Animated Solution for Physics - Current Electricity: The current $I_1$ (in ampere) flowing through $1\Omega$ resistor in the following circuit is

Imagine a river splitting into multiple channels. The water pressure at the start of the split is the same for all channels, regardless of how wide or narrow they are. This is the exact principle governing parallel electrical circuits.

In our problem, we are presented with a circuit containing three distinct branches connected across a $1\text{V}$ battery. Because these branches are connected in parallel, the potential difference (or "electrical pressure") across each individual branch is identical to the battery's voltage. Therefore, the voltage across the upper branch, the middle branch, and the lower branch is exactly $1\text{V}$.

Our objective is to find the current $I_1$, which flows through a specific resistor in the upper branch. Here is a crucial conceptual leap: because the branches are in parallel, the current flowing through the middle branch does not affect the voltage across the upper branch.

We can mentally (or visually) remove the middle $2\Omega$ resistor. It draws its own current from the battery, but it doesn't change the fact that the upper branch still "sees" a full $1\text{V}$ across its terminals, nodes A and B. By isolating the upper branch, we transform a seemingly complex multi-loop circuit into a straightforward series-parallel problem.

Let's zoom in on the upper branch. It consists of a parallel combination of two $1\Omega$ resistors, which is then connected in series with a $2\Omega$ resistor.

First, we resolve the parallel part. When two identical resistors are in parallel, their equivalent resistance is simply half of one resistor's value. Mathematically:

Now, this $0.5\Omega$ equivalent resistance is in series with the $2\Omega$ resistor. To find the total resistance of the upper branch ($R_{eq}$), we simply add them together:

With the total resistance of the upper branch known, we can determine the total current ($I$) entering it from node A. According to Ohm's Law ($V=IR$):

This $0.4\text{A}$ current travels from node A and reaches node C, where it faces a fork in the road. It must split between the two $1\Omega$ resistors.

Current always takes the path of least resistance. However, in this case, both paths offer the exact same resistance ($1\Omega$). Because the paths are perfectly symmetrical, the current divides perfectly in half.

Therefore, the current $I_1$ flowing through the top resistor is:

And just like that, by breaking the circuit down into logical, bite-sized pieces, we arrive at our final answer of $0.2\text{A}$.