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An LR circuit having a time constant of ...

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.

Text Solution

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`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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