MGVCL Exam Paper (30-07-2021 Shift 2)
A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series opposing.
Where, d is distance in m Q is value of variable resistor in ohm P is variable resistance in ohm Here Q/P = 1/3 is given, d = Q/(P + Q)*Loop length = 1/4*(2*600) = 300 m
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (11/05)*cosδ = (11/0.5)*0.5 M = (H*s)/(πf) = 4/(50π)
In single phase energy meter, Current coil is made up from thick wire having few no. of turns. Pressure coil is made up from thin wire having more no. of turns.
EXAMIANS