A current of 6 A enters one corner $$P$$ of an equilateral triangle $$PQR$$ having 3 wires of resistance 2 $$\Omega$$ each and leaves by the corner $$R$$. The currents $$i_1$$ in ampere is ______

We need to find the value of the current $$i_1$$ passing through the branch $$PQ$$ of the equilateral triangular circuit network from the shared page layout.

1. Analyze the Circuit Configuration

A total current of $$I = 6\text{ A}$$ enters corner $$P$$ and exits from corner $$R$$. The triangle $$PQR$$ has three branches, each containing a wire of resistance $$R = 2\ \Omega$$:

• Branch 1 (Direct path): The branch $$PR$$ connects directly from the entering corner to the exiting corner. Its total resistance is:

  $$R_{\text{direct}} = R_{PR} = 2\ \Omega$$

• Branch 2 (Series path): The current can also travel from $$P$$ to $$R$$ by going through corner $$Q$$ (path $$P \rightarrow Q \rightarrow R$$). Since branches $$PQ$$ and $$QR$$ are connected in series, their equivalent resistance is:

  $$R_{\text{series}} = R_{PQ} + R_{QR} = 2\ \Omega + 2\ \Omega = 4\ \Omega$$

2. Apply Current Division Rule

The total current entering node $$P$$ ($$6\text{ A}$$) divides into two parallel paths: $$i_2$$ flowing directly through branch $$PR$$, and $$i_1$$ flowing through the series combination path $$PQR$$.

According to the current division rule, the current in a branch is inversely proportional to its resistance:

$$i_1 = I \times \left( \frac{R_{\text{direct}}}{R_{\text{direct}} + R_{\text{series}}} \right)$$

3. Calculate the Branch Current ($$i_1$$)

Substitute the given parameters into the parallel current division expression:

$$i_1 = 6\text{ A} \times \left( \frac{2\ \Omega}{2\ \Omega + 4\ \Omega} \right)$$

$$i_1 = 6 \times \left( \frac{2}{6} \right) = 2\text{ A}$$

Final Whole Number Answer: $$2$$