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(B) For an $RC$ circuit during charging,the current $i$ at time $t$ is given by $i = \frac{E}{R} e^{-t/RC}$.Taking the natural logarithm on both sides

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$Q$

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(B) For an $RC$ circuit during charging,the current $i$ at time $t$ is given by $i = \frac{E}{R} e^{-t/RC}$.Taking the natural logarithm on both sides,we get $\ln i = -\frac{t}{RC} + \ln(\frac{E}{R})$.This is an equation of a straight line $y = mx + c$,where the slope $m = -\frac{1}{RC}$ and the intercept $c = \ln(\frac{E}{R})$.When the resistance $R$ is doubled $(R' = 2R)$:$1$. The intercept $c' = \ln(\frac{E}{2R}) = \ln(\frac{E}{R}) - \ln 2$. Since $\ln 2 > 0$,the new intercept $c'$ is lower than the original intercept $c$.$2$. The magnitude of the slope $|m'| = \frac{1}{R'C} = \frac{1}{2RC} = \frac{1}{2} |m|$. The new slope is half the magnitude of the original slope,meaning the line becomes less steep.Comparing these changes with the given graph,curve $Q$ starts at a lower intercept and has a smaller magnitude of slope compared to the dotted line. Therefore,curve $Q$ represents the variation.

In an $RC$ circuit while charging,the graph of $\log i$ versus time is as shown by the dotted line in the diagram,where $i$ is the current. When the value of the resistance is doubled,which of the solid curves best represents the variation of $\log i$ versus time?

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