Animated Solution for Physics - Current Electricity: A current of 6 A enters one corner $P$ of an equilateral triangle $PQR$ having three wires of resistance $2\Omega$ each and leaves by the corner $R$. The currents $i_1$ in ampere is ......... .

The beauty of electrical circuits lies in their predictability. Current, much like water flowing down a mountain, always seeks the path of least resistance. But it doesn't only take the easiest path; it divides itself proportionally among all available paths. Let's dive into this elegant problem and see how the current makes its decisions!

Imagine you are standing at node $P$. A total current of $6\text{ A}$ is rushing in behind you. Your destination is node $R$, where the current exits the circuit. Looking ahead, you see two distinct paths to reach $R$.

The first path is a direct bridge from $P$ to $R$, guarded by a single $2\Omega$ resistor. The second path is a bit of a detour: you have to travel from $P$ to $Q$, and then from $Q$ to $R$. Each leg of this detour has a $2\Omega$ resistor.

To understand how the $6\text{ A}$ current splits, we first need to know the total resistance of each path. The direct path is simply $2\Omega$. For the detour path, the current $i_1$ must flow through both the $PQ$ and $QR$ resistors sequentially. Because they are in series, their resistances add up:

Now we have a classic parallel circuit scenario: a $2\Omega$ branch and a $4\Omega$ branch. To find the current $i_1$ flowing through the $4\Omega$ branch, we deploy the Current Divider Rule. This powerful rule states that the current in one branch is equal to the total incoming current multiplied by the resistance of the other branch, divided by the sum of both resistances.

Let's plug in our numbers and watch the math unfold. We know $I_{total}=6\text{ A}$, the other branch $R_{PR}=2\Omega$, and our branch $R_{PQR}=4\Omega$.

The $6$ in the numerator and denominator cancel out beautifully, leaving us with:

This result makes perfect physical sense! The detour path has twice the resistance ($4\Omega$) compared to the direct path ($2\Omega$). Therefore, it should receive exactly half the current. Out of the total $6\text{ A}$, $2\text{ A}$ takes the harder path, while the remaining $4\text{ A}$ rushes through the easier direct path. Physics is perfectly balanced!