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.

To calculate the power supply's open-circuit voltage, internal resistance, and the charging current for two storage batteries connected in series, we need to apply Ohm's Law and Kirchhoff's Laws.

Let's break down the problem step by step:

Step 1: Find the total emf of the system.
Since the two batteries are connected in series, the total emf (E_total) is the sum of the individual emfs:
E_total = emf1 + emf2 = 5.50 V + 5.50 V = 11.00 V

Step 2: Find the total internal resistance of the system.
The total internal resistance (R_total) of the system is the sum of the individual internal resistances:
R_total = internal resistance1 + internal resistance2 = 0.300 Ω + 0.300 Ω = 0.600 Ω

Step 3: Calculate the equivalent resistance of the system.
The equivalent resistance (R_eq) can be calculated using Kirchhoff's Voltage Law (KVL). Since the two batteries are connected in series, the charging current (I) flowing through each component is the same. According to KVL, the emf across the total resistance will be equal to the sum of the emfs across the individual resistances:
E_total - I * R_eq = emf1 + emf2 = 11.00 V

We can rearrange the equation to find R_eq:
R_eq = (E_total - emf1 - emf2) / I
= (11.00 V - 5.50 V - 5.50 V) / 3.30 A
= 0.000 Ω

Step 4: Calculate the charging current.
To calculate the charging current (I), we can use Ohm's Law. The total voltage across the equivalent resistance (R_eq) is the open-circuit voltage of the power supply (V_oc) minus the voltage drop across the internal resistance (V_internal):
V_oc = E_total = 39.0 V
V_internal = I * R_internal = 3.30 A * 2.00 Ω = 6.60 V

V_eq = V_oc - V_internal = 39.0 V - 6.60 V = 32.40 V

Now, using Ohm's Law, we can find I:
I = V_eq / R_eq
= 32.40 V / 0.000 Ω
= undefined

However, since the equivalent resistance came out to be zero, the charging current will be simply the open-circuit voltage divided by the total resistance of the circuit (including the internal resistance of the power supply and the internal resistances of the batteries):
I = V_oc / (R_total + R_internal1 + R_internal2)
= 39.0 V / (0.600 Ω + 2.00 Ω + 0.300 Ω)
= 39.0 V / 2.900 Ω
≈ 13.45 A

Therefore, the charging current for the given system is approximately 13.45 A.