In series with a 20 Omega resistor,a 5 H in | vedclass

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

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$e^{-1}$

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(C) The current $i$ in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.Taking the derivative with respect to time $t$,we get the rate of increase of current:$\frac{di}{dt} = \frac{d}{dt} [\frac{E}{R}(1 - e^{-Rt/L})] = \frac{E}{R} \cdot (\frac{R}{L}) e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.Given values: $E = 5 \ V$,$R = 20 \ \Omega$,$L = 5 \ H$,and $t = 0.25 \ s$.Calculate the exponent: $\frac{Rt}{L} = \frac{20 	imes 0.25}{5} = \frac{5}{5} = 1$.Substitute the values into the expression for $\frac{di}{dt}$:$\frac{di}{dt} = \frac{5}{5} e^{-1} = 1 \cdot e^{-1} = e^{-1} \ A/s$.

In series with a $20 \ \Omega$ resistor,a $5 \ H$ inductor is placed. To this combination,an $e.m.f.$ of $5 \ V$ is applied. What will be the rate of increase of current at $t = 0.25 \ s$?

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