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In a decaying L-R circuit, the time afte...

In a decaying `L-R` circuit, the time after which energy stores in the inductor reduces to one fourth of its initial values is

A

`(ln 2)L/R`

B

`0.5L/R`

C

`sqrt2L/R`

D

`(sqrt2/(sqrt2-1))L/R`

Text Solution

Verified by Experts

The correct Answer is:
A
`1/2Li^2==1/4[1/4Li_0^2]`
So, `i=i_0/2,` half value
`:. t=t_(1//2)=(ln 2)tau_L=(ln 2) (L/R)`
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