Superposition, current IL through 2Ω

Reading the figure

Bottom rail = ground. Node 1: 5 A source (up) into node 1 and a 5 Ω shunt. From node 1, 10 Ω in series with the 10 V battery (+ on the 10 Ω side) to node 2. Node 2: 3 Ω shunt to ground and the 2 Ω load, whose bottom end is joined to ground through the 20 V battery (+ on the ground side). I_L flows downwards in the 2 Ω.

Case 1: 5 A alone (10 V and 20 V shorted)

Node 2 to ground: 3 Ω ∥ 2 Ω = 1.2 Ω From node 1: 5 Ω ∥ (10 + 1.2) = 5 × 11.2/16.2 = 3.457 Ω V₁ = 5 × 3.457 = 17.28 V V₂ = 17.28 × 1.2/11.2 = 1.852 V I′ = 1.852/2 = 0.926 A (downwards)

Case 2: 10 V alone (5 A opened, 20 V shorted)

The 10 V source (with 10 Ω + 5 Ω = 15 Ω) feeds node 2. Its + is on the 10 Ω side, so it pulls node 2 negative:

V₂ (1/3 + 1/2 + 1/15) = −10/15 → V₂ = −0.7407 V I″ = V₂/2 = −0.370 A (upwards)

Case 3: 20 V alone (5 A opened, 10 V shorted)

The 20 V battery has its + on the ground side, so the load's lower end is at −20 V:

V₂/3 + V₂/15 + (V₂ + 20)/2 = 0 → 0.9 V₂ = −10 → V₂ = −11.11 V I‴ = (V₂ + 20)/2 = (−11.11 + 20)/2 = 4.444 A (downwards)

Total

I_L = 0.926 − 0.370 + 4.444 = 5.0 A (downwards through the 2 Ω)

(Check: 25/27 − 10/27 + 120/27 = 135/27 = 5 A.)

Answer: I_L = 5 A