Using source transformation find the current flowing the 8 Ω resistance [Circuit: a common bottom wire. Left: an 18 V battery (longer plate, +, at the top) from the bottom wire up to the top-left corner. Top wire: top-left corner → resistor (label pr…

Reading of the figure: the first series resistor is taken as 6 Ω (its label is over-struck in the print; this is the standard version of the problem).

Step 1: 18 V source with series 6 Ω → current source \( \dfrac{18}{6} = 3 \) A (upward into node 1) in parallel with 6 Ω.

Step 2: two 6 Ω resistors now in parallel at node 1: 6 ∥ 6 = 3 Ω. So 3 A ∥ 3 Ω.

Step 3: convert back to a voltage source: 3 × 3 = 9 V in series with 3 Ω.

Step 4: add the series 6 Ω: 9 V in series with 3 + 6 = 9 Ω (feeding node 2).

Step 5: convert to current source: 9/9 = 1 A (upward into node 2) ∥ 9 Ω.

Step 6: combine with the 2 A source (same direction, into node 2): 1 + 2 = 3 A ∥ 9 Ω.

Step 7: convert to voltage source: 3 × 9 = 27 V in series with 9 Ω; add the series 1 Ω → 27 V in series with 10 Ω, connected to the 8 Ω across X–Y.

   10 ohm
 +-/\/\/-+---- X
 |       |
27 V    8 ohm
 |       |
 +-------+---- Y

\[ I_{8\Omega} = \frac{27}{10+8} = \mathbf{1.5\ A} \] flowing from X to Y (downward) through the 8 Ω.

Check: V_XY = 1.5 × 8 = 12 V, so V₂ = 12 + 1.5 × 1 = 13.5 V; Thevenin form at node 2: (27 − 13.5)/9 = 1.5 A ✓.

(If the first resistor were 8 Ω instead, the same steps give 2.25 A ∥ (8 ∥ 6 = 3.43 Ω) → 7.71 V, 9.43 Ω → 0.818 + 2 = 2.818 A ∥ 9.43 Ω → 26.57 V, 10.43 Ω → I = 26.57/18.43 = 1.44 A.)