posted by Anonymous .
power supply has an open-circuit voltage of 39.0 V and an internal resistance of 2.00 Ù. It is used to charge two storage batteries connected in series, each having an emf of 5.50 V and internal resistance of 0.300 Ù. The charging current is to be 3.30 A.
I=(E-(5.5+5.5)) / (2+0.3+0.3) = 3.30 A.
(E-11) / 2.6 = 3.30.
E-11 = 8.58.
E = 8.58 + 11 = 19.58 Volts=Voltage
setting required for a charging current of 3.30 amps.