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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jun 2008 23:03:54 IST
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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
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5 Jun 2008 23:05:17 IST
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i think it's (c)......R/L seconds
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5 Jun 2008 23:07:48 IST
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but how can i solve it
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5 Jun 2008 23:10:58 IST
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had u solve it
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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.
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5 Jun 2008 23:23:00 IST
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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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