Final answer:
When two 60 W (120 V) lightbulbs are wired in series and connected to a 120 V supply, the combined resistance is 480 ohms. The current through the circuit is 0.25 A, and the total power consumed by both bulbs is 30 W.
Step-by-step explanation:
The student asked about the total power consumed by two 60 W (120 V) lightbulbs wired in series when connected to a 120 V supply. First, we need to calculate the resistance of each bulb using the formula P = V^2 / R, where P is the power in watts, V is the voltage, and R is the resistance. Since a 60 W bulb operates at 120 V, we have R = V^2 / P, and R = 120^2 / 60, which gives us a resistance of 240 ohms for one bulb.
When two resistors (or two lightbulbs in this case) are connected in series, their resistances add up. Therefore, the combined resistance of two lightbulbs in series is Rtotal = R + R = 240 ohms + 240 ohms = 480 ohms. The total current, I, in the circuit can then be calculated using Ohm's law V = IR, which gives us I = V / Rtotal = 120 V / 480 ohms = 0.25 A.
Now, the total power consumed by the two lightbulbs can be found using the formula Ptotal = VI, where V is the supply voltage and I is the total current. The total power consumed is therefore Ptotal = 120 V * 0.25 A = 30 W, making option 4 the correct answer.