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(C) Given that, emf of battery, $E = 6 \, V$.Resistance, $R = 12 \, \Omega$.Inductance, $L = 2 \, H$.The instantaneous current in an $LR$ circuit conn

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$3 \, A \, s^{-1}$

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(C) Given that, emf of battery, $E = 6 \, V$.Resistance, $R = 12 \, \Omega$.Inductance, $L = 2 \, H$.The instantaneous current in an $LR$ circuit connected in series with a battery is given by $I = I_0(1 - e^{-(R/L)t})$, where $I_0 = E/R$ is the steady-state current.The rate of change of current is given by the derivative of $I$ with respect to time $t$:$\frac{dI}{dt} = \frac{d}{dt} \left[ \frac{E}{R} (1 - e^{-(R/L)t}) \right] = \frac{E}{R} \cdot \left( \frac{R}{L} \right) e^{-(R/L)t} = \frac{E}{L} e^{-(R/L)t}$.At the initial instant, $t = 0$, the rate of increase of current is:$\left. \frac{dI}{dt} \right|_{t=0} = \frac{E}{L} e^0 = \frac{E}{L} = \frac{6 \, V}{2 \, H} = 3 \, A \, s^{-1}$.

The initial rate of increase of current when a battery of emf $6 \, V$ is connected in series with an inductance of $2 \, H$ and resistance of $12 \, \Omega$ is

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