In the circuit shown here, the point 'C' is k | vedclass

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(B) $1$. When point '$C$' is connected to '$A$', the current in the circuit becomes constant, $I_0 = E/R$. The inductor acts as a short circuit.$2$. W

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$-1$

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(B) $1$. When point '$C$' is connected to '$A$', the current in the circuit becomes constant, $I_0 = E/R$. The inductor acts as a short circuit.$2$. When '$C$' is connected to '$B$' at $t = 0$, the battery is removed, and the circuit becomes a decaying $LR$ circuit.$3$. The current in the circuit at time $t$ is given by $I(t) = I_0 e^{-Rt/L}$.$4$. At $t = L/R$, the current is $I = I_0 e^{-R(L/R)/L} = I_0 e^{-1} = I_0/e$.$5$. The voltage across the resistance is $V_R = I R = (I_0/e) R = (E/R \cdot 1/e) R = E/e$.$6$. Applying Kirchhoff's voltage law in the closed loop (without battery): $V_R + V_L = 0$, which implies $V_L = -V_R$.$7$. Therefore, the ratio of the voltage across the resistance $(V_R)$ to the voltage across the inductor $(V_L)$ is $V_R / V_L = V_R / (-V_R) = -1$.

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