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(B) The circuit represents the discharging of a capacitor through a resistor $R$. The current $I$ at time $t$ is given by $I = I_0 e^{-t / RC}$,where

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$C$ is increased

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(B) The circuit represents the discharging of a capacitor through a resistor $R$. The current $I$ at time $t$ is given by $I = I_0 e^{-t / RC}$,where $I_0 = V/R$.Taking the natural logarithm on both sides: $\ln I = \ln(V/R) - \frac{t}{RC}$.Converting to base $10$ logarithm: $\log_{10} I = \log_{10}(V/R) - \frac{t}{2.303 RC}$.Comparing this with the equation of a straight line $y = mx + c$,the intercept is $\log_{10}(V/R)$ and the slope is $m = -\frac{1}{2.303 RC}$.From the graph,the intercept for both lines $(1)$ and $(2)$ is the same,which implies that $V$ and $R$ remain constant.The magnitude of the slope of line $(2)$ is less than that of line $(1)$.Since $|m| = \frac{1}{2.303 RC}$,a decrease in the magnitude of the slope implies that the product $RC$ must increase.Since $R$ is constant,$C$ must be increased.

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