IIT JEE , AIEEE Forum - Physics , Magnetism

nitinsharma24

An inductance L and a resistance R are first connected to a battery. After somtime the battery disconnected but L and R remain connected in a closed circuit. Then the current reduces to 37% of its initial value in:

A: RL sec B: R/L sec C: L/R sec D: 1/LRsec

netkid07

i think it's (c)......R/L seconds

nitinsharma24

but how can i solve it

nitinsharma24

had u solve it

ultimator

I think its c. L/R ....... Its called as time constant. It is the time taken for current to reduce by 63% to 37% of its initial value. This is a definition.

rhd92781

yes current reduces to 37% of its initial value in one time constant in an L-R circuit = L/R.

I = Io e^ (-t/(L/R))

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