Animated Solution for Physics - Current Electricity: The circuit shown in the figure consists of a charged capacitor of capacity $3\mu \text{F}$ and a charge of $30\mu \text{C}$. At time $t=0$, when the key is closed, the value of current flowing through the $5\text{ M}\Omega$ resistor is $x\mu \text{A}$. The value of $x$ to the nearest integer is ......... .

Imagine you are standing in front of a dam holding back a massive reservoir of water. The water is calm, but the potential energy is immense. This is exactly what a charged capacitor is like in an electrical circuit. In our problem, we have a capacitor with a capacitance of $C=3\mu \text{F}$ that has been charged up to $q=30\mu \text{C}$. It is sitting there, fully loaded, just waiting for a path to release its stored energy.

Connected in series with this capacitor is a resistor. But this isn't just any resistor; it's a massive $5\text{ M}\Omega$ resistor. Think of this resistor as a very narrow pipe connected to our dam. It will allow water to flow, but it will resist the flow strongly.

Right now, the switch is open. The circuit is broken. The charges on the capacitor plates are trapped. There is a potential difference across the plates, but nowhere for the current to go.

Finding the Voltage

Before we close the switch and let the chaos begin, we need to know exactly how much "pressure" the capacitor is exerting. In electrical terms, this pressure is the voltage or potential difference across the capacitor plates.

The fundamental relationship for a capacitor is given by the equation:

This equation tells us that the voltage is directly proportional to the charge stored and inversely proportional to the capacitance. Let's plug in our given values. We have a charge $q=30\times 10^{-6}\text{ C}$ and a capacitance $C=3\times 10^{-6}\text{ F}$.

Notice how beautifully the $10^{-6}$ terms cancel each other out. This leaves us with a simple division:

So, at this exact moment, our charged capacitor is acting exactly like a $10\text{ V}$ battery.

Closing the Switch

Now, the moment of truth. At time $t=0$, we close the switch. The circuit is now complete. The electrons on the negative plate of the capacitor finally have a path to travel to the positive plate. They surge forward, creating an electrical current.

But here is the crucial conceptual catch: When is this current at its maximum?

As the current flows, charge leaves the capacitor. As the charge $q$ decreases, the voltage $V$ across the capacitor also decreases (since $V=q/C$). If the voltage decreases, the "push" driving the current decreases, which means the current itself must decrease.

Therefore, the current is at its absolute maximum at the exact instant the switch is closed, at $t=0$, before any significant amount of charge has had time to leave the plates.

To find this maximum initial current, we turn to our most trusted tool in circuit analysis: Ohm's Law.

We know the initial voltage driving the circuit is the $10\text{ V}$ from the capacitor. We also know the resistance is $5\text{ M}\Omega$. Here is where many students make a silly mistake—forgetting the "Mega" prefix! Mega means one million, or $10^6$. So, our resistance is $5\times 10^6\Omega$.

Let's substitute these values into Ohm's Law:

First, we divide the numbers: $10/5=2$. Next, we handle the power of ten. Bringing $10^6$ from the denominator to the numerator changes the sign of the exponent, making it $10^{-6}$.

Finding $x$

We have found our maximum current, but the question asks for the value in micro-amperes ($\mu \text{A}$).

Recall that the prefix "micro" ($\mu$) stands for $10^{-6}$. Therefore, $2\times 10^{-6}\text{ A}$ is exactly equal to $2\mu \text{A}$.

The problem states that the current is $x\mu \text{A}$. By comparing our result with this format, it is crystal clear that:

And there we have it! A seemingly complex RC circuit problem dismantled into simple, logical steps.

Where Does It All Go?

Let's take a step back and look at this circuit through the lens of energy. Before the switch is closed, the capacitor is storing electrical potential energy in the electric field between its plates. The formula for the energy stored in a capacitor is:

If we plug in our values, we get:

So, we have $150\mu \text{J}$ of energy waiting to be unleashed.

When the switch is closed, this energy doesn't just vanish. As the current flows, the electrons bump into the atoms of the $5\text{ M}\Omega$ resistor. These collisions transfer kinetic energy to the atoms, causing them to vibrate more vigorously. We perceive this macroscopic vibration as heat.

Over time, as the capacitor fully discharges, every single micro-joule of that stored electrical energy is converted into thermal energy and dissipated by the resistor into the surrounding environment. This is a beautiful demonstration of the Law of Conservation of Energy in action!

Let's dive a bit deeper into that time constant, $\tau =RC$, which we calculated to be 15 seconds. What does this number actually tell us?

The time constant is a measure of the circuit's "inertia" or sluggishness. It tells us how long it takes for the voltage (and the current) to drop to approximately $36.8\%$ of its initial value (since $e^{-1}\approx 0.368$).

If we had used a smaller resistor, say $5\Omega$ instead of $5\text{ M}\Omega$, the time constant would be a mere $15$ microseconds! The capacitor would discharge almost instantaneously in a massive, potentially dangerous spark. The $5\text{ M}\Omega$ resistor acts as a strict traffic controller, ensuring the energy is released slowly and safely over a period of minutes.

By understanding these underlying principles—energy conservation and the physical meaning of the time constant—you aren't just solving a math problem; you are comprehending the very fabric of electrical physics.