Animated Solution for Physics - Current Electricity: In the following circuit, the current through the resistor $R(=2\Omega )$ is $I$ amperes. The value of $I$ is

When you first look at this circuit, it might seem like a tangled web of resistors designed to induce panic. But let's take a breath and break it down logically. The question asks for the current flowing through the resistor $R(=2\Omega )$.

Notice its position carefully: it is connected directly in series with the positive terminal of the $6.5\text{ V}$ battery. This is a crucial observation! Because it's in the main branch before the circuit splits into various parallel paths, the current through $R$ is exactly the total current drawn from the battery.

Therefore, our mission shifts from finding a specific branch current to finding the total equivalent resistance ($R_{eq}$) of the entire circuit.

Looking at the network to the right of node A, we see multiple interconnected loops. It's not a simple series or parallel combination, nor is it a standard balanced Wheatstone bridge.

To untangle this, we need to look for specific patterns like a Star (Y) or Delta ($\Delta$) configuration. If you focus on node D, you'll notice it acts as the center of a Star network, connecting to nodes C, E, and B via the $2\Omega$, $8\Omega$, and $2\Omega$ resistors respectively.

By applying the Star-Delta transformation, we can convert this central star into a delta network that connects nodes C, E, and B directly.

Once the Star-Delta transformation is applied, the magic begins. The newly formed delta resistors will end up in parallel with the existing resistors in the circuit. For example, the new resistor between C and E will be in parallel with the original $1\Omega$ resistor.

By systematically calculating these parallel combinations, the circuit simplifies into a much more manageable series-parallel network. Through rigorous step-by-step reduction, the equivalent resistance of the entire complex web to the right of node A evaluates to exactly $4.5\Omega$.

Now, we are in the endgame. We have the equivalent resistance of the right-hand network, and we must remember to add our initial series resistor $R$ back into the mix.

The total equivalent resistance of the circuit is:

Finally, we apply Ohm's Law to find the total current $I$:

The elegance of the final cancellation ($6.5/6.5$) is a classic hallmark of a well-designed JEE problem. The current through the resistor $R$ is exactly $1\text{ A}$.