$\underline{\text{Using Thevenin’s theorem}} \\ \text{Voc} \\ \text{Equivalent circuit} \\ V= IR \hspace{1.5cm} ƪV= 5 * 6 \\ V= 8 * 15 \hspace{1cm} = 30 V \\ V = 120 V \\ \text{Apply KVL to the loop} \\ 120 = (15 + 5 + 10) I + 20 + 50 \\ I = 120-70/ 30 = 1.667 A \\ ∴ VOC = VAB = 50 + (10 * I) \\ = 50 + (10* 1.667) \\ = 66.67 V \\ RTH: \\ \text{Resistance 5 & 15 are in series} \\ ∴ 5 + 15 = 20 \\ 20 Ω || 10 Ω \\ 1/R = 1/20 + /10 \\ = 6.662 Ω \\ R = RTH \\ \text{Thevenin’s equivalent circuit RTH = 6.62 Ω} \\ \text{From the above figure} \\ IL = VOC/ RTH + RL\\ = 66.67/ 6.67 + 6 \\ = 5.26 A$

$\underline{\text{Using Superposition theorem}} \\ \text{When the 8A source is acting alone} \\ \text{Resistance 10 Ω and 6 Ω are parallel} \\ ∴ 1/R = 1/10+ 1/6 \\ R – 3.75 \\ \text{Resistance R i.e. 3.75 Ω is a series with 5 Ω} \\ \therefore R = 8.75 Ω \\ \text{Equivalent circuit} \\ \therefore \text{By current division rule,} \\ I1 = 15/15 +8.75 * 8A \\ = 5.05 A \\ In = 10/10 + 6 * I1 \\ = 10/10 * 5.05 \\ = 3.10A \\ \text{When 20 V source is acting alone} \\ \text{Total resistance=} R = (10116/ parallel) + (15+5/series) \\ = 3.75 + 5 + 15 = 23.75 Ω \\ \text{Equivalent circuit} \\ \text{By current division rile} \\ I2= 20V/ 23.75 Ω \\ = 0.843A \\ Iᶦᶦ = 10/10+ 6 * I2 \\ = 10/ 16 * 0.842 = 0.53A \\ \text{When 5A acting alone} \\ 15 +5 = 20 \\ 20 ||10/ parallel = 6.67 \\ \text{Equivalent circuit} \\ IM = (6.67)/ (6.67 + 6) * 5 \\ \text{Total current = Iᶦ & Iᶦᶦᶦ are in same direction} \\ \text{While Iᶦᶦ is opposite} \\ ∴ IL = Iᶦ + Iᶦᶦᶦ - Iᶦᶦ \\ = 3.16 – 0.53 + 2.63 = 5.62A$