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(B) For a charging $RC$ circuit,the current is given by $i(t) = \frac{V}{R} e^{-t/RC}$. The initial current is $i_0 = \frac{V}{R}$ and the time consta

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$a-(ii), b-(iv), c-(i), d-(iii)$

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(B) For a charging $RC$ circuit,the current is given by $i(t) = \frac{V}{R} e^{-t/RC}$. The initial current is $i_0 = \frac{V}{R}$ and the time constant is $	au = RC$.$(i)$ $R = R_0, C = C_0$: $i_0 = \frac{V}{R_0}, 	au = R_0 C_0$$(ii)$ $R = 2R_0, C = C_0$: $i_0 = \frac{V}{2R_0}, 	au = 2R_0 C_0$$(iii)$ $R = R_0, C = 2C_0$: $i_0 = \frac{V}{R_0}, 	au = 2R_0 C_0$$(iv)$ $R = 2R_0, C = 2C_0$: $i_0 = \frac{V}{2R_0}, 	au = 4R_0 C_0$Comparing initial currents: $(i)$ and $(iii)$ have $i_0 = \frac{V}{R_0}$ (highest),while $(ii)$ and $(iv)$ have $i_0 = \frac{V}{2R_0}$ (lowest).Comparing time constants: $(iv)$ has the largest $	au = 4R_0 C_0$ (slowest decay),followed by $(ii)$ and $(iii)$ with $	au = 2R_0 C_0$,and $(i)$ has the smallest $	au = R_0 C_0$ (fastest decay).Curve $c$ has high $i_0$ and fast decay $ightarrow (i)$.Curve $a$ has low $i_0$ and fast decay $ightarrow (ii)$.Curve $d$ has high $i_0$ and slow decay $ightarrow (iii)$.Curve $b$ has low $i_0$ and slow decay $ightarrow (iv)$.Thus,the correct matching is $a-(ii), b-(iv), c-(i), d-(iii)$.

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