Step 1: Norton current IN — remove the 3 Ω and short A–B. Take node E (= B) as reference; unknowns VP, VD.
Branch currents leaving P: through 8 Ω–15 V–2 Ω to D: \( \dfrac{V_P - V_D - 15}{10} \) (the 15 V rise from D to T, total resistance 8 + 2 = 10 Ω); through 16 Ω to D: \( \dfrac{V_P-V_D}{16} \); through 5 V–10 Ω to E: \( \dfrac{V_P-5}{10} \); through 2 Ω to the short A–B: \( \dfrac{V_P}{2} \).
KCL at P: \( \dfrac{V_P - V_D - 15}{10} + \dfrac{V_P-V_D}{16} + \dfrac{V_P-5}{10} + \dfrac{V_P}{2} = 0 \)
×80: \( 8V_P - 8V_D - 120 + 5V_P - 5V_D + 8V_P - 40 + 40V_P = 0 \Rightarrow 61V_P - 13V_D = 160 \) … (1)
KCL at D (current arriving from P = current leaving through bottom 2 Ω to E): \( \dfrac{V_P - V_D - 15}{10} + \dfrac{V_P-V_D}{16} = \dfrac{V_D}{2} \)
×80: \( 8V_P - 8V_D - 120 + 5V_P - 5V_D = 40V_D \Rightarrow 13V_P - 53V_D = 120 \) … (2)
Solving: Δ = 61(−53) − (−13)(13) = −3064; \( V_P = \dfrac{160(-53) - (-13)(120)}{-3064} = \dfrac{-6920}{-3064} = 2.2585 \) V; \( V_D = \dfrac{61(120) - 13(160)}{-3064} = -1.7102 \) V.
\( I_N = \dfrac{V_P}{2} = \) 1.129 A (from A to B through the short).
Step 2: Norton resistance RN — replace both voltage sources by shorts, look into A–B:
- Between P and D: (8 + 2) Ω ∥ 16 Ω = 10 × 16/26 = 6.154 Ω.
- In series with bottom 2 Ω (D to E): 6.154 + 2 = 8.154 Ω between P and E.
- In parallel with 10 Ω (P–E): 8.154 × 10/18.154 = 4.4915 Ω.
- Plus top 2 Ω (P to A): RN = 6.4915 Ω ≈ 6.49 Ω.
Step 3: Norton equivalent = 1.129 A current source in parallel with 6.49 Ω, with the 3 Ω connected across A–B:
\[ I_{3\Omega} = I_N\frac{R_N}{R_N+3} = 1.129\times\frac{6.4915}{9.4915} = \mathbf{0.772\ A} \] flowing from A to B through the 3 Ω.
Check (Thevenin): with A–B open, nodal analysis gives VP = Vth = 7.331 V; Vth/RN = 7.331/6.4915 = 1.129 A = IN ✓ and I = 7.331/(6.4915 + 3) = 0.772 A ✓.
