Some light bulbs are connected in parallel to a 120 V source as shown in the figure. Each bulb dissipates an average power of 60 W. The circuit has a fuse F that burns out when the current in the circuit exceeds or equals 9 A. Determine the largest number of bulbs out of the following that can be used in the circuit without burning out the fuse.
Correct answer: B. 17
- A. 9
- B. 17
- C. 36
- D. 34
Explanation
In a parallel circuit, each bulb operates independently with the same voltage across it. Each bulb draws a current of 0.5 A (since P = VI → I = P/V = 60/120 = 0.5 A). To find the maximum number of bulbs that can be used without exceeding the fuse limit of 9 A, we divide the total allowable current by the current per bulb: 9 A / 0.5 A = 18 bulbs. However, since adding one more bulb would reach the fuse limit, the largest safe number is 17 bulbs, which results in a total current of 8.5 A, below the fuse's limit. Hence, option B is correct. Options A, C, and D propose numbers of bulbs that either use less or exceed the allowable total current, making them incorrect.
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