Animated Solution for Physics - Current Electricity: In the given figure, switches $S_1$ and $S_2$ are in open condition. The resistance across $ab$ when the switches $S_1$ and $S_2$ are closed is ...... $\Omega$.

Imagine you are an electrical engineer tasked with simplifying a complex web of resistors. At first glance, the circuit in front of you looks like a tangled mess. We have a top branch, a bottom branch, and two switches, $S_1$ and $S_2$, acting as bridges between them. Our mission is to find the equivalent resistance across terminals $a$ and $b$ when these switches are closed.

Before we flip any switches, let's understand the landscape. The current enters at terminal $a$ and splits into two paths. The top path encounters a $12\Omega$, a $4\Omega$, and a $6\Omega$ resistor in series. The bottom path faces a $6\Omega$, a $4\Omega$, and a $12\Omega$ resistor in series. If the switches remained open, calculating the equivalent resistance would be a simple matter of adding the series resistors and then combining the two parallel branches. But the problem asks us to close the switches. This changes everything!

When we close switches $S_1$ and $S_2$, we are essentially connecting the nodes with ideal, zero-resistance wires. In the world of physics, a zero-resistance wire means there is no potential drop across it. Therefore, the nodes connected by the switch are forced to be at the exact same electrical potential.

Let's look at switch $S_1$. It connects the node between the $12\Omega$ and $4\Omega$ resistors on top to the node between the $6\Omega$ and $4\Omega$ resistors on the bottom. Because these two nodes are now at the same potential, we can conceptually merge them into a single point!

By merging the nodes connected by $S_1$, the $12\Omega$ resistor from the top branch and the $6\Omega$ resistor from the bottom branch are now connected across the exact same two points (from terminal $a$ to the newly merged node). This means they are perfectly in parallel!

Similarly, switch $S_2$ merges the next set of nodes. This forces the two $4\Omega$ resistors to be in parallel with each other. Finally, the remaining $6\Omega$ and $12\Omega$ resistors are also forced into a parallel configuration.

Our complex, tangled circuit has beautifully simplified into three distinct parallel blocks connected one after the other.

Let's call the equivalent resistance of these three parallel blocks $R_1$, $R_2$, and $R_3$. Since these blocks are connected sequentially, they are in series. The total equivalent resistance $R_{ab}$ is simply their sum:

Now, we just need to calculate the resistance of each parallel block. Remember the handy formula for two resistors in parallel: the product over the sum.

Let's crunch the numbers. For the first block ($R_1$), we have $12\Omega$ and $6\Omega$ in parallel:

For the second block ($R_2$), we have two $4\Omega$ resistors in parallel. A quick shortcut: when two identical resistors are in parallel, their equivalent resistance is exactly half!

For the third block ($R_3$), we again have $6\Omega$ and $12\Omega$ in parallel, just like the first block:

Finally, we substitute these values back into our master equation:

And there we have it! By understanding the physical meaning of a closed switch and redrawing the circuit, we transformed a daunting problem into a straightforward calculation. The equivalent resistance is $10\Omega$.