Current I using star-delta transformation

Reading the figure

The 19 V source (+ on top) feeds node T. The current I is the current leaving the source.

• 10 Ω from T to the bottom rail (G).
• 5 Ω from T to the top right node U.
• A star of three 2 Ω resistors: T to O, U to O, O to G.
• 10 Ω from U to G.

Step 1: Star (2, 2, 2 Ω at O) to delta

For equal resistors, R_Δ = 3 R_Y = 6 Ω. The delta joins T, U and G, with 6 Ω between each pair.

  T ---[5 || 6]--- U          T-U : 5 Ω ∥ 6 Ω
  |                |          T-G : 10 Ω ∥ 6 Ω
  [10 || 6]    [10 || 6]      U-G : 10 Ω ∥ 6 Ω
  |                |
  G ---------------G

Step 2: Combine in parallel

• T–U: 5 × 6/(5 + 6) = 2.727 Ω
• T–G: 10 × 6/(10 + 6) = 3.75 Ω
• U–G: 10 × 6/(10 + 6) = 3.75 Ω

Step 3: Total resistance across the source

Path T–U–G = 2.727 + 3.75 = 6.477 Ω, in parallel with T–G = 3.75 Ω:

R_total = 3.75 × 6.477/(3.75 + 6.477) = 2.375 Ω

Step 4: Source current

I = 19/2.375 = 8.0 A

(Check: 19/3.75 = 5.067 A plus 19/6.477 = 2.933 A gives 8.0 A.)

Answer: I = 8 A