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(A) The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$	au_1 = \frac{L}{R} = 3 	imes 10^{-3} 	ext{ s}$ ...$(i)

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$54 	ext{ mH}, 18 \Omega$

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(A) The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$	au_1 = \frac{L}{R} = 3 	imes 10^{-3} 	ext{ s}$ ...$(i)$When a $90 \Omega$ resistor is added in series,the total resistance becomes $(R + 90) \Omega$. The new time constant is $	au_2 = \frac{L}{R + 90} = 0.5 	imes 10^{-3} 	ext{ s}$ ...(ii)Dividing equation $(i)$ by equation (ii):$\frac{3 	imes 10^{-3}}{0.5 	imes 10^{-3}} = \frac{L/R}{L/(R + 90)}$$6 = \frac{R + 90}{R}$$6R = R + 90$$5R = 90 \implies R = 18 \Omega$Substituting $R = 18 \Omega$ into equation $(i)$:$L = 3 	imes 10^{-3} 	imes 18 = 54 	imes 10^{-3} 	ext{ H} = 54 	ext{ mH}$.Thus,the inductance is $54 	ext{ mH}$ and the resistance is $18 \Omega$.

The time constant of an inductance coil is $3 \text{ ms}$. When a $90 \Omega$ resistance is joined in series,the time constant becomes $0.5 \text{ ms}$. The inductance and the resistance of the coil are:

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