A coil of inductance 300, mH and resistance 2 | vedclass

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(D) The current in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady state current.We are given that the current re

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$0.1$

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(D) The current in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady state current.We are given that the current reaches half of its steady state value,so $i = i_0/2$.Substituting this into the equation: $i_0/2 = i_0(1 - e^{-Rt/L})$.This simplifies to $1/2 = 1 - e^{-Rt/L}$,which means $e^{-Rt/L} = 1/2$.Taking the natural logarithm on both sides: $-Rt/L = \ln(1/2) = -\ln(2)$.Thus,$t = (L/R) \ln(2)$.Given $L = 300\, mH = 0.3\, H$ and $R = 2\, \Omega$,we have $t = (0.3 / 2) 	imes 0.693$.$t = 0.15 	imes 0.693 = 0.10395\, s \approx 0.1\, s$.

$A$ coil of inductance $300\, mH$ and resistance $2\, \Omega$ is connected to a source of voltage $2\, V$. The current reaches half of its steady state value in $t$ seconds. Find the value of $t$. (in $, s$)

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