See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 = (i1 + i2) 0.5 + 1i2 i1 = i2 2 = 2i1 (0.5) + i1 For More Material Join: @JEEAdvanced_2024 AITS-FT-IX-PCM(Sol.)-JEE(Main)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 5 = 2i1 i1 = 1 A, i1 = 1 A  the current through 0.5 , resistor i = i1 + i1 = 2 A Maximum power developed across R = i2R = 22 (0.5) = 2 W