mechanical torque increases load angle δ and hence real power output.
Q4. When excitation of one alternator in parallel operation is increased:
A. It supplies more real power
B. It supplies more reactive power
C. Terminal voltage decreases
D. Frequency increases
Correct Option: B
Explanation:
Increasing excitation increases internal EMF, causing the alternator to supply more reactive power (kVAR).
Real power remains almost unchanged as it depends on mechanical input.
Q5. In parallel operation, reactive power sharing between alternators depends on:
A. Speed
B. Load angle
C. Field excitation
D. Mechanical torque
Correct Option: C
Explanation:
Reactive power sharing is controlled by field excitation. Increasing excitation causes the alternator
to supply more lagging VARs, while reducing excitation makes it absorb VARs.
Q6. If the prime mover input of an alternator running in parallel is increased, its:
A. Voltage increases
B. Reactive power increases
C. Real power increases
D. Frequency increases
Correct Option: C
Explanation:
Increasing prime mover input increases rotor torque and load angle δ.
According to P = (EV/Xs) sinδ, real power output increases while frequency remains constant.
Q7. Two alternators of equal rating operate in parallel. If one alternator has higher excitation, it will:
A. Take more kW load
B. Take more kVAR load
C. Run at higher speed
D. Supply less current
Correct Option: B
Explanation:
Higher excitation means higher internal EMF, so that alternator supplies more reactive power.
Real power sharing remains unchanged unless mechanical input is varied.
Q8. Circulating current between two alternators operating in parallel under no-load condition is caused due to:
A. Unequal frequencies
B. Unequal voltages
C. Unequal speeds
D. Unequal loads
Correct Option: B
Explanation:
Under no-load condition, circulating current exists only due to unequal induced voltages.
Unequal frequency or speed would prevent synchronization itself.
Q9. For two alternators operating in parallel, the circulating current is:
A. In phase with terminal voltage
B. 90° lagging with terminal voltage
C. 90° leading with terminal voltage
D. In phase opposition to load current
Correct Option: B
Explanation:
Circulating current flows through synchronous reactance, which is inductive.
Hence, it lags the terminal voltage by 90° and is purely reactive.
Q10. If two identical alternators operate in parallel and excitation of both is same, then:
A. Load shared unequally
B. Load shared equally
C. One alternator trips
D. Circulating current flows
Correct Option: B
Explanation:
If alternators are identical with same excitation and same mechanical input,
no circulating current flows and the total load is shared equally.
Q11. Two alternators are operating in parallel and supplying a total load of 1000 kW. If alternator-1 supplies 620 kW, the load supplied by alternator-2 is:
A. 280 kW
B. 350 kW
C. 380 kW
D. 420 kW
Correct Option: C
Explanation:
Total load = 1000 kW. Load of alternator-2 = 1000 − 620 = 380 kW.
Q12. An alternator connected to infinite bus-bars has J = 400 kg·m² and synchronizing power coefficient Ps = 1600 W/rad. Time period of oscillation is approximately:
A. 1.0 s
B. 1.57 s
C. 3.14 s
D. 6.28 s
Correct Option: C
Explanation:
Time period T = 2Ï€√(J/Ps) = 2Ï€√(400/1600) = 2Ï€×0.5 ≈ 3.14 s.
Q13. Time period of oscillation of an alternator connected to infinite bus-bars depends on:
A. Load current
B. Rotor inertia
C. Armature resistance
D. Terminal voltage only
Correct Option: B
Explanation:
From T = 2Ï€√(J/Ps), the time period depends on rotor inertia and synchronizing power.
Q14. If inertia constant increases, the time period of oscillation will:
A. Decrease
B. Remain same
C. Increase
D. Become zero
Correct Option: C
Explanation:
Time period is proportional to √J. Hence, increasing inertia increases oscillation period.
Q15. The maximum power output of a cylindrical rotor alternator occurs at load angle:
A. 30°
B. 45°
C. 60°
D. 90°
Correct Option: D
Explanation:
From power equation P = (EV/Xs) sinδ, maximum power occurs when sinδ = 1, i.e., δ = 90°.
Q16. An alternator connected to infinite bus-bars has E = 1.2 pu, V = 1 pu, Xs = 0.6 pu. Maximum power output is:
A. 1.2 pu
B. 1.8 pu
C. 2.0 pu
D. 2.4 pu
Correct Option: C
Explanation:
Maximum power Pmax = EV/Xs = (1.2 × 1)/0.6 = 2.0 pu.
Q17. For stable operation of alternator connected to infinite bus-bars, load angle δ must be:
A. Greater than 90°
B. Equal to 90°
C. Less than 90°
D. Equal to 180°
Correct Option: C
Explanation:
For stable equilibrium, load angle must be less than 90°.
Q18. If load angle exceeds 90°, the alternator will:
A. Deliver more power
B. Operate stably
C. Lose synchronism
D. Improve power factor
Correct Option: C
Explanation:
Beyond 90°, restoring torque becomes negative and synchronism is lost.
Q19. Two alternators are in parallel supplying a total load of 1000 kW. If alternator-1 supplies 600 kW, alternator-2 supplies:
A. 200 kW
B. 300 kW
C. 400 kW
D. 500 kW
Correct Option: C
Explanation:
Power supplied by alternator-2 = 1000 − 600 = 400 kW.
Q20. An alternator connected to infinite bus-bars has E = 1.2 pu, V = 1 pu, Xs = 1 pu. Maximum power output is:
A. 0.6 pu
B. 1.0 pu
C. 1.2 pu
D. 2.4 pu
Correct Option: C
Explanation:
Maximum power Pmax = EV/Xs = (1.2 × 1)/1 = 1.2 pu.