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A 20 mH coil is connected in series with...

A 20 mH coil is connected in series with a `2 k Omega` resistor and a 12 V battery.
Calcualte
(i) time constant of the circuit,
(ii) time during which current decrys to 10% of its maximum value.

Text Solution

Verified by Experts

Here, `L = 20 mH = 20 xx 10^(-3) H`,
`R = 2 k Omega = 2 xx 10^(3) ohm, V = 12 V`
(i) Time constant `= (L)/(R ) = (20 xx 10^(-3))/(2 xx 10^(3)) = 10^(-5) s`
(ii) For decay of current, `I = I_(0) e^(-(R )/(L) t)`
`(I_(0))/(10) = I_(0) e^(-10^(5) t)`
`:. e^(10^(5) t) = 10`
`10^(5) t = 2.303 log_(10) 10`
`t = (2.303)/(10^(5)) = 2.303 xx 10^(-5) s`
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