SSC JE Electrical 2019 with solution SET-1
What would the total loss of the 2kVA transformer corresponding to maximum efficiency be, provided the transformer has the iron of 150 W and full-load copper loss of 250W?
When variable loss becomes equal to the constant loss, efficiency is maximum. Losses = Pi + Pc Since copper loss is a variable loss therefore Losses = Pi + Pi = 2pi Thus at a maximum efficiency of this transformer total loss = 150 x 2 = 300 W
The shaded-pole motor is the original type of AC single-phase induction motor. Shaded pole-type single-phase induction motors are provided with shading rings on their poles which are the projected type of poles. The stator of such motors has projected poles like DC machines as shown in Fig. 5.7. The rotor is a squirrel cage type similar to that of split-phase-type motors. The poles are excited by giving a single-phase AC supply. A single-turn thick coil in the form of a ring called the shading ring is fitted on each side of every pole as shown. The portion of the poles where the shading ring is fitted is called the shaded portion, while the other portion is called the unshaded portion. The shaded pole motor has the lowest starting torque as compared to the all single-phase induction motors.
During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
Galvanized steel conductors do not corrode, and possess high resistance. Hence such Wires are used in telecommunications circuits, earth wires, guard wire, stray wire, etc.
In the above option, the wood has the lowest calorific value. Its value is around 17000 – 20000 kilojoule/kg. The calorific value of LPG is 55000 kilojoule/kg. The calorific value of Kerosene oil is 45000 kilojoule/kg. The calorific value of hydrogen oil is 150000 kilojoule/kg.
EXAMIANS
