MGVCL Exam Paper (30-07-2021 Shift 3)
The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil.
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (1.1/0.6)*cosδ = (1.1/0.6)*0.5 = 0.91 M = (H*s)/(πf) = 4/(50π)
Value of power factor is varies from 0 to 1. R→PF = 1 L→PF = 0 lag C→PF = 0 lead R-L→0 lag < PF < 1 R-C→0 lead < PF < 1 R-C-L→depends on value of R, L and C. PF = R/Z
EXAMIANS