A current of 3A enters one vertex P of an equilateral triangle PQR having three resistors of 1 each forming the sides of the equilateral triangle as shown. The value of i₂ in amperes is:

Step-by-step solution

The current of 3 A enters at node P and leaves at node R. The circuit consists of two parallel paths between P and R. The first path is the direct branch PR with resistance R₁ = 1 . The second path is the branch PQR, which has two 1 resistors in series, giving an equivalent resistance R₂ = 1 + 1 = 2 . The current i₂ flows through the branch PQR. Using the current divider rule, the current i₂ is: i₂ = I R₁ R₁ + R₂ Substituting the given values: i₂ = 3 1 1 + 2 = 3 1 3 = 1 A Answer: 1 A