A coil of self-inductance 10,mH and resistanc | vedclass

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(A) Given: Self-inductance $L = 10\,mH = 10 	imes 10^{-3}\,H$,coil resistance $r_1 = 0.1\,\Omega$,internal resistance $r_2 = 0.9\,\Omega$.The total r

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$0.016$

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(A) Given: Self-inductance $L = 10\,mH = 10 	imes 10^{-3}\,H$,coil resistance $r_1 = 0.1\,\Omega$,internal resistance $r_2 = 0.9\,\Omega$.The total resistance of the circuit is $R = r_1 + r_2 = 0.1 + 0.9 = 1.0\,\Omega$.The time constant of the $LR$ circuit is $	au = \frac{L}{R} = \frac{10 	imes 10^{-3}}{1.0} = 10^{-2}\,s$.The current at any time $t$ is given by $i = i_0(1 - e^{-t/	au})$,where $i_0$ is the saturation current.We want to find $t$ when $i = 0.8 i_0$.$0.8 i_0 = i_0(1 - e^{-t/	au})$$0.8 = 1 - e^{-t/	au}$$e^{-t/	au} = 0.2 = \frac{1}{5}$$e^{t/	au} = 5$Taking natural logarithm on both sides: $\frac{t}{	au} = \ln 5$.Given $\ln 5 = 1.6$,so $t = 1.6 	imes 	au = 1.6 	imes 10^{-2}\,s = 0.016\,s$.

$A$ coil of self-inductance $10\,mH$ and resistance $0.1\,\Omega$ is connected through a switch to a battery of internal resistance $0.9\,\Omega$. After the switch is closed,the time taken for the current to attain $80\%$ of its saturation value is: [take $\ln\,5 = 1.6$] (in $,s$)

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