A current of 9A enters point P of an equilateral triangle PQR having three wires of 3 each and leaves by point R. The currents I₁ and I₂ are respectively

Step-by-step solution

The current of 9 A entering at point P splits into two parallel paths to reach point R. Path 1 is along P Q R. The equivalent resistance of this path is R₁ = 3 + 3 = 6 . The current through this path is I₁ . Path 2 is directly from P R. The resistance of this path is R₂ = 3 . The current through this path is I₂ . Since the two paths are in parallel, the potential difference across them is equal: V_ PR = I₁ R₁ = I₂ R₂ I₁(6) = I₂(3) I₂ = 2I₁ According to Kirchhoff's current law at junction P, the total current is the