Animated Solution for Physics - Current Electricity: A parallel combination of $0.1\mathrm{M}\Omega$ resistor and a $10\mu \mathrm{F}$ capacitor is connected across a $1.5\mathrm{V}$ source of negligible resistance. The time required for the capacitor to get charged upto $0.75\mathrm{V}$ is approximately (in second)

Imagine you are standing in front of a circuit board, holding a capacitor, a resistor, and a battery. You wire them up in parallel. The battery is a perfect, ideal voltage source with absolutely zero internal resistance. This is the crux of the problem!

When you connect a capacitor directly across an ideal battery, what happens? The battery acts like an infinite reservoir of charge, ready to pump electrons as fast as the laws of physics allow.

Let's break down the components. We have a parallel combination of a $0.1\mathrm{M}\Omega$ resistor and a $10\mu \mathrm{F}$ capacitor. This entire combination is connected across a $1.5\mathrm{V}$ source.

The most critical piece of information here is that the source has negligible resistance. In the real world, every battery has some internal resistance, which limits how fast it can deliver current. But in this idealized scenario, the internal resistance is exactly zero.

Why does this matter? Think of a capacitor as a water tank and the battery as a massive water tower. The wires connecting them are the pipes. If the pipes are infinitely wide (zero resistance), the water from the tower will instantly fill the tank to the same level. There is no bottleneck, no restriction to the flow.

In our circuit, the $0.1\mathrm{M}\Omega$ resistor is like a separate, very narrow pipe connected to the same water tower. It allows a tiny trickle of water to flow through it constantly. But does this narrow pipe affect how fast the main tank (the capacitor) fills up? Not at all! The tank has its own infinitely wide pipe connecting it directly to the tower.

Many students fall into the trap of calculating an equivalent resistance for the entire circuit. They see a resistor and a capacitor and immediately think of the standard RC circuit formulas. They might even try to use the $0.1\mathrm{M}\Omega$ value to calculate a time constant. But this is a fundamental misunderstanding of how parallel circuits operate. In a parallel circuit, each branch is independent of the others, provided the voltage source is ideal. The battery maintains a strict $1.5\mathrm{V}$ across both the resistor and the capacitor, regardless of what is happening in the other branch.

When a capacitor charges through a resistor, the voltage across it as a function of time is given by the master equation:

where $\tau$ is the time constant of the circuit, defined as $\tau =R_{\mathrm{eq}}C$.

This equation tells a beautiful story of exponential growth. Initially, when the capacitor is empty, it accepts charge eagerly, and the voltage rises rapidly. As it fills up, it starts to push back, and the rate of charging slows down. The time constant $\tau$ dictates the pace of this entire process. A large $\tau$ means a slow, sluggish charge, while a small $\tau$ means a rapid, snappy charge.

Let's delve deeper into where this equation comes from. It arises from Kirchhoff's Voltage Law applied to a simple series RC loop. The sum of the voltage drops across the resistor and the capacitor must equal the battery voltage:

Using Ohm's Law ($V_R=IR$) and the capacitor equation ($V_C=\frac{Q}{C}$), and knowing that current is the rate of change of charge ($I=\frac{dQ}{dt}$), we get a first-order linear differential equation:

Solving this differential equation yields the exponential charging curve we are so familiar with. But notice the crucial role of $R$ in this equation. It is the resistance in series with the capacitor that limits the flow of charge.

Here, $R_{\mathrm{eq}}$ is the equivalent resistance of the charging loop. To find it, we must trace the path that the charging current takes from the battery to the capacitor. Since the capacitor is in parallel with the resistor, and both are connected directly to the battery, the charging path for the capacitor goes straight through the battery and the connecting wires.

Because the battery has zero internal resistance, and we assume ideal wires, the resistance of this charging path is simply $0\Omega$. The parallel resistor of $0.1\mathrm{M}\Omega$ draws its own steady current from the battery, but it does not affect the charging path of the capacitor! The battery, being an ideal voltage source, can supply infinite current to satisfy both branches simultaneously without breaking a sweat.

Let's substitute this zero resistance into our time constant formula:

A time constant of zero is a profound mathematical statement. It means the exponential term $e^{-\frac{t}{0}}$ drops to zero instantly for any $t>0$. The capacitor charges instantaneously. The moment you close the switch, the voltage across the capacitor jumps to the full $1.5\mathrm{V}$ of the battery.

The question asks for the time required for the capacitor to get charged up to $0.75\mathrm{V}$. Since it reaches $1.5\mathrm{V}$ in zero seconds, it also reaches $0.75\mathrm{V}$ in exactly zero seconds.

There is no exponential curve here, no waiting for the charge to build up. It's an instantaneous leap! This is a classic trap in physics problems—giving you extra information (like the $0.1\mathrm{M}\Omega$ resistor) to distract you from the core physical reality. The test-makers want to see if you will blindly plug numbers into a formula or if you will pause and truly understand the physical setup.

Think about the physical implications of this. An instantaneous change in voltage across a capacitor implies an infinite current flowing for an infinitesimally short duration. In mathematical terms, the current is a Dirac delta function. While this is a theoretical abstraction and impossible in the real world (where wires have inductance and resistance), it is the correct conclusion for this idealized problem.

Always look for the path of least resistance! When a capacitor is shorted to an ideal voltage source, the charging time is always zero. It is a beautiful reminder that in physics, the simplest path often holds the most elegant truth. Do not let extraneous numbers cloud your physical intuition. Trust the fundamental principles, and the math will naturally follow.