An inductor coil carries a steady-state curre | vedclass

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(A) Given: Steady-state current $I = 2.0 \, A$,$emf \ E = 4.0 \, V$,and inductance $L = 1.0 \, H$.In steady-state,an inductor acts as a short circuit

Vedclass · Accepted Answer

$0.5$

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(A) Given: Steady-state current $I = 2.0 \, A$,$emf \ E = 4.0 \, V$,and inductance $L = 1.0 \, H$.In steady-state,an inductor acts as a short circuit (zero resistance),but the coil itself has an internal resistance $R$.Using Ohm's law for the steady-state current: $R = \frac{E}{I} = \frac{4.0}{2.0} = 2.0 \, \Omega$.The time constant $	au$ of an $LR$ circuit is given by the formula $	au = \frac{L}{R}$.Substituting the values: $	au = \frac{1.0 \, H}{2.0 \, \Omega} = 0.5 \, s$.

An inductor coil carries a steady-state current of $2.0 \, A$ when connected across an ideal battery of $emf \ 4.0 \, V$. If its inductance is $1.0 \, H$,find the time constant of the circuit in $sec$.

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