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(A) The current in an $LR$ circuit as a function of time is given by $I(t) = \frac{E}{R}(1 - e^{-Rt/L})$.From the graph,we observe that the steady-sta

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$L$ is increased

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(A) The current in an $LR$ circuit as a function of time is given by $I(t) = \frac{E}{R}(1 - e^{-Rt/L})$.From the graph,we observe that the steady-state current $I_{max} = \frac{E}{R}$ is the same for both curves $1$ and $2$. This implies that the ratio $\frac{E}{R}$ remains constant.Since $E$ and $R$ are constant,the change must be in the inductance $L$.The time constant of the circuit is $	au = \frac{L}{R}$.Curve $2$ rises more slowly than curve $1$,which means the time constant $	au$ for curve $2$ is larger than for curve $1$.Since $	au = \frac{L}{R}$ and $R$ is constant,an increase in $	au$ implies an increase in $L$.Therefore,the inductance $L$ was increased.

$A \; (mm \; s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \; (mm)$	$-4$	$-2$	$0$	$2$	$4$

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