A coil of inductance 1,H and resistance 100,O | vedclass

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(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/	au})$,where $	au = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.$(a)$

Vedclass · Accepted Answer

$t = 7\,ms; U = 1\,mJ$

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(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/	au})$,where $	au = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.$(a)$ For the current to reach half its steady-state value $(i = \frac{E}{2R})$:$\frac{E}{2R} = \frac{E}{R}(1 - e^{-t/	au})$$0.5 = 1 - e^{-t/	au} \implies e^{-t/	au} = 0.5$$t = 	au \ln 2 = 10\,ms 	imes 0.693 = 6.93\,ms \approx 7\,ms$.$(b)$ At $t = 15\,ms$,$t/	au = 15/10 = 1.5$:$i = \frac{6}{100}(1 - e^{-1.5}) = 0.06(1 - 0.25) = 0.06 	imes 0.75 = 0.045\,A$.The energy stored is $U = \frac{1}{2}Li^2 = \frac{1}{2} 	imes 1 	imes (0.045)^2 = 0.5 	imes 0.002025 \approx 0.001\,J = 1\,mJ$.

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