SSC JE Electrical 2019 with solution SET-2
While estimating the overhead lines, if the number of poles required is 14, then what is the estimated number of earth sets required?
While estimating the overhead lines, the first and last pole is always earth connected and after every 3rd electrical pole, the fourth pole is earthed. Therefore, the approximate number of required ground set = 4
The size of the conduit is 62 meters. In conduit, there is a 2-phase wire and 1 neutral wire. Therefore the total number of conducting wire is 3 in a conduit Total length of cable = 62 × 3 = 186 meters.
⇒ The electric potential at a point is defined as the work done in bringing the unit positive charge ( +1C) from infinity to that point. The work done is independent of the path taken. Its unit is Volt (V). ⇒ Potential Difference (pd or V) is a measure of the difference in charge between two points in a conductor. Its unit is Volt (V). ⇒ The difference in charge produced by the battery is stored in the battery as electrical potential energy and is called electromotive force (shortened to emf). Electromotive force is also measured in volts. Its unit is Volt (V). ⇒ Electric flux is a measure of how much the electric field vectors penetrate through a given surface. The SI unit of electric flux is N.m2/C. Electrical potential, Potential difference, Electromotive force has the same SI unit i.e Volt (V). Hence Electric flux is alike from others.
A single-phase induction motor with only one winding on the stator cannot produce any starting torque. Hence, some extra arrangement is required to start the motor. In the running condition, the motor is capable of developing the torque with only one winding on the stator. The simplest method of starting is to provide an auxiliary winding on the stator in addition to the main winding. The two windings are placed in the stator with their axes displaced by 90 electrical degrees in space.
Io = V/R = 20/100 Io = 0.2 A Since the given diode is an ideal diode therefore there is no voltage drop across it. Vo = Io × RD Vo = 0.2 × 0 Vo = 0 Hence Io = 0.2 A & Vo = 0
EXAMIANS
