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Suppose that a nearly depleted 12-V lead-acid battery has an open-circuit voltage of 11.7 V and an internal resistance of 0.03 ohm.

a. What voltage would a PV module operate at if it is delivering 6 A to the battery?

b. If 20 A is drawn from a fully charged battery with an open-circuit voltage of 12.7 V, what voltage would the PV module operate at?

Answers:
(a) 11.88 V
(b) 12.1 V

Answer :

(a) To find the voltage a PV module would operate at when delivering 6 A to the battery, we can use Ohm's law. The voltage drop across the internal resistance of the battery is given by:

V_internal = I * R_internal = 6 A * 0.03 ohm = 0.18 V

The voltage delivered by the PV module is the sum of the open-circuit voltage of the battery and the voltage drop across the internal resistance:

V_PV = V_battery + V_internal = 11.7 V + 0.18 V = 11.88 V

Therefore, the PV module would operate at 11.88 V.

(b) When 20 A is drawn from a fully charged battery with an open-circuit voltage of 12.7 V, the voltage drop across the internal resistance is:

V_internal = I * R_internal = 20 A * 0.03 ohm = 0.6 V

The voltage delivered by the PV module is again the sum of the open-circuit voltage of the battery and the voltage drop across the internal resistance:

V_PV = V_battery + V_internal = 12.7 V + 0.6 V = 12.1 V

Therefore, the PV module would operate at 12.1 V.

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