Nodal analysis: current in each branch

Assumptions from the figure: 10 V source (+ at top) with 3 Ω feeds node N1. From N1, the branch 2 Ω in series with a dependent voltage source of value I₂ volts (+ at top) goes to ground. A 5 Ω resistor carries I₂ from N1 to N2. At N2, a 4 Ω resistor goes to ground and a 2 A source pushes current into N2.

Nodal equations (V₁ at N1, V₂ at N2)

Control variable: I₂ = (V₁ − V₂)/5.

Node N2 (current in = current out):

I₂ + 2 = V₂/4
(V₁ − V₂)/5 + 2 = V₂/4        ... (1)

Node N1: current from source branch = current in 2 Ω branch + I₂. The 2 Ω branch current is I_b = (V₁ − I₂)/2:

(10 − V₁)/3 = (V₁ − I₂)/2 + I₂        ... (2)

Solving

From (1): I₂ = V₂/4 − 2, and V₁ = V₂ + 5I₂ = 2.25V₂ − 10.

Put in (2): 2(10 − V₁) = 3(V₁ + I₂) → 20 = 5V₁ + 3I₂

20 = 5(2.25V₂ − 10) + 3(V₂/4 − 2) = 12V₂ − 56
V₂ = 6.333 V
V₁ = 4.25 V
I₂ = 6.333/4 − 2 = −0.4167 A

Branch currents

Check: 1.917 = 2.333 − 0.4167 at N1 ✓; 1.583 = 2 − 0.4167 at N2 ✓.

Answer: I(3 Ω) = 1.917 A, I(2 Ω) = 2.333 A, I₂ = −0.417 A, I(4 Ω) = 1.583 A.