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(A) The current in an $LR$ circuit at time $t$ is given by $I(t) = \frac{E}{R}(1 - e^{-Rt/L})$.As $t 	o \infty$,the steady-state current is $I_{max}

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$L$ is increased

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(A) The current in an $LR$ circuit at time $t$ is given by $I(t) = \frac{E}{R}(1 - e^{-Rt/L})$.As $t 	o \infty$,the steady-state current is $I_{max} = \frac{E}{R}$.From the graph,both curves approach the same steady-state current value,which implies that the resistance $R$ remains unchanged.The time constant of the circuit is $	au = \frac{L}{R}$.Curve $2$ reaches the steady-state value more slowly than curve $1$,which means the time constant for curve $2$ is larger than that for curve $1$ $(	au_2 > 	au_1)$.Since $	au = \frac{L}{R}$ and $R$ is constant,an increase in the time constant $	au$ implies that the inductance $L$ must have been increased.

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