See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 1 2 di di L L dt dt  1 2 i i 1 1 2 2 0 0 L di L di    11 2 2 L i L i  …(i) 1 2 i i i   …(ii) 1 1 2 L i 1 i L          2 1 1 2 L i i L L        and 1 2 1 2 L i i L L         2 2 11 1 1 2 2 2 2 2 1 2 1 2 2 1L i U L L L 2 1 U L L L L i 2    AITS-PT-II-PCM-Sol -JEE(Main)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 9 27. Rms current through resistor ‘R1’ is 1(rms) 50 5 I A 2 2 10 2     Average power developed in the resistor ‘R1’ is 2 2 1 1(rms) 1 5 250 P I R 10