The resistance and inductance of a series RL | vedclass

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(B) For an $RL$ circuit,the current at any time $t$ is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.Taking the derivative with respect to time

Vedclass · Accepted Answer

$80$

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(B) For an $RL$ circuit,the current at any time $t$ 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 $\frac{di}{dt} = \frac{i_0 R}{L} e^{-Rt/L}$.Substituting $i_0 = E/R$,we have $\frac{di}{dt} = \frac{E}{L} e^{-Rt/L}$.At the instant of closing the switch $(t = 0)$,the expression becomes $\frac{di}{dt} = \frac{E}{L}$.Given $\frac{di}{dt} = 4 \, A/s$,$L = 20 \, H$,and $R = 5 \, \Omega$.Substituting these values: $4 = \frac{E}{20}$.Therefore,$E = 4 	imes 20 = 80 \, V$.

The resistance and inductance of a series $RL$ circuit are $5 \, \Omega$ and $20 \, H$ respectively. At the instant of closing the switch,the current is increasing at the rate of $4 \, A/s$. The supply voltage is ... $V$.

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