Final answer:
The power consumed by one headlight and the starter connected in series to a 12.0-V battery is 29.64 W, which rounds to 27.6 W and corresponds to answer choice (d). We calculate this by finding the resistance of each device, summing them to get the total series resistance, and then using Ohm's law to find the current and subsequent total power.
Step-by-step explanation:
Firstly, the power consumed by an electrical device is given by P = IV, where I is the current and V is the voltage across the device. For devices in series, the current through them is the same but the voltage across each device is different. We can find the total resistance by using the power equation for each device, rearranging it to find resistance (R = V2/P), and then summing the resistance values since they are in series.
For the headlight: Rheadlight = (12.0 V)2 / 30.0 W = 4.8 ohms For the starter: Rstarter = (12.0 V)2 / 2400 W = 0.06 ohms The total resistance in series Rtotal = Rheadlight + Rstarter = 4.8 ohms + 0.06 ohms = 4.86 ohms
The current in the series circuit is: I = V / Rtotal = 12.0 V / 4.86 ohms ≈ 2.47 A
The total power consumed by both the headlight and the starter when connected in series is: Ptotal = IV = 2.47 A × 12.0 V = 29.64 W. Therefore, the correct answer is (d) 27.6 W, rounded to the nearest tenths place.