Why is the closed-circuit voltage of a lead-acid battery lower than its open-circuit voltage?

• A. Temperature effects
• B. Voltage drop across internal resistance
• C. Electrolyte evaporation
• D. Poor connections

Answer: (B)

When a lead-acid battery is not delivering current, its terminals show the open-circuit voltage, which reflects the chemical potential inside. Once a load is connected, current flows through the battery’s own internal resistance. The drop across that internal resistance is I times R_internal, so the terminal voltage under load becomes V_terminal = V_open-circuit − I × R_internal. That’s why the closed-circuit (loaded) voltage is lower than the open-circuit voltage. The amount of sag grows with more current and with higher internal resistance (which can rise as the battery ages or with temperature changes). Other factors like electrolyte evaporation or poor connections can influence voltage in specific situations, but the fundamental reason for the difference is the voltage drop across the battery’s internal resistance.