A coil of inductance 5,H is joined to a cell | vedclass

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(A) Given: Inductance $L = 5\,H$,Resistance $R = 10\,\Omega$,$emf$ $V = 6\,V$.The current in an $RL$ circuit at time $t$ is given by $i(t) = i_0(1 - e

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$3$

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(A) Given: Inductance $L = 5\,H$,Resistance $R = 10\,\Omega$,$emf$ $V = 6\,V$.The current in an $RL$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R = 6/10 = 0.6\,A$.The time constant $	au = L/R = 5/10 = 0.5\,s$. Thus,$i(t) = 0.6(1 - e^{-t/0.5}) = 0.6(1 - e^{-2t})$.The $emf$ across the coil is given by $\varepsilon_L = L \frac{di}{dt}$.Calculating $\frac{di}{dt}$: $\frac{di}{dt} = 0.6 	imes (-2) 	imes (-e^{-2t}) = 1.2 e^{-2t}$.Therefore,$\varepsilon_L = 5 	imes 1.2 e^{-2t} = 6 e^{-2t}$.At $t = \ln \sqrt{2} = \frac{1}{2} \ln 2$,we have $e^{-2t} = e^{-2(\frac{1}{2} \ln 2)} = e^{-\ln 2} = \frac{1}{2}$.Substituting this value: $\varepsilon_L = 6 	imes \frac{1}{2} = 3\,V$.

$A$ coil of inductance $5\,H$ is joined to a cell of $emf$ $6\,V$ through a resistance $10\,\Omega$ at time $t = 0$. The $emf$ across the coil at time $t = \ln \sqrt{2}\,s$ is:....$V$

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