An inductor of 2 , H and a resistance of 10 , | vedclass

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(C) The current in an $LR$ series circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.Differentiating with respect to time $t$,we get the r

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$2.5$

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(C) The current in an $LR$ series circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.Differentiating with respect to time $t$,we get the rate of change of current:$\frac{di}{dt} = \frac{d}{dt} [i_0(1 - e^{-Rt/L})] = i_0 \cdot (\frac{R}{L}) e^{-Rt/L} = (\frac{E}{R}) \cdot (\frac{R}{L}) e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.At the initial moment,$t = 0$,the exponential term $e^0 = 1$.Therefore,the initial rate of change of current is $\frac{di}{dt} = \frac{E}{L}$.Given $E = 5 \, V$ and $L = 2 \, H$:$\frac{di}{dt} = \frac{5}{2} = 2.5 \, A/s$.

An inductor of $2 \, H$ and a resistance of $10 \, \Omega$ are connected in series with a battery of $5 \, V$. The initial rate of change of current is ...... $A/s$.

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