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 fullload torque is
A singlephase 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 P_{c} and P_{sc }represent core and fullload 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 singlephase 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 ‘synchronousimpedance 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 noload losses of an electrical machine is represented in its equivalent circuit by a
Noload 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 Vcurves related to Vcurves.
A 6pole, 3phase alternator running at 1000 rpm supplies to a 8pole, 3phase 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 3phase 50 MVA 10kV generator has a reactance of 0.2 ohm per phase. Hence the perunit value of the reactance on a base of 100 MVA 25 kV will be
The results of a ‘SlipTest’ for determining directaxis (X_{d}) and quadratureaxis (X_{q}) reactances of a starconnected, salientpole alternator are given below:
Phase values: V_{max} = 108V; V_{min} = 96V, I_{max} = 12A, I_{min} = 10A. Hence the two reactances will be
A twowinding transformer is used as an autotransformer. The kVA rating of the autotransformer compared to the twowinding transformer will be
Answer would depend upon the voltage ratio
and type of autotransformer
i.e.., setup or stepdown.
For stepup transformer
Where a < 1
Let, N_{1} = N_{2} (Ideal two winding transformer) For stepup transformer
A 20 kVA, 2000/200V, 1phase transformer has nameplate leakage impedance of 8%. Voltage required to be applied on the highvoltage side to circulate fullload current with the lowvoltage winding shortcircuited will be
V = 0.08 × 1 pu
∴ V actual = 0.08 × 2000 = 160V
The fullload copperloss and iron loss of a transformer are 6400W and 5000W respectively. The copperloss 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 singlephase transformer with 2000 turns on the highvoltage side, the opencircuit test result gives 200V, 91 A, 5kW on lowvoltage side. The coreloss component of current is approximately
For a given torque, reducing the diverterresistance of a d.c. series motor
reducing the diverterresistance reduces the ϕ, so for a given load torque, Ia would increase.
would also increase.
Possible threetothree phase transformer connection for parallel operation is
To ensure cophasor primary and secondary windings.
A 4 kVA, 400/200V singlephase 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, 1phase 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
E_{b} = V_{t} – I_{a} R_{a}
∴ 110 = 120 – I_{a} × 0.4 ⇒ I_{a} = 25A
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