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(A) The current in an $LR$ circuit as a function of time is given by $i(t) = \frac{E}{R} (1 - e^{-Rt/L})$.At $t = 0$, the initial rate of change of cu

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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 $i(t) = \frac{E}{R} (1 - e^{-Rt/L})$.At $t = 0$, the initial rate of change of current is $\left( \frac{di}{dt} \right)_{t=0} = \frac{E}{L}$.From the graph, the steady-state current (maximum current $i_{max} = E/R$) is the same for both curves, which implies that $E/R$ remains constant. Thus, $E$ and $R$ are not changed.Comparing the slopes at $t=0$, we see that the slope of curve $2$ is less than the slope of curve $1$, i.e., $\left( \frac{di}{dt} \right)_2 Since $\frac{di}{dt} = \frac{E}{L}$, a decrease in the initial slope implies an increase in inductance $L$.Therefore, $L$ is increased.

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