Explanation: To trickle-charge a battery, a series resistor is used to drop the excess voltage from the power source and limit the charging current. First, determine the voltage that must be dropped across the series resistor. The 110 V line voltage is the source, and the battery's terminal voltage is 12.5 V. Therefore, the resistor must drop: $V_{resistor} = V_{source} - V_{battery} = 110 V - 12.5 V = 97.5 V$ Next, we know the desired charging current is 0.5 A. Since the resistor is in series with the battery, this same current flows through the resistor. Using Ohm's Law ($R = V/I$): $R = V_{resistor} / I = 97.5 V / 0.5 A = 195 \text{ ohms}$ Thus, a 195-ohm resistor is needed. Options B and C would result in different charging currents, either too high or too low, for the given voltage difference.