Final answer:
The currents drawn by an 1800-W toaster, 1400-W electric frying pan, and 75-W lamp in a 120-V circuit are 15 A, 11.67 A, and 0.625 A respectively. Adding them together gives a total current of 27.3 A, which exceeds the 15-A fuse rating, thus the fuse will blow.
Step-by-step explanation:
The subject question deals with the power consumption and current drawn by electrical devices and whether these devices will exceed the current rating of a circuit when used together. Specifically, to answer part (a), the current drawn by each device can be calculated using the formula I = P/V, where I is the current in amperes, P is the power in watts, and V is the voltage in volts. For an 1800-W toaster, 1400-W electric frying pan, and 75-W lamp on a 120-V circuit, the currents are
- Toaster: I = 1800 W / 120 V = 15 A
- Electric frying pan: I = 1400 W / 120 V ≈ 11.67 A
- Lamp: I = 75 W / 120 V ≈ 0.625 A
To answer part (b), we add up the currents to see if they exceed the 15-A fuse rating. The total current is 15 + 11.67 + 0.625 ≈ 27.3 A, which is significantly higher than the fuse rating, meaning the combination will blow the 15-A fuse.