digital 3. Find currents (values and directions) in all resistors of the circuit given Slow. Find total electric power developing in this circuit. Neglect internal resistances of the batteries. Solution As all the Voltage sources are DC (Batteries) there by the capacitor acts as short circuit . I1 = Current through R1 =E1-E2)/R1 = (6-3)/5 = 3/5 Amps (direction from E1 to E2) I2= Current Throuh R3 = E2/R3 = 3/4 Amps (direction from Top to bottom) I3 = Current Through R2 = E3-E2)/R2 = (5-3)/5 = 2/5 Amps (direction from E3 to E2) Power delivered = (I1*E1) + (E3*I3) =(3/5)*6 + (2/5)*5 = 28/5 Watts.