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

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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 $t$ after closing the switch is given by $i(t) = \frac{V}{R} (1 - e^{-Rt/L})$.At $t = 0$, the current in the circuit is $i(0) = \frac{V}{R} (1 - e^0) = 0$.The potential difference across the resistor is $V_R = iR$. Since $i = 0$ at $t = 0$, $V_R = 0 	imes R = 0 \, V$.The potential difference across the inductor is $V_L = L \frac{di}{dt}$. According to Kirchhoff's voltage law, $V = V_R + V_L$. At $t = 0$, $V = 0 + V_L$, which gives $V_L = 25 \, V$.Therefore, at $t = 0$, 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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