Final answer:
Technician A is correct; a replacement fuse must have the same amp rating as the original. Calculations show the 1800-W toaster, 1400-W speaker, and 75-W lamp together draw a total current of 27.295 A, exceeding the 15-A fuse rating. Therefore, this combination will blow the 15-A fuse.
Step-by-step explanation:
Electrical Fuse and Current Draw in a Circuit:
Upon evaluating the statement by Technician A, we can affirm that a replacement fuse must indeed have the same amp rating as the original to ensure the circuit is properly protected. This makes Technician A correct. On the other hand, Technician B's statement that a 20-amp automotive fuse will handle 25 amps of current without blowing is incorrect. Fuses are designed to blow when the current exceeds their rating, typically with a small margin of tolerance, but not as high as 25% over the rating.
To calculate the current drawn by each device, we can use the formula I = P/V, where 'I' is the current, 'P' is the power in watts, and 'V' is the voltage. For a 1800-W toaster, the current 'I' would be 1800 W / 120 V = 15 A. Similarly, for a 1400-W speaker, 'I' = 1400 W / 120 V = 11.67 A, and for a 75-W lamp, 'I' = 75 W / 120 V = 0.625 A.
When we add these currents, the total is 15 A + 11.67 A + 0.625 A = 27.295 A, which exceeds the 15-A fuse rating. Hence, this combination will indeed blow the 15-A fuse, answering (b) of the student's question.