Understand the circuit

The triangle $PQR$ is made of 3 wires, each of resistance $2\,\Omega$.

So the three sides are:

• $PQ = 2\,\Omega$
• $QR = 2\,\Omega$
• $PR = 2\,\Omega$

A current of $6\,\text{A}$ enters at $P$ and leaves at $R$.

Identify the two paths from $P$ to $R$

Current can go from $P$ to $R$ by two paths:

• Direct path: along side $PR$ $R_1 = 2\,\Omega$

• Indirect path: along $P \to Q \to R$ $R_2 = 2 + 2 = 4\,\Omega$

Thus, between $P$ and $R$, we have two parallel branches of resistances $2\,\Omega$ and $4\,\Omega$.

Use current division

Total current entering at $P$ is $6\,\text{A}$.

In parallel branches, current divides inversely in proportion to resistances.

Let current through the branch $PQR$ be $i_1$.

Then, $i_1 = 6 \times \frac{R_1}{R_1 + R_2}$ because current in one branch equals total current multiplied by the resistance of the other branch over the sum.

So for the $4\,\Omega$ branch, $i_1 = 6 \times \frac{2}{2+4} = 6 \times \frac{2}{6} = 2\,\text{A}$

Check by ratio method

Currents divide as: $I_{2\Omega} : I_{4\Omega} = 4 : 2 = 2 : 1$

Total current $= 6\,\text{A}$, so division is: $4\,\text{A} \text{ and } 2\,\text{A}$

Hence current in the path $P \to Q \to R$ is $i_1 = 2\,\text{A}$

Final answer

$\boxed{2}$