Final answer:
To find the maximum voltage for the power supply in a series circuit with two bulbs and a maximum current of 125 mA, the total resistance is calculated and then used in Ohm's Law. The total resistance is twice that of a single bulb (3.2 ohms), hence 6.4 ohms. Multiplying by the current gives a maximum voltage of 0.8 V.
Step-by-step explanation:
The question asked relates to determining the maximum voltage for a power supply in a circuit with two light bulbs connected in series, given a maximum allowable current of 125 mA. To answer this, one must understand Ohm's Law (V = IR) and the concept of series circuits where the current is the same through all components and the total resistance is the sum of individual resistances. We also need information about the resistance of each bulb, which can be deduced from the example that gives a current of 1.25 A for a 4 V supply to a single bulb (hence a resistance of 3.2 ohms per bulb). If we calculate the total resistance in the series, we can find the maximum voltage by multiplying this resistance by the maximum allowable current of 125 mA (0.125 A).
The total resistance of two bulbs in series is double that of one bulb, which gives us 6.4 ohms. Thus, the maximum voltage (V) that the power supply should be set to is V = IR, which in numbers is V = 0.125 A * 6.4 ohms, resulting in a maximum voltage of 0.8 V.