The starting current of a 3ϕ induction motor is 5 times the rated current, while the rated slip is 4%. The ratio of starting torque to full-load torque is
A single-phase transformer when supplied from 220V, 50 Hz has eddy current loss of 50W. If the transformer is connected to a voltage of 330V, 50 Hz, the eddy current loss will be
If Pc and Psc represent core and full-load ohmic losses respectively, the maximum kVA delivered to load corresponding to maximum efficiency is equal to rated kVA multiplied by
A d.c. series motor is accidently connected to single-phase a.c. supply. The torque produced will be
In dc series motor, the ac currents through the field & armature windings will always be in the same direction. So torque will be unidirectional but pulsating due to ac.
The ‘synchronous-impedance method’ of finding the voltage regulation by a cylindrical rotor alternator is generally considered as
As unsaturated value of Zs is more than the saturated value, voltage regulation computed by emf method is much higher than the actual value. It is because of this reason that the emf method is called a pessimistic method.
Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a
No-load losses (essentially core losses not F & W losses) are low and depend upon the applied voltage in an induction machine.
The power factor of a synchronous motor
Refer, inverted V-curves related to V-curves.
A 6-pole, 3-phase alternator running at 1000 rpm supplies to a 8-pole, 3-phase induction motor which has a rotor current of frequency 2 Hz. The speed at which the motor operates is
For maximum current during ‘Slip Test’ on a synchronous machine, the armature mmf aligns along
A 3-phase 50 MVA 10kV generator has a reactance of 0.2 ohm per phase. Hence the per-unit value of the reactance on a base of 100 MVA 25 kV will be
The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:
Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will be
A two-winding transformer is used as an auto-transformer. The kVA rating of the auto-transformer compared to the twowinding transformer will be
Answer would depend upon the voltage ratio
and type of auto-transformer
i.e.., set-up or step-down.
For step-up transformer
Where a < 1
Let, N1 = N2 (Ideal two winding transformer) For step-up transformer
A 20 kVA, 2000/200V, 1-phase transformer has name-plate leakage impedance of 8%. Voltage required to be applied on the high-voltage side to circulate full-load current with the low-voltage winding short-circuited will be
V = 0.08 × 1 pu
∴ V actual = 0.08 × 2000 = 160V
The full-load copper-loss and iron loss of a transformer are 6400W and 5000W respectively. The copper-loss and ironloss at half load will be, respectively
Copper loss = (1/2)2 × 6400 = 1600W Iron loss does not depend upon the load but upon the applied voltage.
In a 100 kVA, 1100/220V, 50 Hz single-phase transformer with 2000 turns on the high-voltage side, the open-circuit test result gives 200V, 91 A, 5kW on low-voltage side. The coreloss component of current is approximately
For a given torque, reducing the diverter-resistance of a d.c. series motor
reducing the diverter-resistance reduces the ϕ, so for a given load torque, Ia would increase.
would also increase.
Possible three-to-three phase transformer connection for parallel operation is
To ensure co-phasor primary and secondary windings.
A 4 kVA, 400/200V single-phase transformer has resistance of 0.02 p.u. and reactance of 0.06 p.u. Its actual resistance and reactance referred to high voltage side are respectively
Two 10 kV/440V, 1-phase transformers of ratings 600 kVA and 350 kVA are connected in parallel to share a load of 800 kVA. The reactances of the transformers, referred to the secondary side are 0.0198Ω and 0.0304Ω respectively (resistances negligible). The load shared by the two transformers will be respectively
The current drawn by a 120V d.c. motor with back e.m.f. of 110V and armature resistance of 0.4 ohm is
Eb = Vt – Ia Ra
∴ 110 = 120 – Ia × 0.4 ⇒ Ia = 25A