Animated Solution for Physics - Current Electricity: The voltage drop across $15\Omega$ resistance in the given figure will be ......... V.

The sight of a sprawling circuit diagram can often send a shiver down a student's spine. With resistors branching out in every direction, it looks less like a physics problem and more like a complex maze. But here is the secret: every complex circuit is just a collection of simple, bite-sized pieces waiting to be unraveled.

Imagine you are an electrician tasked with finding the exact voltage drop across a specific component—in this case, the $15\Omega$ resistor. You can't just guess; you need a systematic approach. Let's embark on this journey together and break down this circuit step by step.

First, let's take a deep breath and look at the big picture. The circuit is divided into two main pathways: an upper branch and a lower branch. These two branches are connected in parallel across the main terminals, labeled 'a' and 'b'.

Powering this entire setup is a $12\text{ V}$ battery. But notice the small detail next to it—a $1\Omega$ resistor. This isn't just a random component; it represents the internal resistance of the battery itself. This means the battery isn't perfect; it consumes a tiny bit of its own energy before delivering it to the rest of the circuit.

Let's zoom in on the upper branch. It looks like a mini-circuit of its own. We have three distinct sections here.

First, we see two $4\Omega$ resistors in parallel. When identical resistors are in parallel, their equivalent resistance is simply half of one resistor's value. So, this part simplifies to $2\Omega$.

Next, this $2\Omega$ equivalent resistance is in series with an actual $2\Omega$ resistor. Adding them up gives us $4\Omega$.

Finally, we encounter a parallel combination of a $15\Omega$ and a $10\Omega$ resistor. Using our trusty parallel resistance formula:

Now, we string these three sections together. The total resistance of the upper branch is:

Now, let's shift our focus to the lower branch. It's even simpler! We have two parallel sections connected in series.

The first section has two $8\Omega$ resistors in parallel. Just like before, two identical resistors in parallel give half the resistance, which is $4\Omega$.

The second section has two $12\Omega$ resistors in parallel. Half of twelve is $6\Omega$.

Adding these two sections together, we find the total resistance of the lower branch:

Look at what we've discovered! Both the upper and lower branches have an identical total resistance of $10\Omega$. This symmetry is a massive advantage.

Since these two $10\Omega$ branches are in parallel across terminals 'a' and 'b', their equivalent resistance is:

This means the entire complex web of resistors can be replaced by a single $5\Omega$ resistor!

Now we can calculate the total current flowing out of the battery. But remember that sneaky internal resistance? We must add it to our equivalent resistance to find the total resistance of the entire circuit.

Using Ohm's Law, we can find the total current $I$:

So, a total of $2\text{ A}$ of current leaves the battery and heads towards terminal 'a'.

When the $2\text{ A}$ current reaches terminal 'a', it faces a choice: go through the upper branch or the lower branch. Because both branches have the exact same resistance ($10\Omega$), the current splits perfectly in half.

Exactly $1\text{ A}$ of current flows through the upper branch.

We are almost there! We need to find the voltage drop across the $15\Omega$ resistor. To do this, we first need to know how much of that $1\text{ A}$ current actually flows through it.

The $1\text{ A}$ current reaches the parallel combination of the $15\Omega$ and $10\Omega$ resistors. We can use the current divider rule to find the exact current through the $15\Omega$ path:

Finally, we apply Ohm's Law one last time to find the voltage drop across the $15\Omega$ resistor:

And there we have it! The voltage drop across the $15\Omega$ resistor is exactly $6\text{ V}$. By breaking the problem down into logical, manageable steps, we turned a daunting circuit into a satisfying puzzle.