In the figure below,the switches S_1 and S_2 | vedclass

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(D) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The current $I$ in the middle wire is $I = I_2 - I_1$.For th

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(D) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The current $I$ in the middle wire is $I = I_2 - I_1$.For the left loop: $V - I_1 R - L \frac{dI_1}{dt} = 0 \implies I_1(t) = \frac{V}{R}(1 - e^{-(R/L)t})$.For the right loop: $V - I_2 R - 2L \frac{dI_2}{dt} = 0 \implies I_2(t) = \frac{V}{R}(1 - e^{-(R/2L)t})$.The current in the middle wire is $I(t) = I_2(t) - I_1(t) = \frac{V}{R} [e^{-(R/L)t} - e^{-(R/2L)t}]$.To find the maximum current,set $\frac{dI}{dt} = 0$:$\frac{dI}{dt} = \frac{V}{R} [-\frac{R}{L} e^{-(R/L)t} + \frac{R}{2L} e^{-(R/2L)t}] = 0$.$\frac{1}{L} e^{-(R/L)t} = \frac{1}{2L} e^{-(R/2L)t} \implies e^{-(R/2L)t} = \frac{1}{2}$.Taking the natural logarithm on both sides: $-\frac{R}{2L} T = \ln(1/2) = -\ln 2 \implies T = \frac{2L}{R} \ln 2$.Substituting $T$ back into the expression for $I(t)$:$I_{\max} = \frac{V}{R} [e^{-(R/L) \cdot (2L/R) \ln 2} - e^{-(R/2L) \cdot (2L/R) \ln 2}] = \frac{V}{R} [e^{-2 \ln 2} - e^{-\ln 2}] = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}] = -\frac{V}{4R}$.The magnitude is $I_{\max} = \frac{V}{4R}$.Thus,statements $(B)$ and $(D)$ are true.

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