Exercise b-n+ = 2 kVL Determine the voltage V and the current / 3-2+1 ZA A オレ 2 A( E ZA (D 42. 42 KLL OA: 2ん = => んール2-2 KVL in Li VAB - V-sV=0 KVL in Lz: 5+ 2- M- hz Z=O 5-212-4-2 iz =0 =>U2 =...

I don't understand how when we did kvl in l2 we added the currents too, isnt kvl the voltage law? How did we include the currents while solving this? And how did we take their sign based on the respective loop drawn?

Extracted text: Exercise b-n+ = 2 kVL Determine the voltage V and the current / 3-2+1 ZA A オレ 2 A( E ZA (D 42. 42 KLL OA: 2ん = => んール2-2 KVL in Li VAB - V-sV=0 KVL in Lz: 5+ 2- M- hz Z=O 5-212-4-2 iz =0 =>U2 = Answers:-9/2 V and 1/84