An inductance coil has a time constant of 4 , | vedclass

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(A) The time constant of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$	au_1 = \frac{L}{R} = 4 \, sec$.When the coil is cut into two eq

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$4$

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(A) The time constant of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$	au_1 = \frac{L}{R} = 4 \, sec$.When the coil is cut into two equal parts,the resistance and inductance of each part become half of the original values,i.e.,$R' = \frac{R}{2}$ and $L' = \frac{L}{2}$.When these two parts are connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'} = \frac{2}{R/2} = \frac{4}{R}$,so $R_{eq} = \frac{R}{4}$.The equivalent inductance $L_{eq}$ for two inductors in parallel is $\frac{1}{L_{eq}} = \frac{1}{L'} + \frac{1}{L'} = \frac{2}{L'} = \frac{2}{L/2} = \frac{4}{L}$,so $L_{eq} = \frac{L}{4}$.The new time constant is $	au_2 = \frac{L_{eq}}{R_{eq}} = \frac{L/4}{R/4} = \frac{L}{R} = 	au_1$.Therefore,the new time constant is $4 \, sec$.

An inductance coil has a time constant of $4 \, sec$. If it is cut into two equal parts and connected in parallel,then the new time constant of the circuit is.....$sec$.

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