Animated Solution for Physics - Current Electricity: A capacitor of capacitance $C=1\mu \text{F}$ is suddenly connected to a battery of $100\text{ V}$ through a resistance $R=100\Omega$. The time taken for the capacitor to be charged to get $50\text{ V}$ is [Take, $\ln 2=0.69$]

Imagine you are tasked with filling a massive water tank using a high-pressure water pump. However, the water must flow through a very narrow, restrictive pipe before it reaches the tank. This physical scenario is the perfect analogy for understanding the RC charging circuit presented in our problem.

In our electrical setup, the $100\text{ V}$ direct current battery acts as the high-pressure water pump, providing a constant electromotive force. The capacitor, with a capacitance of $C=1\mu \text{F}$, is our empty water tank, ready to store electrical charge. Finally, the resistor, with a resistance of $R=100\Omega$, represents the narrow pipe. It acts as a bottleneck, strictly controlling the rate at which the electrical current can flow into the capacitor.

The moment we close the switch, the charging process begins. Initially, the capacitor is completely empty, meaning it offers zero opposition to the incoming current. The current is at its absolute maximum. However, as charge begins to accumulate on the capacitor's plates, a potential difference (voltage) builds up across it.

This newly created voltage across the capacitor acts like a back-pressure. It directly opposes the voltage of the battery. Because the net driving voltage in the circuit is decreasing, the charging current must also decrease over time.

This means the voltage across the capacitor does not rise in a simple, straight linear line. Instead, it rises rapidly at first, and then gradually slows down as it approaches the maximum supply voltage. This beautiful natural phenomenon is described by the exponential charging equation:

Here, $V(t)$ is the instantaneous voltage across the capacitor at any time $t$, $V_0$ is the maximum supply voltage from the battery, and the product $RC$ is known as the time constant ($\tau$). The time constant dictates exactly how sluggish or responsive the circuit is.

Our primary objective in this problem is to find the exact moment in time, $t$, when the voltage across the capacitor reaches exactly $50\text{ V}$. Notice that this is exactly half of our total supply voltage of $100\text{ V}$.

Let's take our known physical values and carefully substitute them into our master exponential equation. We replace the instantaneous voltage $V(t)$ on the left side with $50$. On the right side, we replace the maximum supply voltage $V_0$ with $100$. We leave the exponential term untouched for now, as it contains our unknown variable, time $t$.

This is our raw setup. The physics has now been perfectly translated into a pure mathematical puzzle.

To solve for time $t$, we must isolate the exponential term. We begin by dividing both sides of the equation by $100$. This simplifies the left side to a clean fraction:

Next, we need to gather our terms. We want the exponential term all by itself on one side of the equals sign. Currently, it carries a negative sign. By moving the entire exponential term to the left side of the equation, it becomes positive. Simultaneously, we subtract $\frac{1}{2}$ from $1$ on the right side, leaving us with exactly $\frac{1}{2}$.

We are getting closer, but we still have a negative sign in the exponent. To eliminate it, we apply a neat mathematical trick: taking the reciprocal of both sides. The reciprocal of $e^{-\frac{t}{RC}}$ is simply $e^{\frac{t}{RC}}$, and the reciprocal of $\frac{1}{2}$ is $2$.

Now comes the most powerful tool for dealing with exponentials: the natural logarithm ($\ln$). Because the natural logarithm and the exponential function base $e$ are inverse functions, taking the natural log of both sides perfectly cancels the base $e$, bringing our time variable down from the exponent.

We have successfully freed our time variable! The equation is now completely linear.

Before we can calculate the final time, we must evaluate the denominator, the time constant $RC$. We multiply the resistance ($100\Omega$) by the capacitance ($1\mu \text{F}$). Remember that the prefix 'micro' stands for $10^{-6}$.

This tiny fraction of a second is the characteristic heartbeat of our specific circuit. Now, we substitute this time constant back into our rearranged equation.

The problem kindly provides the value of $\ln 2$ as $0.69$. Multiplying these values together yields our final answer:

This is the exact moment the capacitor hits the $50\text{ V}$ mark. Comparing this result with the given options, we can confidently conclude that option (c) is the correct choice.

Always try to visualize the complete lifecycle of these circuits. If we were to remove the battery now and short the circuit, the capacitor would become the power source, and the voltage would decay exponentially according to $V=V_0{e}^{-\frac{t}{RC}}$. Mastering these exponential relationships builds a profound intuition for electromagnetism!