Electrical Machines The full load copper loss and iron loss of transformer are 6400W and 5000W respectively. The copper loss and iron loss at half load will be respectively? 1600 W and 1250 W 3200 W and 2500 W 3200 W and 5200 W 1600 W and 5000 W 1600 W and 1250 W 3200 W and 2500 W 3200 W and 5200 W 1600 W and 5000 W ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines If iron loss and full load copper loss of any transformer are denoted by Pi and Pc, then the load corresponding to maximum efficiency is given by (Pi / Pc) × full Load. (Pi / Pc)0.5 × Full Load. (Pi / Pc)1.6 × full Load. (Pi / Pc)2 × full Load. (Pi / Pc) × full Load. (Pi / Pc)0.5 × Full Load. (Pi / Pc)1.6 × full Load. (Pi / Pc)2 × full Load. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines The torque angle, in a synchronous motor, is the angle between magnetizing electric current and back emf. None of these the supply voltage and the back emf. the rotating stator flux and rotor flux. magnetizing electric current and back emf. None of these the supply voltage and the back emf. the rotating stator flux and rotor flux. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines The armature reaction mmf in a DC machines has a form of sinusoidal. trapezoidal. triangular. rectangular. sinusoidal. trapezoidal. triangular. rectangular. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines The armature mmf of a DC machine has trapezoidal space distribution and is stationary in space. stepped distribution and rotates at the speed of armature. triangular space distribution and rotates at the speed of armature. triangular space distribution and is stationary in space. trapezoidal space distribution and is stationary in space. stepped distribution and rotates at the speed of armature. triangular space distribution and rotates at the speed of armature. triangular space distribution and is stationary in space. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines In a 3 - Φ IM with line frequency f1 and winding factor kw1 and kw2 for stator and rotor respectively the ratio of per phase stator winding emf E1 to per phase rotor winding emf E2 at standstill is given by E1 / E2 = N1kw1 / N2kw2. E1 / E2 = 1. E1 / E2 = N1 / N2. E1 / E2 = f1kw1 / f2kw2. E1 / E2 = N1kw1 / N2kw2. E1 / E2 = 1. E1 / E2 = N1 / N2. E1 / E2 = f1kw1 / f2kw2. ANSWER DOWNLOAD EXAMIANS APP
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