01․ A 10KVA auto transformer, turn ratio is 0.4. Find the power transferred inductively?
Power transferred inductively = (1-K) × input power = 0.6×10×103 = 6 KVA
02․ A 20KVA auto transformer, turn ratio is 0.8. Find the power transferred conductively?
Power transferred conductively = K × input power = 0.8× 20 KVA = 16 KVA
03․ In an auto transformer, for what value of turn ratio, power transferred by inductively and conductively are same?
Therefore, when turn ratio is 0.5 (K=1-K), the power transferred inductively and conductively are same.04․ The total core loss of a 400 V, 50 Hz transformer is 2400W, when it is supplied by 200 V, 25 Hz loss is 800W. Calculate eddy current loss and hysteresis loss?
Eddy current loss Pe ∝ KeBm²f²
Hysteresis loss Ph ∝ KhBm1.6f
For given problem Bm is constant.
When Bm is constant, Hysteresis loss is proportional to frequency and eddy current loss is proportional to square of the frequency.
Power loss P = Pe+Ph
Bm= 400/50 = 200/25 = 8
From given data,
2400 = Ke×50²+ Kh×50
800 = Ke×25²+Kh×25
From the above equations, Ke = 0.64, Kh = 16
Eddy current loss Pe = 0.64×50² = 1600W
Hysteresis loss Ph = 16×50 = 800 W
05․ In a transformer which of the following losses are zero?
Transformer is a static device and there are no moving parts. Therefore rotational losses are zero in transformer.
06․ When maximum flux density is constant, the eddy current losses are proportional to
Eddy current loss Pe ∝ Ke Bm2f2
When maximum flux density is constant,
Eddy current loss, Pe ∝ f2
07․ When flux density is constant in a transformer, hysteresis loss varies as
08․ In a transformer maximum flux density is not constant, the eddy current loss varies as
09․ To increase starting torque of a 3-phase slip ring induction motor, preferred method is to increase
Induction motor is termed as rotating transformer with short circuited secondary. Rated voltage can not be applied across the stator because as the rotor is stationary high voltage induced in the rotor. It is similar to applying rated voltage across the transformer primary when secondary is short circuited. The motor draws 7 to 10 times the rated current during starting. Therefore, induction motor can not start with rated voltage to get high starting torque.
Starting torque is also proportional to rotor resistance (Tst∝R.), by increasing of rotor resistance starting torque will improve. This technique is used in slip ring induction motors to get high starting torque.10․ In single phase rotary converter the number of slip rings will be
The single phase power supply has two terminals that are phase and neutral. So, two slip rings are enough to provide the power supply to the single phase rotary converter.
<<<4243444546>>>
