The circuit shown in fig. contains three resistors R1 = 100 ohm, R2 = 50 ohm and R3 = 20 ohm and cells of emfs E1 = 2V and E2. The ammeter indicates a current of 50 mA. Determine the currents in the resistors and the emf of the second cell. The internal resistance of the ammeter and that of the cells should be neglected.
Applying Kirchhoff’s voltage law in the loop of ABCD, we get
E1 = (I + 0.05)R1 + IR2 or I = -20 mA
Current through R1 is 30 mA toward right. Current through R2 is 20 mA toward left. Applying KVL in loop BGEF, we get
E2 = (I + 0.05)100 + (0.05)20 = 8V..
Now my question is, since V = IR, it implies that
E1 = IReq.
Or, 2/20mA = Req = 100 ohm, but manually calculating it should be 150 ohm….. where am I mistaken?
Same case is with E2 battery.