Given Power (P) = 20 kW = 20 × 103 W Voltage (V) = 200 V P = VI = I = P/V I = (20 × 103)/200 I = 100A For wave wound No. of Parallel Path = 2 Current in Each parallel Path for wave wound I = 100/2 = 50 A
No of disc required for 11 kV transmission line = 1 disc Since the transmission line is 3-phase, therefore, the number of discs required for a single-pole will be 3. Total no. of poles = 80 No of disc = 3 × 80 = 240
In nuclear power plant the graphite is used as a moderator. The job of the moderator is to absorb some of the kinetic energy of the neutrons to slow them down. This is because slow neutrons are more easily absorbed by uranium-235. A neutron slowed In this way can start the fission process.
EXAMIANS
