Animated Solution for Physics - Current Electricity: A capacitor is charged using an external battery with a resistance $x$ in series. The dashed line shows the variation of $\ln I$ with respect to time. If the resistance is changed to $2x$, the new graph will be NOTE: Here $I$ is the charging current.

You connect it to a battery and a resistor, and the current starts flowing. But it doesn't flow at a constant rate; it starts high and gradually tapers off to zero. This behavior is captured by the charging current equation: where $I_0=\frac{E}{R}$ is the initial maximum current, $E$ is the battery's EMF, $R$ is the resistance, and $C$ is the capacitance.

The problem provides a graph of $\ln I$ versus time $t$. To make sense of this, we need to linearize our equation by taking the natural logarithm on both sides:

It perfectly matches the standard equation of a straight line, $y=mx+c$. - Our y-axis variable is $y=\ln I$. - Our x-axis variable is $x=t$. - The y-intercept is $c=\ln \left(\frac{E}{R}\right)$. - The slope is $m=-\frac{1}{RC}$.

This explains why the original dashed line in the graph is straight and slopes downwards. The intercept tells us where the current starts, and the slope tells us how fast it decays.

Let's break this down into two parts: the intercept and the slope.

1. The New Intercept: If we replace $R$ with $2R$, the new y-intercept becomes: Since we are subtracting $\ln 2$, the new intercept is strictly less than the original one. Visually, this means our new graph must start at a lower point on the y-axis. This immediately eliminates graphs $S$ and $R$, which start at the same high intercept as the original dashed line. We are left with graphs $P$ and $Q$.

2. The New Slope: Now, let's look at the slope. Substituting $2R$ into our slope formula gives: The magnitude of the new slope is exactly half of the original slope. A smaller magnitude means the line is less steep, or flatter, compared to the original dashed line.

Looking at the remaining options, line $P$ is quite steep, while line $Q$ is noticeably flatter.

Graph $Q$ perfectly satisfies both our mathematical conditions. Therefore, Graph Q is the correct representation of the new circuit.

As a quick thought experiment, what if the capacitance $C$ was doubled instead of the resistance? The intercept $\ln (E/R)$ would remain unchanged, but the slope would still be halved. A line with the same intercept but a flatter slope corresponds exactly to graph $S$! Analyzing all options like this is a fantastic way to deepen your intuition.