EXCELLENT PROBLEM SOLVED IN ELECTRICAL CIRCUITS - DIFFERENTIATED TACTIC !

Escola Naval 2017 - Two batteries, with internal resistance r1 = r2 = 1/3 ohm, and a lamp, with resistance RL = 2/3 ohm, are connected in parallel as shown in the circuit in the figure above. The emf of battery 1 is E1 = 1.5 V, but battery 2, with emf E2, is partially discharged so that the ideal ammeter measures a zero current in this cell. Therefore, the value of the emf in volts is worth
a) zero b) 0.50 c) 0.75 d) 1.00 e) 1.25

[@fisicacomchicao]

Ótima tática de resolução..... parabéns, professor Renato ..😊