Animated Solution for Physics - Current Electricity: In the circuit shown in the figure, the current through

Imagine you are a detective trying to trace the flow of water through a complex network of pipes. The battery is your water pump, and the resistors are narrow sections of pipe that slow the water down. Our mission? To find out exactly how much water (current) is flowing through one specific pipe—the $4\Omega$ resistor at the very end of the network.

I know this circuit might look a bit intimidating at first glance with all its loops and branches, but don't worry! We are going to break it down step-by-step, starting from the furthest point and working our way back to the source.

Whenever you face a complex circuit, the best strategy is to start as far away from the battery as possible. In our case, the battery is on the far left, so we will direct our attention to the far right.

Look at the rightmost branch. We have a $2\Omega$ resistor, a $4\Omega$ resistor, and another $2\Omega$ resistor connected end-to-end. Because there are no junctions between them, they are in series.

The equivalent resistance for a series combination is simply the sum of the individual resistances:

Now, imagine replacing that entire right branch with a single $8\Omega$ resistor. This new $8\Omega$ resistor is connected directly across the middle $8\Omega$ resistor. They are in parallel!

When two identical resistors are in parallel, their equivalent resistance is exactly half of their value.

We are making great progress! Now, let's move one step to the left. Our new $4\Omega$ equivalent resistance is in series with the top $2\Omega$ and bottom $2\Omega$ resistors in the middle section.

Adding them up gives us:

Do you see a pattern emerging? This $8\Omega$ equivalent resistance is now in parallel with the leftmost $8\Omega$ resistor. Just like before, two $8\Omega$ resistors in parallel give us $4\Omega$.

Finally, we are left with a very simple circuit: the battery, the $3\Omega$ resistor, our $4\Omega$ equivalent resistance, and the bottom $2\Omega$ resistor, all in series.

The total equivalent resistance of the entire circuit is:

Now that we know the total resistance, we can find the total current drawn from the battery using Ohm's Law ($V=IR$).

This $1\text{A}$ current flows out of the battery and through the $3\Omega$ resistor. But what happens when it reaches the first junction?

At the first junction, the $1\text{A}$ current sees two paths: the left $8\Omega$ resistor, and the entire rest of the circuit. But wait! We already calculated that the entire rest of the circuit also has an equivalent resistance of $8\Omega$.

Because the resistances of the two paths are equal, the current splits exactly in half.

So, $0.5\text{A}$ continues to the right. When it reaches the second junction, it again faces two paths: the middle $8\Omega$ resistor, and the rightmost branch. And guess what? The rightmost branch also has a resistance of $8\Omega$!

Once again, the current splits equally.

This $0.25\text{A}$ current flows through the rightmost branch, which includes our target $4\Omega$ resistor.

Therefore, the current through the $4\Omega$ resistor is $0.25\text{A}$. The elegance of this problem lies in how the circuit perfectly halves the current at each stage. Physics is beautiful, isn't it?