Final answer:
In a series circuit with two 15 W bulbs, each bulb dissipates 3.75 W of power when connected to a 120 V supply.
Step-by-step explanation:
The student is asking how much power is dissipated by each bulb when two 15 W lightbulbs are wired in series and connected to a 120 V supply. To find the power dissipated by each bulb, we need to understand that in a series circuit, the current passing through each component is the same. First, we can calculate the total resistance (Rtotal) for the two bulbs in series.
For a single bulb rated at 15 W on 120 V, the resistance (R) can be calculated using the formula P = V2 / R, where P is power and V is voltage. From this, R = V2 / P = (120 V)2 / 15 W = 960 Ω. For two bulbs in series, their total resistance is Rtotal = R + R = 2 × 960 Ω = 1920 Ω.
Now, using Ohm's Law (V = IR), where I is the current, we find the current in the circuit is I = V / Rtotal = 120 V / 1920 Ω = 0.0625 A. Since both bulbs have the same resistance, each bulb will have the same voltage drop across it, which means each bulb has 60 V across it (half of the supply voltage because there are two bulbs). Finally, we calculate the power dissipated by each bulb using P = IV: P = 0.0625 A × 60 V = 3.75 W.
Therefore, each bulb dissipates 3.75 W of power when they are wired in series across a 120 V supply.