A constant voltage of 25 ~V is applied to a s | vedclass

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(A) In a series $L-R$ circuit,the current $i$ at any time $t$ is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$.At $t = 0$,the current $i = i_0(1

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$0 ~V, 25 ~V$

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(A) In a series $L-R$ circuit,the current $i$ at any time $t$ is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$.At $t = 0$,the current $i = i_0(1 - e^0) = i_0(1 - 1) = 0$.The potential difference across the resistor is $V_R = iR$. Since $i = 0$ at $t = 0$,$V_R = 0 ~V$.The potential difference across the inductor is $V_L = L(di/dt)$. According to Kirchhoff's voltage law,$V = V_R + V_L$. At $t = 0$,$V = 0 + V_L$,so $V_L = 25 ~V$.Therefore,the potential difference across the resistor is $0 ~V$ and across the inductor is $25 ~V$.

$A$ constant voltage of $25 ~V$ is applied to a series $L-R$ circuit at $t=0$,by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$?

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