Given L_1 = 1, mH, R_1 = 1,Omega, L_2 = 2,mH, | vedclass

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(A) The circuit consists of two inductors $L_1$ and $L_2$ in series and two resistors $R_1$ and $R_2$ in series.For inductors in series,the equivalent

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(A) The circuit consists of two inductors $L_1$ and $L_2$ in series and two resistors $R_1$ and $R_2$ in series.For inductors in series,the equivalent inductance is $L = L_1 + L_2 = 1\, mH + 2\, mH = 3\, mH$.For resistors in series,the equivalent resistance is $R = R_1 + R_2 = 1\, \Omega + 2\, \Omega = 3\, \Omega$.The time constant $	au$ for an $RL$ circuit is given by $	au = \frac{L}{R}$.Substituting the values,we get $	au = \frac{3\, mH}{3\, \Omega} = 1\, ms$.

Given $L_1 = 1\, mH, R_1 = 1\,\Omega, L_2 = 2\,mH, R_2 = 2\,\Omega$. Neglecting mutual inductance,find the time constant (in $ms$) for the circuit shown.

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