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An LR circuit having a time constant of 50 ms is connected with an ideal battery of emf epsilon . Find the time taken for the magnetic energy stored in the circuti to change from one fourth of the steady state value to half of the steady stae value. |
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Answer» Solution :`t=(L/R)=50ms=0.05s` (a) `(i_0)/2=(i_0)(1-e^(-t//0.03))` `implies (1/2)=1-e^(-t//0.03)` `implies e^(-t//0.03)=(1/2)` `implies In e^(-t//0.03)=In(1/4)` (take In both sides) `impliest=0.05xx0.693` `=0.03465s` `=34.6ms=35ms` `(B) P=i^(2)R=(E^2)/(R)(1-e^(tR//L))^2` `Maximum POWER =(E^2)/(R)` so, (E^2)/(2R)=(E^2)/(R)(1-e^(tr/L)^2)` `implies1-e^(-R//L)=(1)/(sqrt(2))=0.707` `e^(-R//L)=0.293` `implies t=50xx1.2275ms` `=61.2ms` . |
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