In an R-C charging circuit,the dashed line re | vedclass

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

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$B$

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(B) For an $R-C$ charging circuit,the current at time $t$ is given by $I = \frac{E}{R} e^{-\frac{t}{RC}}$.Taking the natural logarithm on both sides,we get $\ln I = \ln(\frac{E}{R}) - \frac{t}{RC}$.This is a linear equation of the form $y = mx + c$,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 new slope $m' = -\frac{1}{R'C} = -\frac{1}{2RC} = \frac{1}{2}m$. Since the slope becomes half,the line becomes less steep.$2$. The new intercept $c' = \ln(\frac{E}{2R}) = \ln(\frac{E}{R}) - \ln 2$. Since $\ln 2 > 0$,the intercept decreases.Comparing the options,line $B$ has a smaller (less negative) slope and a lower intercept than the dashed line,which matches the derived conditions.

In an $R-C$ charging circuit,the dashed line represents the graph of $\ln I$ versus $t$. If the resistance of the circuit is doubled,which of the following solid lines represents the correct graph of $\ln I$ versus $t$?

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