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.
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
A coil having large number of turns of fine wire is wound on the middle limb of the shunt magnet. This coil is known as pressure or voltage coil and is connected across the supply mains. The voltage coil produces a high ratio of inductance to resistance.
EXAMIANS