Rate of increment of energy in an inductor with time in series RL circuit getting charged with battery of EMF Eis best represented by:

Options

A.
B.
C.
D.

Answer

A

Explanation

Rate of increment of energy in inductor
= `(du)/(dt) = (d)/(dt) ((1)/(2) Li^(2)) = Li(di)/(dt)`
Current in the inductor at time t is :
`i = i_(0)(1 - e^((-t)/(tau))) "and" (di)/(dt) = (i_(0))/(tau) e^((-t)/(tau))`
`(du)/(dt) = (Li_(0)^(2))/(tau)e^((-t)/(tau)) (1-e^((-t)/(tau)))`
`(du)/(dt) = 0 ` at t = 0 and `t = oo`
Hence E is best represent by :